Straight Lines

Let ABCD be the given quadrilateral with vertices A(-4,5) , B(0,7), C(5.-5) and D(-4,-2). Plotting the points on the Cartesian plane and joining AB, BC, CD, AD gives us the required quadrilateral.

To find the area, draw diagonal AC

area (ABCD) = ar ( ∆ABC) + ar(∆ADC)

area ∆ with vertices ( x₁,y₁) , (x₂, y₂) and (x₃,y₃) is

unit²

unit²

Area ∆ ACD =

Since area cannot be negative area ∆ ACD =

Area (ABCD) =

Let ABC be the given equilateral triangle with side 2a

⇒ AB= BC = AC = 2a

Assuming that the base AB lies on the y axis such that the mid-point of AB is at the origin

i.e.BO = OA = a and O is the origin

⇒ Co-ordinates of point A are (0,a) and that of B are (0,-a)

Since the line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular

⇒ Vertex of A lies on the y –axis

On applying Pythagoras theorem

(AC)² = OA² + OC²

⇒(2a)²= a² + OC²

⇒ 4a² – a² = OC²

⇒ 3a²= OC²

⇒ OC =

∴ Co-ordinates of point C =

Thus, the vertices of the given equilateral triangle are (0, a) , (0, -a) , (

Or (0, a), (0, -a) and (

Here the given points are P( x₁, y₁) and Q(x₂, y₂)

(i) When PQ is ∥ to y axis then x₁ = x₂

The distance between P and Q is

(ii) when PQ is parallel to the x-axis then y₁ = y₂

The distance between P and Q is =

=

=

Let (a, 0) be the point on the x-axis that is equidistant from the point (7, 6) and (3, 4).

Accordingly,

Squaring both the sides we get,

⇒ - 8a = - 60

The required point is

The co-ordinates of mid-point of the line segment joining the points P (0, – 4) and B (8, 0) are

The slope (m) of the line non-vertical line passing through the point (x₁, y₁) and

(x₂, y₂) is given by

The slope of the line is passing through (0,0) and (4,-2) is

The required slope is

The vertices of the given triangle are (4, 4), (3, 5) and (–1, –1).

The slope (m) of the line non-vertical line passing through the point (x₁, y₁) and

(x₂, y₂) is given by

∴ the slope of the line AB (m₁) =

the slope of the line BC (m₂) =

the slope of the line CA (m₃) =

Now, m₁ m₃ = -1

⇒ Lines AB and CA are perpendicular to each other

∴ given triangle is right-angled at A (4, 4)

And the vertices of the right-angled ∆ are (4, 4) , (3, 5) and (-1, -1)

If a line makes an angle of 30° with the positive direction of y-axis measured anti-clock-wise , then the angle made by the line with the positive direction of x- axis measure anti-clock-wise is 90° + 30° = 120°

The slope of the given line is tan 120° = tan (180° - 60°) = - tan 60° = -

If the points (x, – 1), (2, 1) and (4, 5) are collinear, then Slope of AB = Slope of BC

⇒

⇒

⇒ 2= 4 - 2x

⇒ x= 1

Thus, required value of x is 1.

Let the point A( -2, -1) , B (4, 0) , C( 3, 3) and D( -3, 2)

The slope of AB =

The slope of CD =

⇒the slope of AB = Slope of CD

⇒ AB ∥ CD

The slope of BC =

The slope of CD =

The slope of BC = Slope of CD

⇒ BC ∥ CD

Thus the pair of opposite sides are quadrilateral are ∥ , hence ABCD is a parallelogram.

Thus A( -2, -1) , B (4, 0) , C( 3, 3) and D( -3, 2) are vertices of a parallelogram.

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

The angle of inclination of line joining the points (3, -1) and (4, -2) is given by

tan θ = -1

θ = (90° + 45°) = 135°

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

let m₁ and m be the slope of the two given lines such that m₁= 2m

We know that if θ is the angle between the lines l1 and l2 with slope m₁ and m₂, then

Given here that the tangent of the angle between the two lines is

∴

⇒

Case 1

⇒

⇒ 1+2m² = -3m

⇒ 2m² +1 +3m = 0

⇒2m(m+1) + 1(m+1)=0

⇒(2m+1)(m+1)= 0

⇒ m =-1 or

If m = -1, then the slope of the lines are -1 and -2

If m = , then the slope of the lines are and -1

Case 2

⇒

⇒ 2m² - 3m + 1 = 0

⇒ m = 1 or 1/2

If m = 1, then the slope of the lines are 1 and 2

If m = , then the slope of the lines are and 1

Hence the slope of the lines are -1 and -2 or and -1 or 1 and 2 or and 1.

The slope of the line passing through (x₁, y₁) and (h, k) is

Given that the slope of the line is m

⇒

_{hence proved.}

If the points A (h, 0), B (a, b) and C (0, k) lie on a line

Slope of AB = Slope of BC

⇒ - ab = (k-b) (a-h)

⇒ - ab = ka- kh –ab +bh

⇒ ka +bh = kh

Dividing both the sides by kh

⇒

⇒ Hence proved.

