Straight Lines

(i) Formula Used:

Distance between any two points A(x₁, y₁) and B(x₂, y₂)=

Distance between A(2, -3) and B(-6, 3)

= 10 units

Therefore, the distance between points A and B is 10 units.

(ii) Distance between C(-1, -1) and D(8, 11) =

= 15 units

Therefore, the distance between points C and D is 10 units.

(iii) Distance between P(-8, -3) and Q(-2, -5)=

units

Therefore, the distance between the points P and Q is units.

(iv) Distance between R(a + b, a - b) and S(a - b, a + b) =

units

Therefore, the distance between the points R and S is units.

Distance of point P(6, -6) from origin (0, 0) =

= 6√2 units

Therefore, the distance of the point P from the origin is units.

Given: Point P(x, y) is equidistant from points A(6, -1) and B(2, 3)

i.e., distance of P from A = distance of P from B

Squaring both sides,

⇒ (x – 6)² + (y – 1)² = (x – 2)² + (y – 3)²

⇒ x² – 12x + 36 + y² – 2y + 1 = x² – 4x + 4 + y² – 6y + 9

⇒ –12x + 36 + 2y + 1 = –4x + 4 – 6y + 9

⇒ –8x + 8y = –24

⇒ x – y = 3

Therefore, x – y = 3 is the required relation.

Let the point on x-axis be P(x, 0).

Given: Point P(x, 0) is equidistant from points A(7, 6) and B(-3, 4)

i.e., distance of P from A = distance of P from B

Squaring both sides,

⇒ (x – 7)² + 36 = (x + 3)² + 16

⇒ x² – 14x + 49 + 36 = x² + 6x + 9 + 16

⇒ –20x = –60

⇒ x = 3

Therefore, the point on the x-axis is (3, 0).

(i) Given: AB is parallel to the x-axis.

When AB is parallel to the x-axis, the y co-ordinate of A and B will be the same.

i.e., y₁ = y₂

Distance

⇒ |x₂ – x₁|

Therefore the distance between A and B when AB is parallel to x-axis is |x₂ – x₁|

(ii) Given: AB is parallel to the y-axis.

When AB is parallel to the y-axis, the x co-ordinate of A and B will be the same.

i.e., x₂ = x₁

Distance

⇒ |y₂ – y₁|

Therefore the distance between A and B when AB is parallel to y-axis is |y₂ – y₁|

Given: The two points are A(-8, 0) and B(0, 15)

Distance between A and B

⇒ √ 289

⇒ 17 units

Therefore, the distance between A and B is 17 units.

Let the point on the y-axis be P(0, y)

Given: P is equidistant from A(-4, 3) and B(5, 2).

i.e., PA = PB

Squaring both sides, we get

⇒ (-4 – 0)² + (3 – y)² = (5 – 0)² + (2 – y)²

⇒ 16 + 9 – 6y + y² = 25 + 4 – 4y + y²

⇒ 25 – 6y = 29 – 4y

⇒ 2y = -4

⇒ y = -2

Therefore, the required point on the y-axis is (0, -2).

Given: The 3 points are A(3, -2), B(5, 2) and C(8, 8).

AB

= 2√5 units …..(1)

BC

= 3√5 units …..(2)

AC

= 5√5 units …..(3)

From equations 1, 2 and 3, we have

⇒ AC = AB + BC

This is possible only if the points are collinear.

Therefore, the points A, B and C are collinear.

Hence, proved.

Given: The 3 points are A(7, 10), B(-2, 5) and C(3, -4)

AB

= √106 units …..(1)

BC

= √106 units …..(2)

AC

= √212 units

From equations 1 and 2, we have

⇒ AB = BC

Therefore, Δ ABC is an isosceles triangle …..(3)

Also, AB² = 106 units …..(4)

BC² = 106 units …..(5)

AC² = 212 units …..(6)

From equations 4, 5 and 6, we have

AB² + BC² = AC²

So, it satisfies the Pythagoras theorem.

Δ ABC is right angled triangle …..(7)

From 3 and 7, we have

Δ ABC is an isosceles right angled triangle.

Hence, proved.

Given: The 3 points are A(1, 1), B(-1, -1) and C(-√3, √3).

AB

= 2√2 units …..(1)

BC

= 2√2 units …..(2)

AC

= 2√2 units …..(3)

From equations 1, 2 and 3, we have

AB = BC = AC = 2√2 units.

Therefore, Δ ABC is an equilateral triangle each of whose sides is 2√2 units.

Hence, proved.

Given: The 4 points are A(2, -2), B(8, 4), C(5, 7) and D(-1, 1).

Note: For a quadrilateral to be a rectangle, the opposite sides of the quadrilateral must be equal and the diagonals must be equal as well.

AB

= 6√2 units …..(1)

BC

= 3√2 units …..(2)

CD

= 6√2 units …..(3)

AD

= 3√2 units …..(4)

From equations 1, 2, 3 and 4, we have

AB = CD and BC = AD …..(5)

Also, AC

= 3√10 units

BD

= 3√10 units

Thus, AC = BD …..(6)

From equations 5 and 6, we can conclude that the opposite sides of quadrilateral ABCD are equal and the diagonals of ABCD are equal as well.

Therefore, point A, B, C and D are the angular points of a rectangle.

Given: The points are A(3, 2), B(0, 5), C(-3, 2) and D(0, -1).

Note: For a quadrilateral to be a square, all the sides of the quadrilateral must be equal in length and the diagonals must be equal in length as well.

AB

= 3√2 units

BC

= 3√2 units

CD

= 3√2 units

DA

= 3√2 units

Therefore, AB = BC = CD = DA …..(1)

AC

= 6 units

BD

= 6 units

Therefore, AC = BD …..(2)

From 1 and 2, we have all the sides of ABCD are equal and the diagonals are equal in length as well.

Therefore, ABCD is a square.

Hence, the points A, B, C and D are the vertices of a square.

Given: Vertices of the quadrilateral are A(1, -2), B(3, 6), C(5, 10) and D(3, 2).

Note: For a quadrilateral to be a parallelogram opposite sides of the quadrilateral must be equal in length, and the diagonals must not be equal.

AB

= 2√17 units

BC

= 2√5 units

CD

= 2√17 units

DA

= 2√5 units

Therefore, AB = CD and BC = DA …..(1)

AC

= 4√10 units

BD

= 4 units

Therefore, AC BD …..(2)

From 1 and 2, we have

Opposite sides of ABCD are equal, and diagonals are not equal. Hence, points A, B, C and D are the vertices of a parallelogram.

Given: Vertices of the quadrilateral are A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2).

Note: For a quadrilateral to be a rhombus, all the sides must be equal in length and the diagonals must not be equal.

AB

= √26 units

BC

= √26 units

CD

= √26 units

DA

= √26 units

Therefore, AB = BC = CD = DA …..(1)

AC

= 4√2 units

BD

= 6√2 units

Also, AC BD …..(2)

From 1 and 2, we have all the sides are equal and diagonals are not equal.

Hence, the points A, B, C and D are the vertices of a rhombus.

Given: Vertices of the parallelogram are A(-2, -1), B(1, 0), C(x, 3) and D(1, y).

To find: values of x and y.

Since, ABCD is a parallelogram, we have AB = CD and BC = DA.

AB

= √10 units

BC

CD

DA

Since AB = CD,

Squaring both sides, we get

⇒ 10 = (1 – x)² + (y – 3)²

⇒ 10 = 1 – 2x + x² + y² – 6y + 9

⇒ x² + y² – 2x – 6y = 0 …..(1)

Since BC = DA,

Squaring both sides,

⇒ (x – 1)² + 9 = 9 + (1 + y)²

⇒ x² – 2x + 1 = 1 + 2y + y²

⇒ x² – y² – 2x – 2y = 0 …..(2)

Equation 1 – Equation 2 gives us,

⇒ 2y² – 4y = 0

⇒ y² – 2y = 0

⇒ y(y – 2) = 0

⇒ y = 0 or y = 2

But y 0 because then point D(1, 0) is same as B(1, 0)

Therefore, y = 2

When y = 2, from equation 1,

⇒ x² + 4 – 2x – 12 = 0

⇒ x² – 2x – 8 = 0

⇒ (x – 4) × (x + 2) = 0

⇒ x = 4 or x = -2

So, the possible set of values for x and y are:

x = 4, y = 2

x = -2, y = 2

But when x = -2, then C(-2, 3). Then ABCD does not form a parallelogram.

Therefore, the only solution is x = 4 and y = 2.

Given: The vertices of the triangle are A(-3, -5), B(5, 2) and C(-9, -3).

Formula: Area of Δ ABC

Here,

x₁ = -3, y₁ = -5

x₂ = 5, y₂ = 2

x₃ = -9, y₃ = -3

Putting the values,

Area of Δ ABC

= 29 square units.

Therefore, the area of Δ ABC is 29 square units.

Given: The points are A(-5, 1), B(5, 5) and C(10, 7).

Note: Three points are collinear if the sum of lengths of any sides is equal to the length of the third side.

AB

= 2√29 units …..(1)

BC

= √29 units …..(2)

AC

= 3√29 units …..(3)

From equations 1, 2 and 3, we have

AB + BC = AC

Therefore, the three points are collinear.

Given: The points are A(-5, 1), B(1, 2) and C(k, 0)

To find: value of k

AB

= √37 units

BC

AC

Since the points are collinear, AB + BC = AC

Squaring both sides and rearranging,

⇒

On simplifying,

Squaring both sides,

⇒ 64 – 96k + 36k² = 37 × (k² – 2k + 5)

⇒ 64 – 96k + 36k² = 37k² – 74k + 185

Rearranging,

⇒ 37k² – 74k + 185 = 36k² – 96k + 64

⇒ k² + 22k + 121 = 0

⇒ (k + 11)² = 0

⇒ k = -11

Therefore, the value of k for which the points A, B and C are collinear is –11.

