Coordinate Geometry

We know that,

x, y is any point on the Cartesian plane in first quadrant.

Then,

x = Perpendicular distance from Y - axis and

x = Perpendicular distance from Y - axis

So, the distance of the point P (2, 3) from the X - axis = 3

By using the distance formula:

d² = (x₂ - x₁)² + (y₂ - y₁)²

Lets calculate the distance between the points (x₁, y₁) and (x₂ , y₂)

We have;

x₁ = 0, x₂ = 0

y₁ = 6, y₂ = - 2

d² = (0 – 0)² + ( - 2 – 6)²

d = √64

d = 8 units

So, the distance between A (0, 6) and B (0, 2) = 8

By using the distance formula:

d² = (x₂ - x₁)² + (y₂ - y₁)²

Lets calculate distance between the points (x₁ , y₁) and (x₂ , y₂)

Here we have;

x₁ = - 6, x₂ = 0

y₂ = 8, y₂ = 0

d² = [0 – ( - 6)]² + [0 – 8]²

⇒

⇒ d = √(36 + 64)

⇒ d = √100

⇒ d = 10

So the distance between P ( - 6, 8) and origin O (0, 0) = 10

By using the formula:

d² = (x₂ - x₁)² + (y₂ - y₁)²

To calculate distance between the points (x₁ , y₁) and (x₂ , y₂)

Here we have;

x₁ = 0, x₂ = - 5

y₂ = 5, y₂ = 0

d² = [( - 5) - 0]² + [0 - 5]²

d= √50= 5√2

So the distance between (0, 5) and ( - 5, 0) = 5√ 2

We have three vertices;

A = (0, 3)

O = (0, 0)

B = (5, 0)

Now,

As we know that diagonals of a rectangle are of equal length,

So,

Length of the diagonal AB = Distance between the points A and B

Calculate the distance between the points (x₁, y₁) and (x₂, y₂);

By the formula;

We have;

d² = (x₂ - x₁)² + (y₂ - y₁)²

x₁ = 0, x₂ = 5

y₂ = 3, y₂ = 0

d² = (5 – 0)² + (0 – 3)²

d = √(25 + 9)= √34

Distance between A (0, 3) and B (5, 0) is √34

Hence, the required length of its diagonal is √34

We plot the vertices of a triangle i.e., (0, 4), (0, 0) and (3, 0) on the paper shown as given below

Now, perimeter of ΔAOB = Sum of the length of all its sides:

Distance between the points

= Distance between A(0, 4) and O(0, 0) + Distance between O(0, 0) and B(3, 0)

+ Distance between A(0, 4) and B(3, 0)

Hence, the required perimeter of triangle is 12

Given,

Vertices of the triangle are,

A (x₁, y₁),

B (x₂, y₂) and

C (x₃, y₃)

By using the formula;

We have;

Area of triangle =| |

=|

=| |

=| |

= 8

Hence, the required area of ΔABC is 8.

Let’s take;

A (- 4, 0),

B (4, 0),

C (0, 3) are the given vertices.

Now,

To calculate the distance between two given points;

Us the formula;

d² = (x₂ - x₁)² + (y₂ - y₁)²

d =

So, the distance between point A and point B;

AB =

AB =

AB = 8

Distance between B (4, 0) and C (0, 3);

BC =

BC =

BC = 5

Distance between A (- 4, 0) and C (0, 3);

AC =

AC =

AC = 5

So, BC = AC

Hence, we can say that ΔABC is an isosceles triangle because an isosceles triangle has two sides equal.

Let’s take A and B the joining point and P is the dividing point;

Let’s assume the co - ordinates of point P = x and y

By using Section formula;

x co - ordinate of point P will be -

x = and

y co - ordinate of point P will be -

∴ x =

Given that,

x₁ = 7, y₁ = - 6,

x₂ = 3, y₂ = 4

m = 1 and

n = 2

So, (x, y) = lies in IV quadrant.

[Since, in IV quadrant, x - coordinate is positive and y - coordinate is negative]

As we know that, the perpendicular bisector of the any line segment divides the line segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the mid - point of the line segment.

As mid - point of any line segment which passes through the points

(x₁, y₁) and (x₂, y₂) is;

=

So mid - point of the line segment joining the points A (- 2, - 5) and B (2, 5) will be;

= = (0, 0)

Hence, (0, 0) is the required point lies on the perpendicular bisector of the lines segment.

Given a parallelogram ABCD whose three vertices are;

A ( - 2, 3),

B (6, 7) and

C (8, 3)

Let the fourth vertex of parallelogram, D = (x₄, y₄) and L, M be the middle points of AC and BD, respectively

L =

Since, mid - point of a line segment having points (x₁, y₁) and (x₂, y₂)

= and

M =

As we know ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other.