Since the line AB passes through points A (1985, 92) and

B (1995, 97), its slope will be

Let y be the population in the year 2010. Then, according to the given graph the AB must pass through point C ( 2010 , y)

Now, slope of AB = Slope of BC

⇒

⇒ y= 7.5 +97 = 104.5

Thus the slope of the line AB is 1/2, while in the year 2010 the population must be 104.5 crores.

The y-coordinate of every point on x-axis is 0.

∴ Equation of x-axis is y = 0.

The x-coordinate of every point on y-axis is 0.

∴ Equation of y-axis is x = 0.

Given point = (-4, 3) and slope, m = 1/2

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

∴ y – 3 = 1/2 (x – (-4))

⇒ y – 3 = 1/2 (x + 4)

⇒ 2(y – 3) = x + 4

⇒ 2y – 6 = x + 4

⇒ x + 4 – (2y – 6) = 0

⇒ x + 4 – 2y + 6 = 0

⇒ x – 2y + 10 = 0

Ans. The equation of the line is x – 2y + 10 = 0.

Given point = (0, 0) and slope = m

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

∴ y – 0 = m (x – 0)

⇒ y = mx

⇒ y – mx = 0

Ans. The equation of the line is y – mx = 0.

Given point = (2, 2√3) and θ = 75°

Equation of line: (y - y₁) = m (x - x₁)
where, m = slope of line = tan θ
and (x₁, y₁) are the points through which line passes

∴ m = tan 75°

75° = 45° + 30°

Applying the formula:




Rationalizing we get,

We know that the point (x, y) lies on the line with slope m through the fixed point (x₁, y1), if and only if, its coordinates satisfy the equation y – y₁ = m (x – x₁)

∴ y – 2√3 = (2 + √3) (x – 2)

⇒ y – 2√3 = 2 x - 4 + √3 x - 2 √3

⇒ y = 2 x - 4 + √3 x

⇒ (2 + √3) x – y - 4 = 0

Ans. The equation of the line is (2 + √3) x – y - 4 = 0.

We know that if a line L with slope m makes x-intercept d, then equation of L is

y = m(x − d).

If the distance is 3 units to the left of origin then d = -3

Given, slope, m = -2

∴ y = (-2) (x – (-3))

⇒ y = (-2) (x + 3)

⇒ y = -2x – 6

⇒ 2x + y + 6 = 0

Ans. The equation of the line is 2x + y + 6 = 0.

We know that the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c.

If distance is 2 units above the origin, c = +2

Given θ = 30°

We know that slope, m = tanθ

∴ m = tan30° = (1/√3)

∴ y = (1/√3)x + 2

⇒ y = (x + 2√3) / √3

⇒ √3 y = x + 2√3

⇒ x - √3 y + 2√3 = 0

Ans. The equation of the line is x - √3 y + 2√3 = 0.

Given points (-1, 1) and (2, -4)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

⇒ 3 (y – 1) = (-5) (x + 1)

⇒ 3y – 3 = -5x – 5

⇒ 3y – 3 + 5x + 5 = 0

⇒ 5x + 3y + 2 = 0

Ans. The equation of the line is 5x + 3y + 2 = 0.

We know that the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.

Given p = 5 and ω = 30°

Substituting the values in the equation, we get

⇒ x cos30° + y sin30° = 5

⇒ x(√3 / 2) + y( 1/2 ) = 5

⇒ √3 x + y = 5(2) = 10

⇒ √3 x + y – 10 = 0

Ans. The equation of the line is √3 x + y – 10 = 0.

Given vertices of ΔPQR i.e. P (2, 1), Q (-2, 3) and R (4, 5)

Let RL be the median of vertex R.

So, L is a midpoint of PQ.

We know that the midpoint formula is given by .

∴ L = = (0, 2)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

⇒ (-4) (y – 5) = (-3) (x – 4)

⇒ -4y + 20 = -3x + 12

⇒ -4y + 20 + 3x – 12 = 0

⇒ 3x – 4y + 8 = 0

Ans. The equation of median through the vertex R is 3x – 4y + 8 = 0.

Given points are (2, 5) and (-3, 6).

We know that slope, m =

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

∴ y – 5 = 5(x – (-3))

⇒ y – 5 = 5x + 15

⇒ 5x + 15 – y + 5 = 0

⇒ 5x – y + 20 = 0

Ans. The equation of the line is 5x – y + 20 = 0.

We know that the coordinates of a point dividing the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m: n are .

∴

We know that slope, m =

∴ m =

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

∴ m = (-1/m) =

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

Here, the point is.