Given: The vertices of the quadrilateral are A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2).

Formula: Area of a triangle

Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ ADC

Area of Δ ABC

= –29

Taking modulus (∵ area is always positive),

Area of Δ ABC = 29 sq. units …..(1)

Area of Δ ADC

= 31.5 sq. units …..(2)

From 1 and 2,

Area of quadrilateral ABCD = 29 + 31.5

= 60.5 square units.

Therefore, the area of quadrilateral ABCD is 60.5 square units.

The figure is as shown above.

x₁ + x₂ = 2 × 3 = 6 …..(1)

x₁ + x₃ = 2 × 1 = 2 …..(2)

x₂ + x₃ = 2 × 5 = 10 …..(3)

Equation 1 – Equation 2 gives us

x₂ – x₃ = 4 …..(4)

Equation 3 + Equation 4,

2x₂ = 14 ⇒ x₂ = 7

∴ x₁ = -1 and x₃ = 3

Similarly,

y₁ + y₂ = 2 × -1 = -2 …..(5)

y₁ + y₃ = 2 × -3 = -6 …..(6)

y₂ + y₃ = 2 × 3 = 6 …..(7)

Equation 5 – Equation 6 gives us

y₂ – y₃ = 4 …..(8)

Equation 7 + Equation 8,

2y₂ = 10 ⇒ y₂ = 5

∴ y₁ = -7 and y₃ = 1

Area of Δ ABC

= 8 square units

Therefore, the area of Δ ABC is 8 square units.

Let P(x, y) be the point that divides the join of A(-5, 11) and B(4, -7) in the ratio 2 : 7

Formula: If m₁ : m₂ is the ratio in which the join of two points is divided by another point (x, y), then

Here, x₁ = -5, x₂ = 4, y₁ = 11 and y₂ = -7

Substituting,

⇒ x = -3

⇒ y = 8

Therefore, the coordinates of the point which divided the join of A(-5, 11) and B(4, -7) in the ratio 2 : 7 is (-3, 8).

Let the point which cuts the join of A(4, 5), and B(-10, -2) in the ratio k : 1 be P(x, 0)

Formula: If k : 1 is the ratio in which the join of two points is divided by another point (x, y), then

Taking for the y co-ordinate,

⇒ 2k = 5

Therefore,

x = -6

Therefore, the ratio in which x-axis cuts the join of the points A(4, 5) and B(-10, -2) is 5 : 2and the point of intersection is (-6, 0).

Let the point which cuts the join of A(-4, 2) and B(8, 3) in the ratio k : 1 be P(0, y)

Formula: If k : 1 is the ratio in which the join of two points are divided by another point (x, y), then

Taking for the x co-ordinate,

⇒ 8k = 4

Therefore,

Therefore, the ratio in which the line segment joining the points A(-4, 2) and B(8, 3) divided by the y-axis is 1 : 2 and the point of intersection is

Let the new origin be (h, k) = (1, 2) and (x, y) = (3, -4) be the given point.

Let the new coordinates be (X, Y)

We use the transformation formula:

x = X + h and y = Y + k

⇒ 3 = X + 1 and -4 = Y + 2

⇒ X = 2 and Y = -6

Thus, the new coordinates are (2, -6)

Let the new origin be (h, k) = (-3, -2) and (x, y) = (3, -5) be the given point.

Let the new coordinates be (X, Y)

We use the transformation formula:

x = X + h and y = Y + k

⇒ 3 = X – 3 and -5 = Y – 2

⇒ X = 6 and Y = -3

Thus, the new coordinates are (6, -3)

Let the new origin be (h, k) = (0, -2) and (x, y) = (3, 2) be the given point.

Let the new coordinates be (X, Y)

We use the transformation formula:

x = X + h and y = Y + k

⇒ 3 = X + 0 and 2 = Y + (-2)

⇒ X = 3 and Y = 4

Thus, the new coordinates are (3, 4)

Let the new origin be (h, k) = (2, -1) and (x, y) = (-3, 5) be the given point.

Let the new coordinates be (X, Y)

We use the transformation formula:

x = X + h and y = Y + k

⇒ -3 = X + 2 and 5 = Y + (-1)

⇒ X = -5 and Y = 6

Thus, the new coordinates are (-5, 6)

Let (h, k) be the point to which the origin is shifted. Then,

x = -4, y = 2, X = 3 and Y = -2

∴ x = X + h and y = Y + k

⇒ -4 = 3 + h and 2 = -2 + k

⇒ h = -7 and k = 4

Hence, the origin must be shifted to (-7, 4)

Let the new origin be (h, k) = (1, 1)

Then, the transformation formula become:

x = X + 1 and y = Y + 1

Substituting the value of x and y in the given equation, we get

x² + xy – 3x – y + 2 = 0

Thus,

(X + 1)² + (X + 1)(Y + 1) – 3(X + 1) – (Y + 1) + 2 = 0

⇒ (X² + 1 + 2X) + XY + X + Y + 1 – 3X – 3 – Y – 1 + 2 = 0

⇒ X² + 1 + 2X + XY – 2X – 1 = 0

⇒ X² + XY = 0

Hence, the transformed equation is X² + XY = 0

Let the new origin be (h, k) = (1, 1)

Then, the transformation formula become:

x = X + 1 and y = Y + 1

Substituting the value of x and y in the given equation, we get

xy – y² – x + y = 0

Thus,

(X + 1)(Y + 1) – (Y + 1)² – (X + 1) + (Y + 1) = 0

⇒ XY + X + Y + 1 – (Y² + 1 + 2Y) – X – 1 + Y + 1 = 0

⇒ XY + X + Y + 1 – Y² – 1 – 2Y – X + Y = 0

⇒ XY – Y² = 0

Hence, the transformed equation is XY – Y² = 0

Let the new origin be (h, k) = (1, 1)

Then, the transformation formula become:

x = X + 1 and y = Y + 1

Substituting the value of x and y in the given equation, we get

x² – y² – 2x + 2y = 0

Thus,

(X + 1)² – (Y + 1)² – 2(X + 1) + 2(Y + 1) = 0

⇒ (X² + 1 + 2X) – (Y² + 1 + 2Y) – 2X – 2 + 2Y + 2 = 0

⇒ X² + 1 + 2X – Y² – 1 – 2Y – 2X + 2Y = 0

⇒ X² – Y² = 0

Hence, the transformed equation is X² – Y² = 0

Let the new origin be (h, k) = (1, 1)

Then, the transformation formula become:

x = X + 1 and y = Y + 1

Substituting the value of x and y in the given equation, we get

xy – x – y + 1 = 0

Thus,

(X + 1)(Y + 1) – (X + 1) – (Y + 1) + 1 = 0

⇒ XY + X + Y + 1 – X – 1 – Y – 1 + 1 = 0

⇒ XY = 0

Hence, the transformed equation is XY = 0

Let the new origin be (h, k) = (1, -2)

Then, the transformation formula become:

x = X + 1 and y = Y + (-2) = Y – 2

Substituting the value of x and y in the given equation, we get

2x² + y² – 4x + 4y = 0

Thus,

2(X + 1)² + (Y – 2)² – 4(X + 1) + 4(Y – 2) = 0

⇒ 2(X² + 1 + 2X) + (Y² + 4 – 4Y) – 4X – 4 + 4Y – 8 = 0

⇒ 2X² + 2 + 4X + Y² + 4 – 4Y – 4X + 4Y – 12 = 0

⇒ 2X² + Y² – 6 = 0

⇒ 2X² + Y² = 6

Hence, the transformed equation is 2X² + Y² = 6

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

x – 2y + 3 = 0 …(i)

2x – 3y + 4 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 2, we get

2x – 4y + 6 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 3y + 4 – 2x + 4y – 6 = 0

⇒ y – 2 = 0

⇒ y = 2

Putting the value of y in eq. (i), we get

x – 2(2) + 3 = 0

⇒ x – 4 + 3 = 0

⇒ x – 1 = 0

⇒ x = 1

Hence, the point of intersection P(x₁, y₁) is (1, 2)

Let AB is the line drawn from the point of intersection (1, 2) and passing through the point (4, -5)

Firstly, we find the slope of the line joining the points (1, 2) and (4, -5)

Now, we have to find the equation of line passing through point (4, -5)

Equation of line: y – y₁ = m(x – x₁)

⇒ 3y + 15 = -7x + 28

⇒ 7x + 3y + 15 – 28 = 0

⇒ 7x + 3y – 13 = 0

Hence, the equation of line passing through the point (4, -5) is 7x + 3y – 13 = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

x – y = 7 …(i)

2x + y = 2 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 2, we get

2x – 2y = 14 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 2y – 2x – y = 14 – 2

⇒ – 3y = 12

⇒ y = -4

Putting the value of y in eq. (i), we get

x – (-4) = 7

⇒ x + 4 = 7

⇒ x = 7 – 4

⇒ x = 3

Hence, the point of intersection P(x₁, y₁) is (3, -4)

Let AB is the line drawn from the point of intersection (3, -4) and passing through the origin.