So, L and M are the same points

3 = and

3 =

→ 6 = 6 + x₄ and 6 = 7 + y₄

→ x₄ = 0 and y₄ = 6 – 7

∴ x₄ = 0 and y₄ = - 1

Hence, the fourth vertex of parallelogram is D = (x4, y4) = (0, 1)

Given,

Point P (2, 1) lies on the line segment AB, joining the points A (4, 2) and B (8, 4);

Now,

The distance between A (4, 2) and P (2, 1),

By using the distance formula;

∵ Distance between two points (x₁, y₁) and (x₂, y₂),

d =

AP =

Distance between A (4, 2) and B (8, 4)

AB

Distance between B (8, 4) and P (2, 1),

BP =

∴ AB =

2AP = AB

Hence, required condition is AP =

Given,

P is the mid - point of the line segment joining the points Q and R

Where;

P =

Q = ( - 6, 5)

R = ( - 2, 3)

Shown in the figure given below;

∴ Mid - point of QR = P

P = ( - 4, 4)

Since, mid - point of line segment having points (x₁, y₁) and (x₂, y₂);

=

But given coordinates of mid - point P is ;

∴ ( - 4, 4)

On comparing the coordinates, we get

= - 4

∴ a = - 12

Hence, the required value of a = - 12

First, we have to plot the points of the line segment on the paper and join them.

As we know that the perpendicular bisector of line segment AB, perpendicular at AB and passes through the mid - point of AB.

Let P be the mid - point of AB

Now find the mid - point,

Mid - point of AB =

_∵ Mid - point of line segment passes through the points (x₁, y₁) and (x₂, y₂)

=

Find the slope of the bisector:

Slop of the given line =

Slope =
Slope of given line multiplied by slope of bisector = - 1
Slope of bisector =
= - 3
Find out the bisector's formula by using the point slope form;

Which is;
- 3 =

- 3(2.5 – x) = 5.5 - y
- 7.5 + 3x = 5.5 - y
3x + y - 13 = 0
Transform the formula into slope - intercept form
3x + y - 13 = 0
y = - 3x + 13
because slope - intercept form is y = mx + c,

Where, m is the slope and c is the y - intercept
Thus, perpendicular bisector cuts the y - axis at (0, 13)

So, the required point is (0, 13).

Let the coordinate of the point P is (h, k).

Given,

Point P is equidistant from the three vertices O, A and B

Where;

O (0, 0),

A (0, 2y) and

B (2x, 0)

Then,

PO = PA = PB

→ (PO)² = (PA)² = (PB)² ….(i)

By using distance formula,

=

=

= h² + k² = h² + (k – 2y)² …..(ii)

= (h – 2x)² + k² ….. (iii)

Taking first two equations, we get

h² + k² = h² + (k – 2y)²

→ k² = k² + 4y² – 4yk → 4y(y – k) = 0

→ y = k [∵ y ≠ 0]

Taking first and third equations, we get

h² + k² = (h – 2x)² + k²

→ h² + h² + 4x² – 4xh

→ 4x (x – h) = 0

→ x = h [∵ x ≠ 0]

∴ Required points = (h, k) = (x, y)

Given,

Centre of circle in (0, 0) and passes through the point

So,

Radius of circle = Distance between origin (0, 0) and point

=

A point lie outside on the circle or inside the circle of the distance of it from the center of the circle is greater than equal to or less than radius of the circle.

Now, compare all the options one by one.

(a) Distance between (0, 0) and

So, the point lies interior to the circle.

(b) Distance between (0, 0) and

So, the point lies inside the circle.

(c) Distance between (0, 0) and

= 5.02 < 6.5

So, the point lies inside the circle.

(d) Distance between (0, 0) and

= 6.5

So, the point does not lie in the circle.

Let the coordinates of P (0, y) and Q (x, 0).

So, the mid - point of P (0, y) and Q (x, 0) = M

Coordinates of M =

_∵ Mid - point of a line segment having points (x₁, y₁) and (x₂, y₂)

=

Given,

Mid - point of PQ is (2, - 5)

= 4 = x + 0

x = 4

= - 10 = y + 0

– 10 = y

So,

x = 4 and

y = - 10

Thus, the coordinates of P and Q are (0, - 10) and (4, 0)

Let the vertices of a triangle = A, B and C

Where;

A = (x₁, y₁) = (a, b + c)

B = (x₂, y₂) = (b, c + a) and

C = (x₃, y₃) = (c, a + b)

∵ Area of

[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

[a(c - b) + b(a - c) + c(b - a)]

[ac - ab + ab - bc + bc - ac] = (0) = 0

Hence, the area of triangle is 0.