∴

⇒ 3((1 + n) y – 3) = (-(1 + n) x + 2 + n)

⇒ 3(1 + n) y – 9 = - (1 + n) x + 2 + n

⇒ (1 + n) x + 3(1 + n) y – n – 9 – 2 = 0

⇒ (1 + n) x + 3(1 + n) y – n – 11 = 0

Ans. The equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0.

We know that equation of the line making intercepts a and b on x-and y-axis, respectively, is

Given that the line cuts off equal intercepts on the coordinate axes i.e. a = b.

So,

⇒ x + y = a … (1)

Given point = (2, 3)

∴ 2 + 3 = a

⇒ a = 5

Substituting a value in (1),

⇒ x + y = 5

⇒ x + y – 5 = 0

Ans. The equation of the line is x + y – 5 = 0.

We know that equation of the line making intercepts a and b on x-and y-axis, respectively, is . … (1)

Given, sum of intercepts = 9

∴ a + b = 9

⇒ b = 9 - a

Substituting b value in the above equation,

Given, the line passes through the point (2, 2),

So,

⇒ 18 = a (9 – a)

⇒ 18 = 9a – a²

⇒ a² – 9a + 18 = 0

Factorizing,

⇒ a² – 3a – 6a + 18 = 0

⇒ a (a – 3) – 6 (a – 3) = 0

⇒ (a – 3) (a – 6) = 0

∴ a = 3 or a = 6

Substituting in (1),

Case 1 (a = 3):

Then b = 9 – 3 = 6

⇒

⇒ 2x + y = 6

⇒ 2x + y – 6 = 0

Case 2 (a = 6):

Then b = 9 – 6 = 3

⇒

⇒ x + 2y = 6

⇒ x + 2y – 6 = 0

Ans. The equation of the line is 2x + y – 6 = 0 or x + 2y – 6 = 0.

Given point = (0, 2) and θ = 2π/3

We know that m = tanθ

∴ m = tan (2π/3) = -√3

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

∴ y – 2 = -√3 (x – 0)

⇒ y – 2 = -√3 x

⇒ √3 x + y – 2 = 0

Given, equation of line parallel to above obtained equation crosses the y-axis at a distance of 2 units below the origin.

So, the point = (0, -2) and m = -√3

From point slope form equation,

⇒ y – (-2) = -√3 (x – 0)

⇒ y + 2 = -√3 x

⇒ √3 x + y + 2 = 0

Ans. The equation of line is √3 x + y – 2 = 0 and the line parallel to it is √3 x + y + 2 = 0.

Given points are origin (0, 0) and (-2, 9).

We know that slope, m =

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

∴ m = (-1/m) =

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

∴ y – 9 = (2/9) (x – (-2))

⇒ 9(y – 9) = 2(x + 2)

⇒ 9y – 81 = 2x + 4

⇒ 2x + 4 – 9y + 81 = 0

⇒ 2x – 9y + 85 = 0

Ans. The equation of line is 2x – 9y + 85 = 0.

Assuming L along X-axis and C along Y-axis, we have two points (124.942, 20) and (125.134, 110) in XY-plane.

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

∴

⇒

⇒ 0.192 (C – 20) = 90 (L – 124.942)

⇒ L =

Ans. The required relation is L = .

The relationship between selling price and demand is linear.

Assuming selling price per litre along X-axis and demand along Y-axis, we have two points (14, 980) and (16, 1220) in XY-plane.

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

∴

⇒

⇒ y – 980 = 120 (x – 14)

⇒ y = 120 (x – 14) + 980

When x = Rs 17/litre,

⇒ y = 120 (17 – 14) + 980

⇒ y = 120(3) + 980

⇒ y = 360 + 980 = 1340

Ans. The owner can sell 1340 litres weekly at Rs 17/litre.

Let AB be a line segment whose midpoint is P (a, b).

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

We know that the midpoint formula is given by .

Since P is the midpoint of (a, b),

⇒

⇒ a = x/2 and b = y/2

⇒ x = 2a and y = 2b

∴ A = (0, 2b) and B = (2a, 0)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

∴

⇒

⇒

⇒ a (y – 2b) = -bx

⇒ ay – 2ab = -bx

⇒ bx + ay = 2ab

Dividing by ab on both sides,

⇒

⇒

Ans. The equation of line is .

Let AB be the line segment such that R (h, k) divides it in the ratio 1: 2.

Let the coordinates of A and B are (0, y) and (x, 0) respectively.

We know that the coordinates of a point dividing the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m: n are .

∴

⇒

∴ h = 2x/3 and k = y/3

⇒ x = 3h/2 and y = 3k

∴ A = (0, 3k) and B = (3h/2, 0)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

If we have to show that the given three points (3, 0), (– 2, – 2) and (8, 2) are collinear, we have to show that the line passing through the points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

∴

⇒

⇒ -5y = -2 (x – 3)

⇒ -5y = -2x + 6

⇒ 2x – 5y = 6

If 2x – 5y = 6 passes through (8, 2),

LHS = 2x – 5y = 2(8) – 5(2) = 16 – 10 = 6 = RHS

The line passing through the points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

Hence proved.