Firstly, we find the slope of the line joining the points (3, -4) and (0, 0)

Now, we have to find the equation of the line passing through the origin

Equation of line: y – y₁ = m(x – x₁)

⇒ 4x + 3y = 0

Hence, the equation of the line passing through the origin is 4x + 3y = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

x + y = 9 …(i)

2x – 3y + 7 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 2, we get

2x + 2y = 18

or 2x + 2y – 18 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 3y + 7 – 2x – 2y + 18 = 0

⇒ -5y + 25 = 0

⇒ -5y = -25

⇒ y = 5

Putting the value of y in eq. (i), we get

x + 5 = 9

⇒ x = 9 – 5

⇒ x = 4

Hence, the point of intersection P(x₁, y₁) is (4, 5)

Now, we have to find the equation of the line passing through the point (4, 5) and having slope

Equation of line: y – y₁ = m(x – x₁)

⇒ 2x + 3y – 15 – 8 = 0

⇒ 2x + 3y – 23 = 0

Hence, the equation of line having slope -2/3 is 2x + 3y – 23 = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

x – y = 1 …(i)

2x – 3y + 1 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 2, we get

2x – 2y = 2

or 2x – 2y – 2 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 3y + 1 – 2x + 2y + 2 = 0

⇒ -y + 3 = 0

⇒ y = 3

Putting the value of y in eq. (i), we get

x – 3 = 1

⇒ x = 1 + 3

⇒ x = 4

Hence, the point of intersection P(x₁, y₁) is (4, 3)

Now, we find the slope of the given equation 3x + 4y = 12

We know that the slope of an equation is

So, the slope of a line which is parallel to this line is also

Then the equation of the line passing through the point (4, 3) having a slope is:

y – y₁ = m (x – x₁)

⇒ y – 3 = - 3x + 12

⇒ 4y – 12 = -3x + 12

⇒3x + 4y – 12 – 12=0

⇒ 3x + 4y – 24 = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

5x – 3y = 1 …(i)

2x + 3y = 23 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Adding eq. (i) and (ii) we get

5x – 3y + 2x + 3y = 1 + 23

⇒ 7x = 24

Putting the value of x in eq. (i), we get

Hence, the point of intersection P(x₁, y₁) is

Now, we know that, when two lines are perpendicular, then the product of their slope is equal to -1

m₁ × m₂ = -1

⇒ Slope of the given line × Slope of the perpendicular line = -1

Slope of the perpendicular line

⇒The slope of the perpendicular line

So, the slope of a line which is perpendicular to the given line is

Then the equation of the line passing through the point having slope

y – y₁ = m (x – x₁)

⇒ 63x + 105y – 781 = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

2x – 3y = 0 …(i)

4x – 5y = 2 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 2, we get

4x – 6y = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

4x – 5y – 4x + 6y = 2 – 0

⇒ y = 2

Putting the value of y in eq. (i), we get

2x – 3(2) = 0

⇒ 2x – 6 = 0

⇒ 2x = 6

⇒ x = 3

Hence, the point of intersection P(x₁, y₁) is (3, 2)

Now, we know that, when two lines are perpendicular, then the product of their slope is equal to -1

m₁ × m₂ = -1

⇒ Slope of the given line × Slope of the perpendicular line = -1

⇒ The slope of the perpendicular line = 2

So, the slope of a line which is perpendicular to the given line is 2

Then the equation of the line passing through the point (3, 2) having slope 2 is:

y – y₁ = m (x – x₁)

⇒ y – 2 = 2(x – 3)

⇒ y – 2 = 2x – 6

⇒ 2x – y – 6 + 2 = 0

⇒ 2x – y – 4 = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

x – 7y + 5 = 0 …(i)

3x + y – 7 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 3, we get

3x – 21y + 15 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

3x + y – 7 – 3x + 21y – 15 = 0

⇒ 22y - 22 = 0

⇒ 22y = 22

⇒ y = 1

Putting the value of y in eq. (i), we get

x – 7(1) + 5 = 0

⇒ x – 7 + 5 = 0

⇒ x – 2 = 0

⇒ x = 2

Hence, the point of intersection P(x₁, y₁) is (2, 1)

The equation of line parallel to x – axis is of the form

y = b where b is some constant

Given that this equation of the line passing through the point of intersection (2, 1)

Hence, point (2, 1) will satisfy the equation of a line.

Putting y = 1 in the equation y = b, we get

y = b

⇒ 1 = b

or b = 1

Now, the required equation of a line is y = 1

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

2x – 3y + 1 = 0 …(i)

x + y – 2 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 2, we get

2x + 2y – 4 = 0 …(iii)

On subtracting eq. (iii) from (i), we get

2x – 3y + 1 – 2x – 2y + 4 = 0

⇒ -5y + 5 = 0

⇒ -5y = -5

⇒ y = 1

Putting the value of y in eq. (ii), we get

x + 1 – 2 = 0

⇒ x – 1 = 0

⇒ x = 1

Hence, the point of intersection P(x₁, y₁) is (1, 1)

The equation of a line parallel to y – axis is of the form

x = a where a is some constant

Given that this equation of the line passing through the point of intersection (1, 1)

Hence, point (1, 1) will satisfy the equation of a line.

Putting x = 1 in the equation y = b, we get

x = a

⇒ 1 = a

or a = 1

Now, required equation of line is x = 1

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

2x + 3y – 2 = 0 …(i)

x – 2y + 1 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 2, we get

2x – 4y + 2 = 0 …(iii)

On subtracting eq. (iii) from (i), we get

2x + 3y – 2 – 2x + 4y – 2 = 0

⇒ 7y – 4 = 0

⇒ 7y = 4

Putting the value of y in eq. (ii), we get

Hence, the point of intersection P(x₁, y₁) is

Now, the equation of a line in intercept form is:

where a and b are the intercepts on the axis.

Given that: a = 3

⇒ bx + 3y = 3b …(i)

⇒ b + 12 = 21b

⇒ b – 21b = - 12

⇒ 20b = 12

Putting the value of ‘b’ in eq. (i), we get

⇒ 3x + 15y = 9

⇒ x + 5y = 3

Hence, the required equation of line is x + 5y = 3

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

3x – 4y + 1 = 0 …(i)

5x + y – 1 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 4, we get

20x + 4y – 4 = 0 …(iii)

On adding eq. (iii) and (i), we get

20x + 4y – 4 + 3x – 4y + 1 = 0

⇒ 23x – 3 = 0

⇒ 23x = 3

Putting the value of x in eq. (ii), we get

Hence, the point of intersection P(x₁, y₁) is

Now, the equation of line in intercept form is:

where a and b are the intercepts on the axis.

Given that: a = b

⇒ x + y = a …(i)

Putting the value of ‘a’ in eq. (i), we get

⇒ 23x + 23y = 11

Hence, the required line is 23x + 23y = 11

We know that the slope of a given line is given by

Slope = Where angle of inclination

(i) Given that θ = 30°

Slope = tan(30°) =

(ii) Given that θ = 120°

Slope = tan(120°) = tan(90°+30°) = -cot(30°) =

(iii) Given that θ = 135°

Slope = tan(135°) = tan(90°+45°) = -cot(45°)

(iv) Given that θ = 90°

Slope = tan(90°) = ∞

We know that the slope of a given line is given by

Slope = Where angle of inclination

If a line passing through then slope of the line is given by

(i) Given points are (0,0) and (4,-2)

(ii) Given points are (0, -3) and (2, 1)

(iii) Given points are (2, 5) and (-4, -4)

(iv) Given points are (-2, 3) and (4, -6)

If a line passing through then slope of the line is given by .

Given points are A(x,2) and B(6,-8), and the slope is

⇒-40 = -30+5x
⇒5x = -10
⇒x = -2

We know that for two lines to be parallel, their slope must be the same.

Given points are A(5,6),B(2,3) and C(9,-2),D(6,-5)

⇒1 = 1

Hence proved.

We know that for two lines to be parallel, their slope must be the same. The given points are A(3,x),B(2,7) and C(-1,4),D(0,6)

⇒-2 = 7-x
⇒x = 9

For two lines to be perpendicular, their product of slope must be equal to -1.

Given points are A(-2,6),B(4,8) and C(3,-3),D(5,-9)

Slope of line AB×slope of line CD = -1

⇒-1 = -1
⇒LHS = RHS

For two lines to be perpendicular, their product of slope must be equal to -1.

Given points are A(2, -5),B(-2, 5) and C(x, 3),D(1, 1)

Slope of line AB is equal to

And the slope of line CD is equal to

Their product must be equal to -1

the slope of line AB×Slope of line CD = -1

⇒5 = x-1
⇒x = 6

The is made up of three lines, AB,BC and CA

For a right angle triangle, two lines must be at 90° so they are perpendicular to each other.

Checking for lines AB and BC

For two lines to be perpendicular, their product of slope must be equal to -1.

Given points A(1, 2), B(4, 5) and C(6, 3)

Slope of AB =

Slope of BC =

Slope of CA =

Checking slopes of line AB and BC

1×-1 = -1

So AB is Perpendicular to BC .

So it is a right angle triangle.

For three points to be collinear, the slope of all pairs must be equal, that is the slope of AB = slope of BC = slope of CA

Given points are A(6, -1), B(5, 0) and C(2, 3)

Slope of AB =

Slope of BC =

Slope of CA =

Therefore slopes of AB, BC and CA are equal, so Points A,B,C are collinear.

For three points to be collinear, the slope of all pairs must be equal, that is the slope of AB = slope of BC = slope of CA

Given points are A(5, 1), B(1, -1) and C(x, 4)

Slope of AB =

The slope of BC =

Slope of CA =

The slope of all lines must be the same

Note:- We can use any two points to get the value of “x”.

A rectangle has all sides perpendicular to each other, so the product of slope of every adjacent line is equal to -1.

Given point in order are A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)

Slope of AB =

Slope of BC =

The slope of CD =

Slope of DA =

⇒slopeofAB × slopeofBC

Hence AB is perpendicular to BC

Slope of BC × slope of CD

Hence BC is perpendicular to CD

Slope of CD × slope of DA

Hence CD is perpendicular to DA

Slope of DA × slope of AB

Hence DA is perpendicular to AB.

All angles are 90°.

So this is a rectangle ABCD.

The property of parallelogram states that opposite sides are equal.

We have 4 sides as AB,BC,CD,DA

Given points are A(-2,-1),B(1,0),C(4,3) and D(1,2)

AB and CD are opposite sides, and BC and DA are the other two opposite sides.