Given,

The distance between the points (4, p) and (1, 0) = 5

By using the formula of distance,

We get;

∵ Distance between the points (x₁, y₁) and (x₂, y₂)

→

On squaring both the sides, we get

9 + p² = 25

→ p² = 16 → p = ± 4

Hence, the required value of p is ± 4

Let the points are;

A = (x₁, y₁) = (1, 2)

B = (x₂, y₂) = (0, 0)

C = (x₃, y₃) = (a, b)

∵ Area of ∆ABC= ∆ = 1/2

[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

∴ ∆ =1/2 [1(0 - b) + 0(b - 2) + a(2 - 0)]

⇒Δ=1/2 ( - b + 0 + 2a)=1/2(2a - b)

As, the points A (1, 2), B (0, 0) and C (a, b) are collinear, then area of ΔABC will be equals to the zero

Area of ΔABC = 0

⇒1/2 (2a - b)

→ 2a - b = 0

→ 2a = b

Hence, the required relation is 2a = b

True

Given,

ΔABC = ΔDEF

Now,

Calculate the distance between A (2, 0) and B (2, 0) in ΔABC;

∵ Distance between the points (x₁, y₁) and (x₂, y₂);

So,

Similarly, distance between B (2, 0) and C (0, 2)

Distance between C (0, 2) and A (2, 0)

Now,

Calculate the distance between F (0, 4) and D ( - 4, 0) in ΔDEF;

Distance between F (0, 4) and E ( - 4, 0);

Distance between E (4, 0) and D ( - 4, 0);

Now,

Here, we see that sides of ΔABC and ΔFDE are proportional.

Hence, both the triangles are similar by SSS rule.

True

First plot all the points A, B and P on the graph paper.

As given, coordinates of point P is ( - 4, 2), A ( - 4, 6) and B ( - 4, - 6);

Now draw a line connecting point A and B;

As we can see from the figure, point P ( - 4, 2) lies on the line segment joining the points A ( - 4, 6) and B ( - 4, - 6),

False

The points are collinear if area of a triangle, formed by its points is equals to the zero.

Given,

x₁ = 0, x₂ = 0, x₃ = 3 and

y₁ = 5, y₂ = - 9, y₃ = 6

∵ Area of triangle = [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

∆ = [0( - 9 – 6) + 0(6 – 5) + 4(5 + 9)]

∆ = (0 + 0 + 3 × 14

∆ = 42/2 = 21 ≠ 0

As we can see the area of triangle formed by the points (0, 5), (0 - 9) and (3, 6) is not zero, and the points are only be collinear if area of a triangle, formed by its points is equals to the zero.

Hence, the points are non - collinear.

False

We know that, the points lies on perpendicular bisector of the line segment joining the two points is equidistant from these two points.

So,

PA should be equals to the PB.

Now, with the help of distance formula;

∵ PA ≠ PB

As we can see that PA ≠ PB

So, the point P does not lie on the perpendicular bisector of AB.

True

Let the coordinates of A = (x₁, y₁) = (3, 1)

Coordinates of B = (x₂, y₂) = (12, - 2)

Coordinates of C = (x₃, y₃) = (0, 2)

Area of ∆ABC = ∆ = 1/2 [x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

Δ= 1/2 [3 - (2 - 2) + 12(2 - 1) + 0{1 - ( - 2)}]

Δ =1/2 [3( - 4) + 12(1) + 0]

Δ =1/2 ( - 12 + 12)=0

Area of ΔABC = 0

Hence, the points A (3, 1), B (12, - 2) and C (0, 2) are collinear.

So, the points A (3, 1), B (12, - 2) and C (0, 2) can’t be the vertices of a triangle.

False

As we know that in a parallelogram, opposite sides are equal.

By using the distance formula calculates the distance between the sides;

As distance between the points (x₁, y₁) and (x₂, y₂);

Distance between A (4, 3) and B (6, 4);

Distance between B (6, 4) and C (5, - 6),

Distance between C (5, 6) and D ( - 3, 5),

Distance between D ( - 3, 5) and A (4, 3),

Here, as we can see that all sides AB, BC, CD and DA are different.

Hence, given vertices can’t be the vertices of a parallelogram.

True

Let’s draw a circle and mark a point P on it as shown in the figure.

Now, mark a point Q outside the circle.