Ans. The given three points are collinear.

Slope – intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y intercept

Given equation is x + 7y = 0

The above equation is of the form y = mx + c, where m = -1/7 and c = 0.

Slope – intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y intercept

Given equation is 6x + 3y – 5 = 0

⇒ 3y = -6x + 5

The above equation is of the form y = mx + c, where m = -2 and c = 5/3.

Equation of line in slope – intercept form is given by ‘y = mx + c’, where m is the slope and c is the y intercept

Given equation is y = 0

⇒ y = 0 × x + 0

The above equation is of the form y = mx + c, where m = 0 and c = 0.

Equation of line in intercept form is given by, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

Given equation is 3x + 2y – 12 = 0

Dividing both sides by 12, we get

⇒

The above equation is of the form, where a = 4, b = 6

Intercept on x – axis = 4

Intercept on y – axis = 6

Equation of line in intercept form is given by, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

Given equation is 4x – 3y = 6

Dividing both sides by 6, we get

⇒

⇒

The above equation is of the form, where a = 3/2, b = -2

Intercept on x – axis = 3/2

Intercept on y – axis = -2

Equation of line in intercept form is given by, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

Given equation is 3y + 2 = 0

Dividing both sides by -2, we get

⇒

⇒

The above equation is of the form, where a = 0, b = -2/3

Intercept on x – axis = 0

Intercept on y – axis = -2/3

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

Given equation is x – y + 8 = 0

Dividing both sides by

The above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4.

Perpendicular distance of line from origin = 4

Angle between perpendicular and positive x – axis = 120°

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

Given equation is y – 2 = 0

0 × x + 1 × y = 2

Dividing both sides by

The above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2.

Perpendicular distance of line from origin = 2

Angle between perpendicular and positive x – axis = 90°

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

Given equation is x – y + 4 = 0

Dividing both sides by

The above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = 2√2.

Perpendicular distance of line from origin = 2√2

Angle between perpendicular and positive x – axis = 315°

Given equation of the line 12(x + 6) = 5(y – 2).

⇒ 12x + 72 = 5y – 10

⇒12x – 5y + 82 = 0 … (1)

Comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 12, B = –5, and C = 82

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Given point (x₁, y₁) = (-1, 1)

∴ distance of point (–1, 1) from the given line

d = 5 units

Given equation of line

⇒ 4x + 3y = 12

⇒ 4x + 3y – 12 = 0

Comparing above equation with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

⇒ |4a – 12| = 4 × 5

⇒ (4a – 12) = 20

⇒ 4a – 12 = 20 or - (4a – 12) = 20

⇒ 4a = 20 + 12 or 4a = -20 + 12

⇒ a = 32/4 or a = -8/4

⇒ a = 8 or a = -2

∴ The required points on the x – axis are (-2, 0) and (8, 0)

Given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

Distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by

Here A = 15, B = 8, C₁ = -34, C₂ = 31

Distance between parallel lines is

∴ d = 65/17

Given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.

⇒ lx + ly + p = 0 and lx + ly – r = 0

Distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by

Here A = l, B = l, C₁ = p, C₂ = -r

Distance between parallel lines is

∴

Given line is 3x – 4y + 2 = 0

⇒

⇒

Which is of the form y = mx + c, where m is the slope of the given line.

∴ slope of the given line is 3/4

We know that parallel line have same slope.

∴ slope of other line = m = 3/4

Equation of line having slope m and passing through (x₁, y₁) is given by

∴ equation of line having slope 3/4 and passing through (-2, 3) is

⇒ 4y – 3 × 4 = 3x + 3 × 2

⇒ 3x – 4y = 18

Given equation of line is x – 7y + 5 = 0

⇒

Which is of the form y = mx + c, where m is the slope of the given line.

∴ slope of the given line is 1/7

Slope of the line perpendicular to the line having slope m is -1/m

Slope of the line perpendicular to the line having a slope of 1/7 is = -7

Equation of line with slope -7 and x intercept 3 is given by y = m(x – d)

⇒ y = -7 (x – 3)

⇒ y = -7x + 21

⇒ 7x + y = 21

Given lines are x + y = 1 and x + y = 1.

Slope of line (1) is m₁ = -√3, while the slope of line (2) is m₂ = -1/√3

Let θ be the angle between two lines

∴ θ = 30°

Thus the angle between the given lines is either 30° or 180° - 30° = 150°

Let the slope of the line passing through (h, 3) and (4, 1) be m₁

Then,

Let the slope of line 7x – 9y – 19 = 0 be m₂

⇒ 7x – 9y – 19 = 0

m₂ = 7/9

Since, the given lines are perpendicular

Therefore, m₁ × m₂ = -1

⇒ -14 = -1 × (36 – 9h)

⇒ 36 – 9h = 14

⇒ 9h = 36 – 14

⇒ h = 22/9

∴ h = 22/9

Let the slope of line Ax + By + C = 0 be m

Ax + By + C = 0

∴ m = -A/B

Equation of the line passing through point (x₁, y₁) and having slope is

∴ A(x – x₁) + B(y – y₁) = 0

So, the line through point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x – x₁) + B (y – y₁) = 0

Let the slope of the first line be m₁

Given, m₁ = 2

Let the slope of the other line be m₂.