So slopes of AB = CD and slopes BC = DA

Slope of AB =

The slope of BC =

The slope of CD =

Slope of DA =

Therefore the Slope of AB = Slope of CD and

The slope of BC = Slope of DA

Also, the product of slope of two adjacent sides is not equal to -1, therefore it is not a rectangle.

Hence ABCD is a parallelogram.

For the lines to be in a line, the slope of the adjacent lines should be the same.

Given points are A(h,k),B(x₁,y₁) and C(x_2,y₂)

So slope of AB = BC = CA

Slope of AB =

Slope of BC =

Slope of CA =

=

Now Cross multiplying the first two equality,

Hence proved.

Given points are A(a,0),B(0,b) and P(x,y)

For three points to be collinear, the slope of all pairs must be equal, that is the slope of AB = slope of BP = slope of PA.

Slope of AB =

Slope of BP =

Slope of PA =

Now Slope of AB = BP = PA

Using the first two equality

Dividing the equation by “ab”, We get

Hence proved.

For the line to make an obtuse angle with X-axis, the angle of the line should be greater than 90

For the angle to be greater than 90°, tanθ must be negative

Where tanθ is the slope of the line.

Given points are A(4, -6) and B(-2, -5)

The slope of line AB is

Which is less than 0, hence negative.

,tanθ is negative in 2^nd quadrant whose angle is >90°.

So line AB makes obtuse angle(>90) with the X-axis.

The vertices of the given quadrilateral are A(-4,-2) B(2, 6), C(8, 5) and D(9, -7)

The mid point of a line A(x₁,y₁) and B(x₂,y₂) is found out by

Now midpoint of AB =

The midpoint of BC =

The midpoint of CD =

Midpoint of DA =

So now we have four points

P(-1,2),Q(5,5.5),R(8.5,-1),S(2.5,-4.5)

Slope of PQ =

Slope of QR =

Slope of RS =

Slope of SP =

Now we can observe that slope of PQ = RS and slope of QR = SP

Which shows that line PQ is parallel to RS and line QR is parallel to SP

Also, the product of two adjacent lines is not equal to -1

Therefore PQRS is a parallelogram.

According to the given figure, the angle made by the line from X-axis is 90+30 = 120°

We also know that slope of a line is equal to tanθ, Where

θ = 120°

tan(120°) = tan(90°+30°) = -cot(30°) = -√3

Therefor the slope of the given line is -√3.

To find out the angle between two lines, the angle is equal to the difference in θ.

The slope of a line = tanθ =

So slope of the first line =

The slope of the second line =

Now the difference between the two lines is θ₁-θ₂

= 60°-30°

= 30°

We know that if slope of two lines are m1 and m2 respectively, then the angle between them is given by

Here and

Where θ is the angle between two lines.

The points are represented as below:



Now let us find the slope of line AC,


The line AB is parallel to x-axis.

We can see that, slope of AC = - tan (A)

-2 = -tan A

tan A = 2

Now, finding the slope of BC,


The line AB is parallel to x-axis.

Therefore, slope of AB = - tan (B)

-2/3 = -tan B

tan B = 2/3

Now,


Putting the values, we will get,

tan C = 8.

The given points are A(0,0),B(2,3) and C(2,-2),D(3,5).

The slope of line AB is

And the slope of line CD is

We know that angle between two lines with their slopes as m₁ and m₂ is given by

Hence proved.

Given points of the parallelogram are A(0, 2), B(2,-1), C(4, 0) and D(2, 3)

The slope of diagonal AC =

The slope of diagonal BD =

So diagonal BD is perpendicular to X-axis. Hence it is parallel to Y-axis.

Product of slope of two diagonals is equal to -1.

Hence proved.

In a square, all sides are perpendicular to the adjacent side, so the product of slope of two adjacent sides is -1.

Let the position of point D(a,b).

Given points of the square are A(0, 6),B(2, 1),C(7, 3) and D(a,b).

The slope of line AB =

The slope of line BC =

The slope of line CD =

The slope of line DA =

The slope of diagonal AC =

The slope of diagonal BD = m₅

(i) We know that in a square, two diagonals are perpendicular to each other, therefore

The slope of diagonal AC×slope of diagonal BD = -1

So the slope of diagonal BD is 7/3.

(ii) We know that midpoint of diagonal AC = midpoint of diagonal BD

and comparing x and y coordinates respectively.

So coordinate of the point D(5,8).

Given points are

A(1, 1), B(7, 3) and C(3, 6)

Slope of line BC =

(i) As D is the midpoint of BC, coordinate of D are

Now the slope of AD =

(ii) As AL is perpendicular to BC

The slope of AL×slope of BC = -1

Let slope of AL be m₁

So Slope of line AL is .

(i) Equation of line parallel to x - axis is given by y = constant, as the y - coordinate of every point on the line parallel to x - axis is 4,i.e. constant. Now the point lies above x - axis means in positive direction of y - axis,

So, the equation of line is given as y = 4.

(ii) Equation of line parallel to x - axis is given by y = constant, as the y - coordinate of every point on the line parallel to x - axis is - 5 i.e. constant. Now the point lies below x - axis means in negative direction of y - axis,

So, the equation of line is given as y = - 5.

(i) Equation of line parallel to y - axis is given by x = constant, as the x - coordinate of every point on the line parallel to y - axis is 6 i.e. constant. Now the point lies to the right of y - axis means in the positive direction of x - axis,

So, required equation of line is x = 6.

(ii) Equation of line parallel to y - axis is given by x = constant, as the x - coordinate of every point on the line parallel to y - axis is - 3. Now point lies to the left of y - axis means in the negative direction of x - axis,

So, required equation of line is given as x = - 3.

Equation of line parallel to x - axis is given by y = constant, as x - coordinate of every point on the line parallel to y - axis is - 3 i.e. constant.

So, the required equation of line is y = - 3.

Equation of line parallel to x - axis (horizontal) is y = constant, as y - coordinate of every point on the line parallel to x - axis is - 2 i.e. constant. Therefore equation of the line parallel to x - axis and passing through (4, - 2) is y = - 2.

Equation of line parallel to y - axis (vertical) is given by x = constant, as x - coordinate is constant for every point lying on line i.e. 6.

So, the required equation of line is given as x = 6.

For the equation of line equidistant from both lines, we will find point through which line passes and is equidistant from both line.

As any point lying on x = - 2 line is ( - 2, 0) and on x = 6 is (6, 0), so mid - point is

(x, y) = (2,0)

So, equation of line is x = 2.

For the equation of line equidistant from both lines, we will find point through which line passes and is equidistant from both line.

As any point lying on y = 8 line is (0, 8) and on y = - 2 is (0, - 2), so mid - point is

(x, y) = (0, 3)

So, equation of line is y = 3.

As slope is given m = 4 and passing through (5, - 7).using slope - intercept form of equation of line, we will find value of intercept first

y = mx + c ………..(1)

- 7 = 4(5) + c

- 7 = 20 + c

c = - 7 - 20

c = - 27

Putting the value of c in equation (1), we have

y = 4x + ( - 27)

y = 4x - 27

4x - y - 27 = 0

So, the required equation of line is 4x - y - 27 = 0.

As slope is given m = - 3 and line is passing through point ( - 2, 3).Using slope - intercept form of equation of line, we will find intercept first

y = mx + c ……………….(1)

3 = - 3( - 2) + c

3 = 6 + c

c = 3 - 6

c = - 3

Putting the value of c in equation (1), we have

y = - 3x + ( - 3)

y = - 3x - 3

3x + y + 3 = 0

So, the required equation of line is 3x + y + 3 = 0.

We have given angle so we have to find slope first given by m = tanθ.

(tanx is negative in II quadrant)

Now the line is passing through the point (0, 2).Using the slope - intercept form of the equation of the line, we will find intercept

y = mx + c ……………….(1)

Putting the value of c in equation(1),we have

As angle is given so we have to find slope first given by m = tanθ

m = tan30°

Now the line is passing through the point (0, 5).using slope - intercept form of the equation of the line, we will find the intercept

y = mx + c ……………….(1)

Putting the value of c in equation (1),we have

As angle is given so we have to find slope first give by m = tanθ

m = tan150°

(tan (180° - θ) is in II quadrant, tanx is negative)

Now the line is passing through the point (3, - 5).Using the slope - intercept form of the equation of the line, we will find the intercept

y = mx + c ……………….(1)

Putting the value of c in equation (1),we have

As angle is given so we have to find slope first give by m = tanθ

m = tan120°

)

(tan (180° - θ) is in II quadrant, tanx is negative)

Now equation of line passing through origin is given as y = mx

As intercept is given i.e. c = 5 and angle given so first we will find slope of line.

m = tanθ

Now using slope intercept form of the equation of a line

y = mx + c

.

As two points passing through a line parallel to the line are given, we will calculate slope using two points(slope of parallel lines is equal).

Now using the slope - intercept form, we will find intercept for a line passing through (4, - 5)

y = mx + c ………………….(1)

Putting value in equation (1)

.

As two points passing through line perpendicular to the line are given, we will calculate slope using two points. Let slopes of the two lines be m1 and m2.

Now the slope of the equation can be found using

Using slope - intercept form we will find intercept for line passing through ( - 3, 5)

y = mx + c ………………….(1)

5 = 5( - 3) + c

c = 5 + 15

c = 20

Putting value in equation (1)

y = 5x + 20

5x - y + 20 = 0

So, the required equation of line 5x - y + 20 = 0.

Slope of equation can be calculated using

Now using two point form of the equation of a line

8y + 16 = 5x - 15

5x - 8y - 16 - 15 = 0

5x - 8y - 31 = 0

So, required equation of line is 5x - 8y - 31 = 0.

The slope of the equation can be calculated using

Now using two point form of the equation of a line

3y - 3 + 5x + 5 = 0

5x + 3y + 2 = 0

So, required equation of line is 5x - 8y - 31 = 0.