Angle between the two lines is 60°.

⇒

Case 1: When

The equation of the line passing through point (2, 3) and having a slope m₂ is

⇒

∴ Equation of the other line is

Case 2: When

The equation of the line passing through point (2, 3) and having a slope m₂ is

Equation of the other line is

The right bisector of a line segment bisects the line segment at 90°.

End-points of the line segment AB are given as A (3, 4) and B (–1, 2).

Let mid-point of AB be (x, y)

(x, y) = (7/2, 1/2)

Let slope of line AB be m₁

m₁ = (2 – 4)/(-1 – 3) = -2/(-4)

m₁ = 1/2

Let slope of the line perpendicular to AB be m₂

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = -2 (x – 1)

⇒ y – 3 = - 2x + 2

⇒ 2x + y = 5

Thus, the required equation of the line is 2x + y = 5.

Let the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)

Let the slope of the line joining (-1, 3) and (a, b) be m₁

Let the slope of the line 3x – 4y – 16 = 0 be m₂

∴

Since these two lines are perpendicular m₁ × m₂ = -1

⇒ 3b – 9 = -4a – 4

⇒ 4a + 3b = 5 …….(1)

Point (a, b) lies on the line 3x – 4y = 16

∴ 3a – 4b = 16 ……..(2)

Solving equations (1) and (2), we obtain

a = 68/25 and b = -49/25

Co-ordinates of the foot of perpendicular is (68/25, -49/25)

Given equation of line is y = mx + c

Given that the perpendicular from the origin meets the given line at (–1, 2).

Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

Slope of the line joining (0, 0) and (–1, 2) = 2/(-1) = -2

Slope of the given line is m.

m × (-2) = -1

∴ m = 1/2

Since, point (-1, 2) lies on the given line so,

y = mx + c

⇒ 2 = 1/2 × (-1) + c

⇒ c = 2 + 1/2 = 5/2

The respective values of m and c are 1/2 and 5/2 respectively.

The equations of given lines are

x cos θ – y sin θ = k cos 2θ …………………… (1)

x sec θ + y cosec θ = k ……………….… (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain

A = cos θ, B = -sin θ, and C = -k cos 2θ

Given that p is the length of the perpendicular from (0, 0) to line (1).

∴

p = k cos 2θ

Squaring both sides

P² = k² cos²2θ …………………(3)

Comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain

A = sec θ, B = cosec θ, and C = -k

Given that q is the length of the perpendicular from (0, 0) to line (2)

Multiplying both sides by 2

2q = 2k cos θ sin θ = k × 2sinθ cosθ

2q = k sin 2θ

Squaring both sides

4q² = k² sin²2θ …………………(4)

Adding (3) and (4) we get

p² + 4q² = k² cos² 2θ + k² sin² 2θ

⇒ p² + 4q² = k² (cos² 2θ + sin² 2θ) [∵cos² 2θ + sin² 2θ = 1]

∴ p² + 4q² = k²

Let AD be the altitude of triangle ABC from vertex A.

Accordingly, AD is perpendicular to BC

Given vertices A (2, 3), B (4, –1) and C (1, 2)

Slope of line BC = m₁

m₁ = (- 1 - 2)/(4 - 1)

m₁ = -1

Let slope of line AD be m₂

AD is perpendicular to BC

∴ m₁ × m₂ = -1

⇒ -1 × m₂ = -1

∴ m₂ = 1

The equation of the line passing through point (2, 3) and having a slope of 1 is

⇒ y – 3 = 1 × (x – 2)

⇒ y – 3 = x – 2

⇒ y – x = 1

Equation of the altitude from vertex A = y – x = 1

Length of AD = Length of the perpendicular from A (2, 3) to BC

Equation of BC is

y + 1 = -1 × (x – 4)

⇒ y + 1 = -x + 4

⇒ x + y – 3 = 0 …………………(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain

A = 1, B = 1 and C = -3

Length of AD =

The equation and the length of the altitude from vertex A are y – x = 1 and

√2 units respectively.

Equation of a line whose intercepts on the axes are a and b is

⇒ bx + ay = ab

⇒ bx + ay – ab = 0 ………………..(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain

A = b, B = a and C = -ab

If p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (1), we obtain

⇒

Squaring both sides we get

⇒

⇒

∴

The given equation of the

(k–3) x – (4 – k²) y + 6

If the given line is parallel to the x- axis, then

Slope of the given line = Slope of the x- axis.