The slope of the equation can be calculated using

Now using two point form of the equation of a line

3x - 15 - 5y + 15 = 0

3x - 5y = 0

So, required equation of line is 3x - 5y = 0.

The slope of the equation can be calculated using

m = 0 (Horizontal line)

Now using two point form of the equation of a line

y - b = 0(x - a)

y = b

So, required equation of line is y = b.

To find angle, we will find slope using two points

Now as we have m = tanθ

So, angle line makes with the positive x - axis is 60°.

If two lines having the same slope pass through a common point, then two lines will coincide. Hence, if A, B and C are three points in the XY - plane, then they will lie on a line, i.e., three points are collinear if and only if slope of AB = slope of BC.

Slope of AB = slope of BC

- 3 = - 3

Hence verified, i.e. points are collinear. Now using two point form of the equation

y - 4 = - 3(x - 1)

y - 4 + 3x - 3 = 0

3x + y - 7 = 0

So, required equation of line is 3x + y - 7.

Using two point form equation of lines AB, BC and AC can be find. Now A is origin so the lines passing through A (origin) are simply y = mx so we have to find slope of AB and AC.

For line AB,

m = 2

So, the equation of line AB is y = 2x.

For line AC,

Now using y = mx

So, the equation of line AC is 2x - 3y = 0.

Now for line BC, the y coordinate of both is same means horizontal line (parallel to the x - axis) then the equation of line BC is given as

y = 4

So, the required equations of lines for AB: y = 2x

AC: 2x - 3y = 0

BC: y = 4

Construction: - Draw median from vertices A, B and C on lines BC,AC and AC respectively .Let the mid - points of lines BC,AC and AB be L,M and N respectively.

Now find the coordinate of L, M and N using mid - point theorem.

Now equation of medians AL, BM and CN using two point form

For median AL,

4(y - 6) = - 29(x + 1)

4y - 24 + 29x + 29 = 0

29x + y + 5 = 0

For median BM,

5y + 45 = 8x + 24

8x - 5y + 24 - 45 = 0

8x - 5y - 21 = 0

For median CN,

4(y + 8) = 13(x - 5)

4y + 32 = 13x - 65

13x - 4y - 65 - 32 = 0

13x - 4y - 97 = 0

So, the required line of equations for medians are for AL: 29x + y + 5 = 0

For BM: 8x - 5y - 21 = 0

For CN: 13x - 4y - 97 = 0

Perpendicular bisector: A perpendicular bisector is a line segment which is perpendicular to the given line segment and passes through its mid - point (or we can say bisects the line segment).

Now to find the equation of perpendicular bisector first, we will find mid - point of the given line using mid - point formula (call it midpoint as M),

Now we will calculate the slope of the given line and since lines are perpendicular, so the slope of two is related as m1.m2 = - 1.

Now the slope of perpendicular bisector is

Now equation of perpendicular bisector using two point form,

10y - 65 = 28x - 84

28x - 10y - 84 + 65 = 0

28x - 10y - 19 = 0

So, required equation of perpendicular bisector 28x - 10y - 19 = 0.

Altitude: A line drawn from the vertex that meets the opposite side at right angles. It determines the height of the triangle.

In triangle ABC, let the altitudes from vertices A, B and C are AL, BM and CN on sides BC,AC and AB respectively.

Now we will find slope of sides and using the relation between the slopes of perpendicular lines i.e. m1.m2 = - 1 we will find the slopes of altitudes.

Now equation of altitudes using two point form

For altitude AL,

y - ( - 2) = - 2(x - 2)

y + 2 + 2x - 4 = 0

2x + y - 2 = 0

For altitude BM,

y - 1 = - 1(x - 1)

y - 1 + x - 1 = 0

x + y - 2 = 0

For altitude CN,

3y = x + 1

x - 3y + 1 = 0

So, the required equations of altitudes are for AL: 2x + y - 2 = 0

For BM: x + y - 2 = 0

For CN: x - 3y + 1 = 0

Construction: Draw a line from vertex A intersecting side BC of the triangle at D (as there is one bisector for exterior angle also but it is the default that we have to find interior angle bisector).

As the angle between the sides AB and angle bisector AD and side AC and angle bisector AD is equal.

Then using the angle between two lines, if the slope of AD be m and slope of AB

Putting the values in the equation

……………..(1)

…………………(2)

Again for side AC slope

Putting in equation (1)

..……………….(3)

From equation (2) and (3),we have

tanθ = - 3 as tanx is negative in II and IV quadrant means it is obtuse angle either way(exterior here)we require interior angle so will consider the positive value of m.

As we obtained the slope of angle bisector which passes through A vertex so using slope intercept form first calculate the value of the intercept

y = mx + c ……………….(4)

Putting the value of c in equation (4),we have

So, the required equation of angle bisector is x - 3y + 5 = 0.

Let us consider the coordinates of vertices of triangle A, B, C be (a, b), (c, d) and (e, f). Now using mid - point formula

Now from above equations, we have

c + e = 4, d + f = 2 (i)

a + e = - 10, b + f = 14 (ii)

a + c = - 10, b + d = - 10 (iii)

From subtract (i) from (ii),we get

a - c = - 14, b - d = 12 (iv)

Adding (iii) and (iv)

Putting values of a, b in equation (iii)

Again putting values in (i)

So coordinates of A ( - 12,1), B(2, - 11) and C(2,13).

Using two point form of the equation

Equation of side AB:

14(y - 1) = - 12(x + 12)

14y - 14 + 12x + 144 = 0

12x + 14y + 130 = 0

6x + 7y + 65 = 0

Equation of side BC:

y = - 11(slope is not defined i.e. line is vertical)

Equation of side CA:

14(y - 13) = 12(x - 2)

12x - 24 - 14y + 182 = 0

12x - 14y + 158 = 0

6x - 7y + 79 = 0

So, the required equations of sides for AB: 6x + 7y + 65 = 0

For BC: y = - 11

For CA: 6x - 7y + 79 = 0

Construction: Draw a line segment from vertex A intersecting BC at the midpoint (D).

(i)Equation of median AD, we will find the midpoint of side BC

Now using two point form of the equation of the line, we have

Equation of side AD:

13x - y - 13 + 4 = 0

13x - y - 9 = 0

So, required equation of altitude is 3x - y - 9 = 0.

(ii) For the equation of altitude, we will need slope as we have a point through which line passes (A).

Now we will find the slope of side BC and using the relation between the slopes of perpendicular lines, i.e. m₁ .m₂ = - 1 we will find the slopes of altitude.

Using slope intercept form, we will first calculate intercept,

y = mx + c ……………………(1)

Putting in equation (1)

So, required equation of altitude is 3x + 5y - 23 = 0.

(iii)We have a slope of perpendicular and a mid point from the previous solution

,

Now for perpendicular bisector, it passes through the midpoint of BC, i.e. we have a slope of the equation and a point through which it passes so we can use the slope - intercept form and calculate intercept,

y = mx + c …………………(i)

Putting in equation (i) value of c,

So, the required equation of perpendicular bisector is 3x + y + 11 = 0.

(i) Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

Here, m = 3 and c = 5.

Hence, y = (3)x + (5)

i.e. y = 3x + 5

(ii) Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

Here, m = - 1 and c = 4.

Hence, y = ( - 1)x + (4)

i.e. x + y = 4

(iii) Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

Here, m = and c = - 3.

Hence, y = ()x + ( - 3)

Or, 5y = - 2x - 3 i.e. 2x + 5y + 3 = 0

Given : The given line makes an angle of 30° with the x - axis. The y - intercept = - 4.

So, the slope of the line is m = tan = tan30° = .

Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

The equation of the line is y =

Or,√3y = x - 4√3 i.e. x - √3y = 4√3

Given:

slope, m =

The y - intercept is 6 units.

Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

The equation of the line is

i.e. √3y + x + 6√3 = 0

Given: The line is equally inclined to both the axes.

The angle between the coordinate axes = 90°

If the inclination to both the axes is then

i.e.

slope of the line, m = tan = tan45° = 1

The y - intercept = - 2 units

Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

The equation of the line is y = 1.x + (-2) = x - 2

i.e. x - y = 2

Given: The straight lines are x = 0 and y = 0.

Formula to be used: If is the angle between two straight lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the equation of their angle bisector is

the equation of the angle bisectors is

i.e.

Given: The y - intercept = - 2.

The line passes through ( - 1,5).

Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

The equation of the line is y = mx + ( - 2) = mx - 2.

Now, this line passes through ( - 1,5).

5 = m( - 1) - 2 = - m - 2 i.e. m = - (5 + 2) = - 7

y = ( - 7)x + ( - 2) = - 7x - 2 i.e.7x + y + 2 = 0

Given: The given line is 2x - 3y = 8. The line parallel to this line has a y - intercept of 5units.

Formula to be used: If ax + by = c is a straight line then the line parallel to the given line is of the form ax + by = d, where a,b,c,d are arbitrary real constants.

A line parallel to the given line has a slope of and is of the form 2x - 3y = k , where k is any arbitrary real constant.

Now, 2x - 3y = k

or, 3y = 2x - k

which is of the form y = mx + c , where c is the y - intercept.

So, k = ( - 3)x5 = - 15

Equation of the required line is 2x - 3y = - 15

i.e. 2x - 3y + 15 = 0

Given: The given line is x - 2y + 5 = 0. The line perpendicular to this given line passes through (0,3)

Formula to be used: The product of slopes of two perpendicular lines = - 1.

The slope of this line is 1/2 .

the slope of the perpendicular line =

The equation of the line can be written in the form y = ( - 2)x + c

(c is the y - intercept)

This line passes through (0,3) so the point will satisfy the equation of the line.

3 = ( - 2)x0 + c i.e. c = 3

The required equation is y = - 2x + 3

i.e. 2x + y = 3

Given: The given line is 4x + 3y = 10. The line perpendicular to this line passes through (2,3).