The given line can be written as

(4 – k²) = (k–3) x + k² –7k + 6 = 0

⇒ y = which is the form y = mx+c

Thus, Slope of the given line =

Slope of x- axis = 0

Then,

⇒ k – 3 = 0

⇒ k = 3

Therefore, if the given line is parallel to the x – axis, then the value of k is 3.

The given equation of the

(k–3) x – (4 – k²) y + 6

If the given line is parallel to the y- axis, then its slope will be undefined.

The slope of the line is =

Now, is undefined at k² = 4

⇒ k² = 4

⇒ k = 2

Therefore, if the given line is parallel to the y – axis, then the value of k is 2.

The given equation of the

(k–3) x – (4 – k²) y + 6

If the given line is passing through the origin, then point (0,0) satisfies the given equation of line.

(k -3) (0) – (4-k²)(0) + k² -7k +6 = 0

⇒ k² -7k +6 = 0

⇒ k² -6k-k +6 = 0

⇒ (k-6)(k-1)=0

⇒ k = 1or 6

Therefore, if the given line is passing through the origin, then the value of k is either 1 or 6.

The equation of the given line is

This equation can be reduced as

⇒

On dividing both sides , we obtain,

…………….(1)

On comparing equation (1) with xcosƟ + ysinƟ = p, we get,

cosƟ = and sinƟ = and p = 1

Since, the value of sinƟ and cosƟ are negative,

Therefore, the respective values of Ɵ and p are and 1.

Let the intercepts cut by he given lines on the axes be a and b.

It is given that

a + b = 1 ---------(1)

ab = -6 ---------(2)

On solveing equations (1) and (2), we get,

a = 3 and b = -2 or a = -2 and b = 3

We know that the equation of the line whose intercepts on the axes are a and b is

=> bx + ay – ab = 0

Now, when a = 3 and b = -2 , we get

The equation of the line is -2x + 3y + 6 = 0

=> 2x – 3y = 6

And when a = -2 and b = 3, we get,

The equation of the line is 3x – 2y +6 = 0

=> -3x + 2y = 6

Therefore, the equation of the lines are 2x – 3y = 6 or -3x + 2y = 6.

The given line can be written as 4x + 3y -12 = 0

On comparing equation (1) to the general equation of line Ax + By + C = 0, we get,

A = 4 , B = 3 and C = -12.

It is known that the perpendicular distance (d) of the line Ax + By + C = 0 from a point

(x₁, y₁) is given by

Therefore, if (0,b) is the point on the y – axis whose distance from line is 4 units, then,

⇒ 20 = |3b – 12|

⇒ 20 = (3b – 12)

⇒ 20 = (3b – 12) or 20 = -(3b-12)

⇒ 3b = 20 +12 or 3b = -20 +12

⇒

Therefore, the required points are

The equation of the line joining the points (cosƟ, sinƟ) and (cosɸ, sinɸ) is given by

⇒ y(cosɸ-cosƟ) – sinƟ(cosɸ-cosƟ) = x(sinɸ-sinƟ) – cosƟ(sinɸ-sinƟ)

⇒ x(sinƟ- sinɸ) + y(cosɸ- cosƟ) + cosƟsinɸ - cosƟsinƟ – sinƟcosɸ + sinƟcosƟ = 0

⇒ x(sinƟ- sinɸ) + y(cosɸ- cosƟ) + sin(ɸ-Ɵ) = 0

Ax + By + C = 0, where A = sinƟ- sinɸ, B = cosɸ- cosƟ, and C = sin(ɸ-Ɵ)

It is known that the perpendicular distance (d) of the line Ax + By + C = 0 from a point

(x₁, y₁) is given by

Thus, the perpendicular distance (d) of the given line from point (x₁, y₁) =(0,0) is

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

x = a -------- (1)

The two given lines are

x – 7y + 5 = 0 ---------(2)

3x + y = 0 ------- (3)

Thus,

is the point of intersection of lines (2) and (3).

Since, line x = a pass through point a =

Therefore, the required equation of the line is x =

The equation of the given line is

This equation can also be written as 3x + 2y -12 = 0

which is of the form y = mx + c

Thus, Slope of the given line =

and Slope of line perpendicular to the given line =

Let the given intersects the y – axis at (0,6)

The equation of the line has a slope of and passes through point (0,6) is

=> 3y – 18 = 2x

=> 2x – 3y + 18 = 0

Therefore, the required equation of the line is 2x – 3y + 18 = 0.

The equations of the given lines are

y – x = 0 ------(1)

x + y = 0------(2)

x – k = 0------(3)

The point of intersection of lines (1) and (2) is given by x = 0 and y = 0.

The point of intersection of lines (2) and (3) is given by x = k and y = -k.

The point of intersection of lines (3) and (1) is given by x = k and y = k.

Then, the vertices of the triangle formed by the three given lines are (0, 0), (k, -k) and (k,k).