Formula to be used: The product of slopes of two perpendicular lines = - 1

Slope of this line is .

the slope of the perpendicular line =

The equation of the line can be written in the form

(c is the y - intercept)

This line passes through (2,3), so the point will satisfy the equation of the line.

The required equation is

or, 4y = 3x + 6 i.e. 3x - 4y + 6 = 0.

Given: The line is perpendicular to x - axis and passes through (2,4)

The equation of the line perpendicular to the x - axis (y = 0) can be represented as x = c, where c is a real constant.

Now, this line passes through (2,4).

c = 2

The required equation is x = 2

Given: The given line is 3x + 5y = 4. The perpendicular line has an x - intercept of - 3.

Formula to be used: The product of slopes of two perpendicular lines = - 1.

The slope of this line is .

the slope of the perpendicular line =

The equation of the line can be written in the form

(c is the y - intercept)

This line intercepts the x - axis when y = 0.

So, the x - intercept:

Now, it is given that the x - intercept is - 3.

i.e. c = 5

The required equation of the line is

i.e. 5x - 3y + 15 = 0

Given: The given line is 3x + 2y = 8. The perpendicular line passes through the midpoint of (6,4) and (4, - 2).

Formulae to be used: The product of slopes of two perpendicular lines = - 1.

If (a,b) and (c,d) be two points, then their midpoint is given by

The slope of this line is .

the slope of the perpendicular line =

The equation of the line can be written in the form

(c is the y - intercept)

This line passes through the midpoint of (6,4) and (4, - 2).

The co - ordinates of the midpoint of the line joining the given points is

(5,1) satisfies the equation

The required equation is

i.e. 2x - 3y = 7

Given: The line perpendicular to the line passing through ( - 2,3) and (4, - 5) has the y - intercept of - 3.

Formula to be used: If (a,b) and (c,d) are two points then the equation of the line passing through them is

Product of slopes of two perpendicular lines = - 1

The equation of the line joining points ( - 2,3) and (4, - 5) is

or,3y + 15 = - 4x + 16 or, 4x + 3y = 1

Slope of this line is .

the slope of the perpendicular line =

The equation of the line can be written in the form

(c is the y - intercept)

But, the y - intercept is - 3.

The required line is

i.e. 3x - 4y = 12

Given: The line perpendicular to the line passing through (2,5) and ( - 3,6) passes through ( - 3,5).

Formula to be used: If (a,b) and (c,d) are two points then the equation of the line passing through them is

Product of slopes of two perpendicular lines = - 1

The equation of the line joining points (2,5) and ( - 3,6) is

Or, 5y - 25 = - x + 2

i.e. the given line is x + 5y = 27.

The slope of this line is .

the slope of the perpendicular line =

The equation of the line can be written in the form y = 5x + c.

(c is the y - intercept)

This line passes through ( - 3,5).

Hence, 5 = 5x( - 3) + c or, c = 20

The required equation of the line will be y = 5x + 20

i.e. 5x - y + 20 = 0

Given: A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : 2.

Formula to be used: If (a,b) and (c,d) are two points then the equation of the line passing through them is

If (a₁,b₁) and (a₂,b₂) be two points , then the co - ordinates of the point dividing their join in the ratio a:b is given by

The equation of the line joining points (1,0) and (2,3) is

or,

i.e. the given line is 3x - y = 3.

Accordingly, the required co - ordinates of the point dividing the join of (1,0) and (2,3) in the ratio 1:2 are

The given line is 3x - y = 3.

The slope of this line is .

the slope of the perpendicular line =

The equation of the line can be written in the form

(c is the y - intercept)

This line will pass through .

The required equation is

i.e. 3x + 9y = 13

To Find: The equation of a line with intercepts -3 and 5 on the x-axis and y-axis respectively.

Given :Let a and b be the intercepts on x-axis and y-axis respectively.

Then, the x-intercept is a = -3

y-intercept is b = 5

Formula used:

we know that intercept form of a line is given by:

= 1

= 1

5x-3y = -15

5x - 3y + 15 = 0

Hence 5x - 3y + 15 = 0 is the required equation of the given line.

To Find:The equation of the line with intercepts 4 and -6 on the x-axis and y-axis respectively.

Given : Let a and b be the intercepts on x-axis and y-axis respectively.

Then,x-intercept be a = 4

y-intercept be b = -6

Formula used:

we know that intercept form of a line is given by:

= 1

= 1

-3x + 2y = -12

3x – 2y -12 = 0

Hence 3x – 2y -12 = 0 is the required equation of the given line.

To Find: The equation of the line with equal intercepts on the coordinate axes and that passes through the point (4,7).

Given : Let a and b be two intercepts of x-axis and y-axis respectively.

Also,given that two intercepts are equal, i.e., a=b

And (4, 7) passes through the point (x, y)

Formula used:

Now since intercept form of a line is given:

= 1

= 1

= 1

a= 11 = b

Therefore, The required Equation of the line is = 1

x + y = 11

To Find: The equation of the line passing through (3, -5) and cuts off intercepts on the axes which are equal in magnitude but opposite in sign.

Given : Let a and b be two intercepts of x-axis and y-axis respectively.

According to the question a = -b or b = -a

And (3 , -5) passes through the point (x, y), thus satisfies the equation

Formula used:

Now since intercept form of the line is given by , = 1

= 1

= 1

a = 8 and b= -8

Equation of the line is = 1

Hence ,the required equation of the line is = 1 x– y = 8

To Find: The equation of the line passing through the point (2, 2) and cutting off intercepts on the axes, whose sum is 9.

Given : Let a and b be two intercepts of x-axis and y-axis respectively.

sum of the intercepts is 9,i.e,a+b = 9

a = 9 – b or b = 9 – a

Formula used:

The equation of a line is given by:

= 1

The given point (2, 2) passing through the line and satisfies the equation of the line.

= 1

2(9 – a) + 2a = 9a – a²

18 – 2a +2a = 9a – a²

a² – 9a + 18 = 0

a² – 6a – 3a + 18 =0

a(a - 6) - 3(a - 6) = 0

(a - 3) (a - 6) = 0

a = 3, a = 6

when a = 3, b=6 and a=6, b=3

case 1 : when a=3 and b=6

Equation of the line : = 1

Hence, 2x + y = 6 is the required equation of the line.

case 2 : when a=6 and b=3

Equation of the line : = 1

= 1

Hence , x + 2y = 6 is the required equation of the line.

Therefore, 2x + y = 6 is the required equation of the line when a=3 and b=6.And , x + 2y = 6 is the required equation of the line when a=6 and b=3.

To Find:The equation of the line that passes through the point (22, -6) and intercepts on the x-axis exceeds the intercept on the y-axis by 5.

Given : let x-intercept be a and y-intercept be b.

According to the question : a = b + 5

Formula used:

And the given point satisfies the equation of the line, so

= 1

= 1

22b – 6b -30 = b² + 5b

11b – 30 = b²

b² -11b +30 = 0

b² -6b -5b +30 =0

b(b-6) -5(b-6) =0

(b-5) (b-6) =0

The values are b=5 ,b=6

When b=5 then a=10

and b=6 then a=11

case 1 : when b=5 and a=10

Equation of the line : = 1

= 1

= 1

Hence, x + 2y = 10 is the required equation of the line.

case 2 : when b=6 and a=11

Equation of the line : = 1

= 1

= 1

Hence, 6x + 11y = 66 is the required equation of the line.

Therefore, x + 2y = 10 is the required equation of the line when b=5 and a=10 .And 6x + 11y = 66 is the required equation of the line when b=6 and a=11.

To Find: The equation of the line whose portion intercepted between the axes is bisected at the point (3, -2).

Formula used:

Let the equation of the line be

= 1

Since it is given that this equation , whose portion is intercepted between the axes is bisected i.e.; is divided into ratio 1:1 .

Let A(a,0) and B(0,b) be the points foring the coordinate axis.

a and b are intercepts of x and y-axis respectively.

By using mid-point formula (m:n = 1:1)

(x, y) =( =

Since given point (3 , -2) divides coordinate axis in 1:1 ratio

(x , y) = (3 , -2)

=3 and = -2

a=6 b=-4

equation of the line : = 1

= 1

-4x +6y = -24

-2x +3y = -12

Hence the required equation of the line is 2x -3y = 12.

To Find: The equation of the line whose portion intercepted between the coordinate axes is divided at the point (5, 6) in the ratio 3 : 1.

Given : The coordinate axes is divided in the ratio 3 : 1

(x₁ , y₁) = A(a,0)

(x₂ , y₂) = B(0, b)

Where a and b are intercepts of the line.

Formula used:

The equation of the line is : = 1

And the co-ordinate axis is divided at (5,6) , thus by using Section formula

(x ,y) =

= =

(5,6) divides the co-ordinate axis, thus (x,y)= (5,6).

= 5 a = 20 , = 6 b= 8

Equation of the line becomes = 1

8x +20y = 160

2x +5y = 40

Hence the required equation of the line is 2x +5y = 40.

Given : The ratio of the line intercepted between the axes is 2 :3

Let (x₁ , y₁) = A(a,0)

And (x₂ , y₂) = B(0, b)

Where a and b are intercepts of the line.

Formula used:

The equation of the line is : = 1

And the co-ordinate axis is divided at (5,-2) , thus by using Section formula

(x ,y) =

= =

(5,-2) divides the co-ordinate axis, thus (x,y)= (5,-2).

= 5 a = 25/3 , = -2 b= -5

Equation of the line becomes = 1

= 1

= 1

= 1

Hence ,3x – 5y = 25 is the required equation of the line.

To Find: The values of a and b when the line = 1 passes through the points (8, -9) and (12, -15).