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃,y₃) is

Thus, area of the triangle formed by the three given lines

sqare units

square units

square units

square units

The equations of the given lines are

3x + y -2 = 0 ------ (1)

px + 2y – 3 = 0------(2)

2x – y – 3 = 0 ------ (3)

On solving equations (1) and (3), we get,

x = 1 and y = -1

Since, the three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).

p(1) + 2(-1) -3 = 0

=> p – 2 -3 = 0

=> p =5

Therefore, the required value of p is 5.

The equation of the given lines are

y = m₁x + c₁ ---------(1)

y = m₂x + c₂ ---------(2)

y = m₃x + c₃ ---------(3)

On subtracting equation (1) from (2), we get,

0 = (m₂ - m₁)x + (c₂ – c₁)

⇒ (m₂ - m₁)x = (c₂ – c₁)

⇒ x =

On substituting this value of x in (1), we get,

Thus,

is the point of intersection of lines (1) and (2).

It is given that lines (1), (2) and (3) are concurrent.

Thus, the point of intersection of lines (1) and (2) will also satisfy equation (3).

Let the slope of the required line be m₁.

The given line can be written as , which is of the from y = mx + c

Thus, Slope of the given line = m₂ =

It is given that the angle between the required line and line x – 2y = 3 is 45^o.

We know that if Ɵ is the acute angle between lines l₁ and l₂ with slopes m₁ and m₂ respectively, then

and

⇒ 2 + m₁ = 1- 2m₁ or 2 + m₁ = -1 + 2m₁

⇒ m₁ = or m₁ = 3

Now, when m₁ = 3, then,

The equation of the line passing through (3,2) and having a slope of 3 is:

y – 2 = 3(x – 3)

⇒ y -2 = 3x – 9

⇒ 3x – y = 7

And, when m₁ =

The equation of the line passing through (3,2) and having a slope of is:

3y – 6 = -x + 3

x + 3y = 9

Therefore, the equations of the lines are 3x – y = 7 and x + 3y = 9.

Let the equation of the line having equal intercepts on the axes be

=> x + y = 1 ---------------- (1)

On solving equations 4x + 7y – 3 = 0 and 2x – 3y +1 = 0, we get

Thus,

is the point of intersection of the two given lines.

Since, equation (1) passes through point

⇒ a =

Thus, Equation (1) becomes

, i.e., 13x + 13y = 6

Therefore, the required equation of the line 13x + 13y = 6.

Let the equation of the line passing through the origin be y = m₁x.

If this line makes an angle of Ɵ with line y = mx + c, then angle Ɵ is given by

Thus,

Case I :

Case II :

Therefore, the required line is given by

The equation of the line joining the points are (-1,1) and (5, 7) is given by

x – y + 2 = 0 ---------- (1)

The equation of the given line by

x + y – 4= 0 -----------(2)

The point of intersection of lines (1) and (2) is given by x = 1 and y = 3.

Let points (1,3) divide the line segment joining (-1,1) and (5,7) in the ratio 1: k.

Thus, by using section formula, we get,

Thus,

⇒ -k + 5 = 1+ k

⇒ 2k = 4

⇒ k = 2

Therefore, the line joining the points (-1, 1) and (5, 7) is divided by line

x + y = 4 in the ratio 1: 2.

The given lines are

2x – y = 0 ------------- (1)

4x + 7y + 5 = 0 --------------- (2)

A(1,2) is a point on line (1).

Let B be the point of intersection of lines (1) and (2).

So, solving equations (1) and (2), we get,

x = and y =

Thus, coordinates of point B are

So, according to distance formula, the distance between points A and B can be obtained as

AB = units

= units

= units

= units

= units

= units

= units

= units

Therefore, the required distance is units

Let y = mx + c be the line through points (-1, 2).

2 = m (-1) + c

⇒ c = m + 2

Thus, y = mx + m + 2 --------------- (1)

The given line as

x + y = 4 ------------ (2)

On solving these equations, we get,

and

Thus, is the point of intersection of lines (1) and (2).

Since, this point is at a distance of 3 units from point (-1, 2),

So, now using distance formula,

⇒ 1 + m² = m² + 1 + 2m

⇒ 2m = 0

⇒ m = 0

Therefore, the slope of the required line must be zero,

Hence, the line must be parallel to the x – axis.

Let A (1, 3) and B (-4, 1) be the coordinates of the end points of the hypotenuse.

Now, potting the line segment joining the points A (1, 3) and B (-4, 1) on the coordinates plane, we will get two right triangles with AB as the hypotenuse. Now, from the graph, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can either P or Q.

Now, when ΔAPB is taken.

The perpendicular sides in ΔAPB are AP and PB.

Now, sides PB is parallel to the x-axis and at a distance of 1 unit above the x-axis.

So, equation of PB is, y = 1 or y -1 = 0.

The side AP is parallel to y – axis and at a distance of 1 units on the right of y – axis.

So, equation of AP is x = 1 or x - 1 = 0.

And when ΔAQB is taken.

The perpendicular sides in ΔAQB are AQ and QB.

Now, sides AQ is parallel to the x-axis and at a distance of 3 units above x-axis.

So, equation of AQ is, y = 3 or y -3 = 0

The side QB is parallel to the y-axis and at a distance of 4 units on the left of the y-axis.

So, equation of QB is x = -4 or x +4 = 0.

The equation of the given line is

x +3y = 7

Let point B (a, b) be the image of point A (3, 8)

So, line (1) is the perpendicular bisector of AB.

Slope of AB = , while the slope of the line (1) =

Since, line (1) is perpendicular to AB.

=> b – 8 = 3a – 9

=> 3a – b = 1 --------------- (2)

Mid – point of AB =

The mid – point of line segment AB will also satisfy line (1).

Thus, from equation (1), we get,

⇒ a + 3 + 3b + 24 = 14

⇒ a + 3b = -13 -------------(3)

On solving equations (2) and (3), we get, a = -1 and b = -4

Therefore, the image of the given point with respect to the given line is (-1, -4).

The equation of the given lines are

y = 3x +1 ------------ (1)

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

y = mx + 4 ------------- (3)

Slope of line (1), m₁ = 3

Slope of line (2), m₂ =

Slope of line (3), m₃ = m

It is given that lines (1) and (2) are equally inclined to line (3).

So, the angle between lines (1) and (3) equals between lines (2) and (3).

Thus,

or

Now, if , we get,

(3 – m)(m +2) = (1 – 2m)(1 +3m)

⇒ -m² + m + 6 = 1 + m – 6m²

⇒ 5m² + 5 = 0

⇒ (m + 1) = 0

⇒ m = , which is not real

If , then

(3 – m)(m +2) = - (1 – 2m) (1 +3m)

⇒ -m² + m + 6 = -(1 + m – 6m²)

⇒ 7m² + 2m - 7 = 0

⇒ m =

⇒ m =

⇒ m =

Therefore, the required value of m is.

The equations of the given lines are:

x + y – 5 = 0 ---------------- (1)

3x – 2y +7 = 0 ------------ (2)

The perpendicular distance of P(x, y) from line (1) is given by

and

i.e, and

It is given that

Thus,

(Assuming (x+y-5) and (3x-2y+7))

Which is the equation of the line.

Similarly, we can get the equation of line for any signs of (x + y -5) and (3x – 2y + 7).

Therefore, point P must move on a line.

The equations of the given lines are

9x + 6y – 7 = 0 ---------------- (1)

3x + 2y + 6 = 0 ---------------- (2)

Let P (h, k) be the arbitrary point that is equidistant from (1) and (2).

The perpendicular distance of P(h, k) from line (1) is given by

The perpendicular distance of P(h, k) from line (2) is given by

Since, P (h, k) is equidistant from lines (1) and (2),

Thus,

or

So, when is not possible as

which is not at all possible

And when

⇒ 9h + 6k – 7 = -9h – 6k – 18

⇒ 18h + 12k +11 = 0

Therefore, the required equation of the line is 18h + 12k +11 = 0.

Let the coordinate of point A be (a, 0)

Draw a line (AL) perpendicular to the x –axis.

We know that angle of incidence is equal to angle of reflection. Hence, let

∠BAL =∠CAL = ɸ

Let ∠CAX = Ɵ

Therefore,

∠OAB = 180⁰ –(Ɵ+2ɸ) = 180⁰ –[Ɵ+2(90⁰ - Ɵ)]

= 180⁰ – Ɵ+180⁰ +2Ɵ

= Ɵ

Thus, ∠BAX = 180⁰ – Ɵ

Now, slope of line AC =

------------ (1)

Slope of line AB =

----------------- (2)

From equations (1) and (2), we get,

⇒ 3a – 3 = 10 – 2a

⇒ a =

Therefore, the coordinates of point A are .

The equation of the given line is

⇒ bxcosƟ + aysinƟ – ab = 0 ……………….(1)

Length of the perpendicular from point to line (1) is

--------------- (2)

Length of the perpendicular from point to line (2) is

--------------- (3)

On multiplying equations (2) and (3), we get,

Hence Proved.

The equations of the given lines are

2x-3y+4 = 0 --------------- (1)

3x+4y- 5 = 0 ---------------- (2)

6x – 7y + 8 = 0

The person is standing at the junction of the paths represented by lines (1) and (2).

So, on solving equations (1) and (2), we get,

x = and y =

Then, the person is standing at point

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point

Thus, slope of the line (3) =

Thus, Slope of the line perpendicular to line (3) =

The equation of the line passing through and having a slope of is given by

⇒ 6(17y-22) = -7(17x+1)

⇒ 102y – 132 = 119x -7

⇒ 119x + 102y = 125

Therefore, the path that the person should follow is 119x + 102y = 125.