Given : the equation of the line : = 1 equation 1

Also (8, -9) passes through equation 1

= 1

8b - 9a = ab equation 2

And (12, -15) passes through equation 1

= 1

12b - 15a = ab equation 3

Solving equation 2 and 3

a= 2.

Put a=2 in equation 2

8b- 9a = ab

8b -18 = 2b

6b =18 b=3

Hence the values of a and b are 2 and 3 respectively.

To Find:The equation of the line.

Given: p = 3 and ∝ = 450

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

X cos ∝ + y sin ∝ =p

X cos 450 + y sin 450 =3

i.e; cos 450 = cos (360 +90) = cos 90 [ cos(360+ x) = cos x]

similarly, sin 450 = sin (360 +90) = sin 90 [ sin(360+ x) = sin x]

hence, x cos 90 + y sin 90 =3

x (0)+ y 1=3

Hence the required equation of the line is y=3.

Given: p = 5 and ∝ = 1350

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ =p

x cos 1350 + y sin 1350 =5

i.e; cos 1350 = cos ((4360) -90) = cos((42 )- 90)= cos 90

similarly, sin 1350 = sin ((4360) - 90) = sin((42 ) - 90)= -sin 90

hence, x cos 90 + y (–sin 90) =5

x (0)- y1=5

Hence The required equation of the line is y=-5.

Given: p = 8 and ∝ = 1500

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ =p

x cos 1500 + y sin 1500 =8

i.e; cos 1500 = cos ((4360) +60) = cos((42 )+60)= cos 60

similarly, sin 1500 = sin ((4360) +60) = sin((42 )+60)= sin 60

hence, x cos 60 + y sin 60 =8

x (1/2)+ y (/2)=8

Hence The Required equation of the line is x + y = 16.

Given: p = 3 and ∝ = 2250

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ =p

x cos 2250 + y sin 2250 =3

i.e; cos 2250 = cos ((6360) +90) = cos((62 )+ 90)= cos 90

similarly, sin 2250 = sin ((660) + 90) = sin((62 ) +90)= sin 90

hence, x cos 90 + y sin 90 =3

x (0)+y1=3

Hence The required equation of the line is y=3.

Given: p = 2 and ∝ = 3000

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

X cos ∝ + y sin ∝ =p

X cos 3000 + y sin 3000 =2

i.e; cos 3000 = cos ((8360) +120) = cos((82 )+120)= cos 120 =cos(180-60)=cos60

similarly, sin 3000 = sin ((8360) +120) = sin((82 )+120)= sin 120

= sin(180-60) = -sin 60

hence, x cos 60 + y (–sin 60) =2

x (1/2)- y (/2)=2

Hence The required equation of the line is x - y = 4

Given: p = 4 and ∝ = 1800

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ =p

x cos 1800 + y sin 1800 =4

i.e; cos 1800 = cos (5360) = cos(52 )= cos 360 = 1

similarly, sin 1800 = sin (5360) = sin(52 )= sin 360 =0

hence, x 1+ y0=4

Hence The required equation of the line is x=4.

To Find: The equation of the line .

Given : p=2 units and = .

Since sin ∝ = =

Using Pythagoras theorem :

adj = = = 2 units.

i.e; cos ∝ = =

Formula used:

equation of the line: x cos ∝ +y sin ∝ =p

x () +y ( ) = 2

Hence ,2 x + y = 6 Or x +y =6 is the required equation of the line.

To Find:The equation of the line.

Given : tan ∝ = and p =3 units.

Since tan ∝ = =

Using Pythagoras theorem :

hyp = = = 13 units.

From the figure: cos ∝ = = and sin ∝ = =

Formula used:

equation of the line: x cos ∝+y sin ∝ =p

x () +y × ( ) = 5

Hence ,12x + 5y = 65 is the required equation of the line.

Given equation is 2x – 3y – 5 = 0

We can rewrite it as 2x – 5 = 3y

This equation is in the slope-intercept form i.e. it is the form of

, where m is the slope of the line and c is y-intercept of the line

Therefore, and

Conclusion: Slope is and y-intercept is

Given equation is 5x + 7y -35 = 0

We can rewrite it as 7y = 35 - 5x

7y = - 5x + 35

This equation is in the slope-intercept form i.e. it is the form of

, where m is the slope of the line and c is y-intercept of the line

Therefore, and c = 5

Conclusion: Slope is and y-intercept is 5

Given equation is y + 5 = 0

We can rewrite it as y = -5

This equation is in the slope-intercept form, i.e. it is the form of

, where m is the slope of the line and c is y-intercept of the line

Therefore, m = 0 and c = - 5

Conclusion: Slope is 0 and y-intercept is -5

Given equation is 3x - 4y +12 = 0

We can rewrite it as 3x - 4y = -12

This equation is in the slope intercept form i.e. in the form + = 1

Where, x-intercept = -4 and y-intercept = 3

Two points are: (-4, 0) on the x-axis and (0, 3) on y-axis

We know distance between two points is

Length of the line

= 5

Conclusion: Length of the portion of the line intercepted between the axes is 5

Given equation is 5x - 12y = 60

We can rewrite it as

This equation is in the slope intercept form i.e. in the form + = 1

Where, x-intercept = 12 and y-intercept = -5

Two points are: (12, 0) on the x-axis and (0,-5) on y-axis

We know the distance between two points is

Length of the line

= 13

Conclusion: length of the portion of the line intercepted between the axes is 13

(i) Given equation is x + y + 6 =0

We can rewrite it as

It is in the form of

Where and

The inclination of the line is

Therefore

Conclusion: Inclination of the line is

(ii) Given equation is

We can rewrite it as

It is in the form of

Where and

Therefore

Conclusion: Inclination of line 3x + 3y + 8 = 0 is

(iii) Given equation is

We can rewrite it as

It is in the form of

Where and c = -4

)

Conclusion: Inclination of the line is

Given equation is = 0

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by , so now

Divide by

This is in the form of

And

Conclusion: and p =1

Given equation is

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by, so now

Divide by

Now we get

This is in the form of

Where

And p = 1

Conclusion: and p =1

If the equation is in the form of ax + by = c, to get into the normal form we should divide it by , so now

Divide by =

This is in the form of , where and

Conclusion: + = is the normal form of x + y - 2 = 0

(ii)

⇒x + y = - √2

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by, so now

Divide by

Our new equation is

This is in the form of , where and p =1

Conclusion: is the normal form of

(iii)

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by, so now

Divide the equation by

Our new equation is

This is in the form of , where and p = 5

Conclusion: is the normal form of

(iv)

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by, so now

Divide by

Our new equation is

This is in the form of , where and

Conclusion: is the normal form of

(v)

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by , so now

Divide by

Our new equation is

This is in the form of , where

or and

Conclusion: is the normal form of

Given: Point (3,-5) and line 3x – 4y = 27

To find: The distance of the point (3, -5) from the line 3x – 4y = 27

Formula used:

We know that the distance between a point P(m,n) and a line ax + by + c = 0 is given by,

D

The equation of the line is 3x – 4y - 27=0

Here m= 3 and n = -5 , a = 3 , b = -4 , c = -27

D

D

D

The distance of the point (3, -5) from the line 3x – 4y = 27 is units

Given: Point (-2,3) and line 12x – 5y = 13

To find: The distance of the point (-2, 3) from the line 12x – 5y = 13

Formula used:

We know that the distance between a point P (m,n) and a line ax + by + c = 0 is given by,

D

The given equation of the line is 12x – 5y - 13=0

Here m= -2 and n = 3 , a = 12 , b = -5 , c = -13

D

D

D

The distance of the point (-2, 3) from the line 12x = 5y + 13 is 4 units

Given: Point (-4,3) and line 4(x + 5) = 3(y – 6)

To find: The distance of the point (-4, 3) from the line 4(x + 5) = 3(y – 6)

Formula used:

We know that the distance between a point P (m,n) and a line ax + by + c = 0 is given by,

D

The equation of the line is 4x + 20 = 3y – 18

4x – 3y + 38 = 0

Here m= -4 and n = 3 , a = 4 , b = -3 , c = 38

D

D

D

The distance of the point (-4, 3) from the line 4(x + 5) = 3(y – 6) is units

Given: Point (2,3) and line y = 4

To find: The distance of the point (2, 3) from the line y = 4

Formula used:

We know that the distance between a point P (m,n) and a line ax + by + c = 0 is given by,

D

The equation of the line is y - 4 = 0

Here m= 2 and n = 3 , a = 0 , b = 1 , c = -4

D

D

D

The distance of the point (2, 3) from the line y = 4 is units

Given: Point (4,2) and the line joining the points (4, 1) and (2, 3)

To find: The distance of the point (4,2) from the line joining the points (4, 1) and (2, 3)

Formula used:

The equation of the line joining the points (x₁,y₁) and (x₂,y₂) is given by

Here x₁ =4 y₁ =1 and x₂ =2 y₂ =3

The equation of the line is

Formula used:

We know that the distance between a point P (m,n) and a line ax + by + c = 0 is given by,

D

The equation of the line is

Here m = 4 and n = 2 , a = 1 , b = 1 , c = -5

D

D

D

The distance of the point (4,2) from the line joining the points (4, 1) and (2, 3) is units

Given: Point (0,0) and line 7x + 24y = 50

To find: The length of the perpendicular from the origin to the line 7x + 24y = 50

Formula used:

We know that the length of the perpendicular from P (m,n) to the line ax + by + c = 0 is given by,

D

The given equation of the line is 7x + 24y - 50=0

Here m= 0 and n = 0 , a = 7 , b = 24 , c = -50

D

D

D

The length of perpendicular from the origin to the line 7x + 24y = 50 is units

(ii) Given: Point (0,0) and line 4x + 3y = 9

To find: The length of perpendicular from the origin to the line 4x + 3y = 9

Formula used:

We know that the length of perpendicular from P (m,n) to the line ax + by + c = 0 is given by,

D

The given equation of the line is 4x + 3y - 9=0

Here m= 0 and n = 0 , a = 4 , b = 3 , c = -9

D

D

D

The length of perpendicular from the origin to the line 4x + 3y = 9 is units

(iii) Given: Point (0,0) and line x = 4

To find: The length of perpendicular from the origin to the line x = 4

Formula used:

We know that the length of perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

The given equation of the line is x - 4=0

Here m= 0 and n = 0 , a = 1 , b = 0 , c = -4

D

D

D

The length of perpendicular from the origin to the line x = 4 is units

Given: Point , and line

To Prove: The product of the lengths of perpendiculars drawn from the points

and to the line , is b²

Formula used:

We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

The equation of the line is

At point A, m= and n= ,

D₁

D₁

At point B, m= and n= ,

D₂

D₂

Product of the lengths of perpendiculars drawn from the points A and B is D₁D₂

D₁D₂

(In the numerator we have and )

D₁D₂

D₁D₂

D₁D₂

Product of the lengths of perpendiculars drawn from the points A and B is

Given: Point (4,1) , line 3x – 4y + k = 0 and length of perpendicular is 2 units

To find: The values of k

Formula used:

We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

The equation of the line is 3x – 4y + k =0

Here m= 4 and n= 1 , a = 3 , b = -4 , c = k and D=2 units

D

D

. or

or

or

The values of k are 2 and -18

Given: Points (7,0) and (2,1) , line 5x + 12y – 9 = 0

To Prove : length of the perpendicular from the point (7, 0) to the line 5x + 12y – 9 = 0 is double the length of perpendicular to it from the point (2, 1)

Formula used:

We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

Let D₁ be the length of perpendicular from the point (7, 0) to the line 5x + 12y – 9 = 0

The given equation of the line is 5x + 12y – 9 = 0

Here at point (7,0) m= 7 and n= 0 , a = 5 , b = 12 , c = -9

D₁

D₁

D₁

Let D₂ be the length of perpendicular from the point (2, 1) to the line 5x + 12y – 9 = 0

The given equation of the line is 5x + 12y – 9 = 0

Here at point (2,1) m= 2 and n= 1 , a = 5 , b = 12 , c = -9

D₂

D₂

D₂

D₁=2D₂=2

Thus the length of the perpendicular from the point (7, 0) to the line 5x + 12y – 9 = 0 is double the length of perpendicular to it from the point (2, 1)

Given: points A(2, 3), B(4, -1) and C(-1, 2) are the vertices of ΔABC

To find : length of the perpendicular from C on AB and the area of ΔABC

Formula used:

We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

The equation of the line joining the points (x₁,y₁) and (x₂,y₂) is given by

The equation of the line joining the points A(2,3) and B(4,-1) is

Here x₁ =2 y₁ =3 and x₂=4 y₂=-1

The equation of the line is 2x + y – 7 =0

The length of perpendicular from C(-1, 2) to the line AB

The given equation of the line is 2x + y – 7 =0

Here m= -1 and n= 2 , a = 2 , b = 1 , c = -7

D

D

D

The length of the perpendicular from C on AB is units.

Height of the triangle is units

The distance between points A(x₁, y₁) and B(x₂, y₂) is given by

AB =

Here x1=2 and y1=3 ,x2=4 and y2=-1

AB =

Base AB = units

Area of the triangle =

Area of the triangle ABC =

Area of the triangle ABC = 7 square units

Given: perpendicular distance is 4 units and line

To find : points on the x-axis

Formula used:

We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

The equation of the line is 4x + 3y - 12=0

Any point on the x-axis is given by (x,0)

Here m= x and n= 0 , a = 4 , b = 3 , c = -12 and D=4 units

D

D

|4x-12| = 4 × 5 = 20

4x - 12 = 20 or 4x - 12 = -20

4x = 20 + 12 or 4x = -20 + 12

4x = 32 or 4x = -8

x = 32/4 = 8 or x = (-8)/4 = -2

(8,0) and (2,0)are the points on the x-axis whose perpendicular distance from the line is 4 units

Given: points lie on the line x + y = 4 , perpendicular distance = 1 units

To find : points on the line x + y = 4

Formula used:

We know that the distance between a point (m,n) and a line ax + by + c = 0 is given by,

D

The equation of the line is 4x + 3y – 10 = 0 and D=1 units

Here m= x and n= 4 - x (from the equation x + y = 4 ),a = 4 , b = 3 , c = -10

D

D

or

or

or

We know that the points lie on the line x + y = 4

y = 4 – 7 = -3 or y=4-(-3) =7

(7,-3) and (-3,7) are the points on the line x + y = 4 that lie at a unit distance from

4x + 3y = 10.

Given: ABCD is a square and equation of BC is 3x – 4y – 10 = 0

To find : Area of the square

Formula used:

We know that the length of perpendicular from (m,n) to the line ax + by + c = 0 is given by,

D

The given equation of the line is 3x – 4y – 10 = 0

Here m= 0 and n= 0 , a = 3 , b = -4 , c = -10

D

D

D

Side of the square=D=2

Area of the square = 22 = 4 square units

Area of the square = 4 square units

Given: parallel lines 4x – 3y + 5 = 0 and 4x – 3y + 7 = 0

To find : distance between the parallel lines

Formula used :

The distance between the parallel lines ax + by + c =0 and ax + by + d =0 is,

D

Here a = 4 ,b = -3 ,c = 5 ,d = 7

D

The distance between the parallel lines 4x – 3y + 5 = 0 and 4x – 3y + 7 = 0 is units

Given: parallel lines 8x + 15y – 36 = 0 and 8x + 15y + 32 = 0.

To find : distance between the parallel lines

Formula used :

The distance between the parallel lines ax + by + c =0 and ax + by + d =0 is,

D

Here a = 8 ,b = 15 ,c = -36 ,d = 32

D

The distance between the parallel lines 8x + 15y – 36 = 0 and 8x + 15y + 32 =0 is 4 Units

Given: parallel lines y = mx + c and y = mx + d

To find : distance between the parallel lines

Formula used :

The distance between the parallel lines ax + by + c =0 and ax + by + d =0 is,

D

The parallel lines are mx – y + c = 0 and mx – y + d=0

Here a = m ,b = -1 ,c = c ,d = d

D

The distance between the parallel lines y = mx + c and y = mx + d is units

Given: parallel lines p(x + y) = q = 0 and p(x + y) – r =0

To find : distance between the parallel lines p(x + y) - q = 0 and p(x + y) – r =0

Formula used :

The distance between the parallel lines ax + by + c =0 and ax + by + d =0 is,

D

The parallel lines are p(x + y) - q = 0 and p(x + y) – r =0

The parallel lines are px + py - q = 0 and px + py – r = 0

Here a = p ,b = p ,c = -q ,d = -r

D

The distance between the parallel lines p(x + y) =q = 0 and p(x + y) – r = 0 is units

Given: parallel lines 12x – 5y – 3 = 0, 12x – 5y + 7 = 0, 12x – 5y – 13 = 0

To Prove : line 12x – 5y – 3 = 0 is mid-parallel to the lines 12x – 5y + 7 = 0 and 12x – 5y – 13 = 0

Formula used :

The distance between the parallel lines ax + by + c =0 and ax + by + d =0 is,

D

The equation of line l is 12x – 5y + 7 = 0

The equation of line m is 12x – 5y – 3 = 0

The equation of line n is 12x – 5y – 13 = 0

Let D₁ be the distance between the lines l and m .

Here a = 12 ,b = -5 ,c = 7 ,d = -3

D_1..

The distance between the parallel lines l and m is units

Let D₂ be the distance between the lines m and n .

Here a = 12 ,b = -5 ,c = 7 ,d = -3

D₂

The distance between the parallel lines m and n is units

Distance between the parallel lines l and m = Distance between the parallel lines m and n

Thus the line 12x – 5y – 3 = 0 is mid-parallel to the lines 12x – 5y + 7 = 0 and 12x – 5y – 13 = 0

Given: perpendicular distance from orgin is 5 units, and the slope is -1

To find : the equation of the line

Formula used :

We know that the perpendicular distance from a point (x_0,y₀) to the line ax + by + c = 0 is given by

The equation of a straight line is given by y=mx+c where m denotes the slope of the line.

The equation of the line is mx – y + c =0

Here x₀ = 0 and y₀= 0 , a = m , b = -y , c = c and D=5 units

D

Slope of the line = m = -1 ,Substituting in the above equation we get,

Thus the equation of the straight line is y = x + 5 or x + y - 5 = 0

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

∴ 4x + 3y = 5

or 4x + 3y – 5 = 0 …(i)

and x = 2y – 7

or x – 2y + 7 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 4, we get

4x – 8y + 28 = 0 …(iii)

On subtracting eq. (iii) from (i), we get

4x – 8y + 28 – 4x – 3y + 5 = 0

⇒ -11y + 33 = 0

⇒ -11y = -33

Putting the value of y in eq. (i), we get

4x + 3(3) – 5 = 0

⇒ 4x + 9 – 5 = 0

⇒ 4x + 4 = 0

⇒ 4x = -4

⇒ x = -1

Hence, the point of intersection P(x₁, y₁) is (-1, 3)

Suppose the given two lines intersect at a point P(2, 3). Then, (2, 3) satisfies each of the given equations.

So, taking equation x + 7y = 23

Substituting x = 2 and y = 3

Lhs = x + 7y

= 2 + 7(3)

= 2 + 21

= 23

= RHS

Now, taking equation 5x + 2y = 16

Substituting x = 2 and y = 3

LHS = 5x + 2y

= 5(2) + 2(3)

= 10 + 6

= 16

= RHS

In both the equations pair (2, 3) for (x, y) satisfies the given equations, therefore both lines pass through (2, 3).