By using Distance formula;

Distance between two points (x₁, y₁) and (x₂, y₂);

d =

Calculate the distance between origin O (0, 0) and P (5, 0),

OP =

= Radius of circle

Distance between origin O (0, 0) and Q (6, 8),

OQ =

= 10

As we know that, if the distance of any point from the center is less than the radius, then the point is inside.

If the distance of any point from the center is equal to the radius, then the point is on the circle. And if the distance of any point from the center is more than the radius, then the point is outside the circle.

As we can see here that, OQ>OP

Hence, it is true that point Q (6, 8) lies outside the circle.

False

As we know that a perpendicular bisector divides a line segment into two equal parts.

So, if A (2, 7) lies on perpendicular bisector of P (6, 5) and Q (0, - 4),

Then AP = AQ

By distance formula;

AQ =

As we can see AP ≠ AQ

Hence, the point A (2, 7) does not lie on the perpendicular bisector of the line segment PQ.

True

Let P (5, - 3) be the point which divides the line segment joining the points A (7, - 2) and B (1, - 5) in the ratio k:1 internally.

By using the section formula,

The coordinate of point

Given coordinate of P = (5, - 3)

So,

⇒ k + 7 = 5k + 5

⇒ - 4k = - 2

∴ k = 1/2

So the point P divides the line segment AB in ratio 1:2

Hence, Point P in the point of trisection of AB.

True

We know by the rule that, area of a triangle formed by the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is zero, then the points are collinear.

Area of the triangle = 1/2 [x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

Here we have;

x₁ = - 6

x₂ = - 4

x₃ = 3

and

y₁ = 10

y₂ = 6

y₃ = - 8

Area of the triangle ∆ ABC = 1/2 [ - 6 {6 – ( - 8)} + ( - 4) ( - 8 – 10) + 3(10 – 6)]

= 1/2 [ - 6(14) + ( - 4) ( - 18) + 3(4)]

= 1/2 ( - 84 + 72 + 12) = 0

So, given points are collinear.

Now,

Calculate the distance between A ( - 6, 10) and B ( - 4, 6);

∵ Distance between the points (x₁, y₁) and (x₂, y₂);

Distance between A ( - 6, 10) and C (3, - 8);

∴ AB=2/9 AC

False

We know that if the distance between the center and any point is equal to the radius, then we say that point lie on the circle.

Now,

By using the distance formula;

∵ Distance between the points (x₁, y₁) and (x₂, y₂);

Calculate the distance between P ( - 2, 4) and center (3, 5)

As we can see that distance between P and center is , which is not equal to the radius of the circle.

Hence, the point P ( - 2, 4) does not lies on the circle.

True

We know that, in a rectangle, opposite sides and diagonals are equal and bisect each other

Calculate the distance between A ( - 1, - 2) and B (4, 3);

By using the distance formula;

∵ Distance between the points (x₁, y₁) and (x₂, y₂);

So,

Distance between C (2, 5) and D ( - 3, 0);

Distance between A ( - 1, - 2) and D ( - 3, 0);

AD =

Distance between B (4, 3) and C (2, 5);

BC =

Since, AB = CD and AD = BC

Also, Calculate the distance of diagonals;

Distance between A ( - 1, - 2) and C (2, 5);

AC =

Distance between D ( - 3, 0) and B (4, 3);

DB =

Since, diagonals AC and BD are equal

Hence, the points A ( - 1, - 2), B (4, 3), C (2, 5) and D ( - 3, 0) form a rectangle.

By using the distance formula lets determine the length of all three sides.

∵ Distance between the points (x₁, y₁) and (x₂, y₂);

d =

So,

AB =

BC =

CA =

By calculating the distance between two points, now we can find out the type of triangle.

As we can see;

AB≠BC≠CA

ΔABC also not satisfies the condition Pythagoras i.e.,

(Hypotenuse)² = (Base)² + (Perpendicular)²

Hence, the triangle is scalene because all of its sides are not equal i.e., different from each other.

We know that, every point on the X - axis in the form (x, 0).

Let P(x, 0) the point on the X - axis have distance from the point Q(7, - 4).

As we know;

Distance between the points (x₁, y₁) and (x₂, y₂);

d =

By given condition,

Distance of PQ =

(PQ)² = 4 × 5

(x – 7)² + (0 + 4)² = 20

x² + 49 - 14x + 16 = 20

x² - 14x + 65 - 20 = 0

x² - 14x + 45 = 0

x² - 9x – 5x + 45 = 0

By Factorization method:

x(x - 9) – 5(x – 9) = 0

(x – 9)(x – 5) = 0

∴ x = 5, 9

Hence, there are two points lies on the axis, which are (5, 0) and (9, 0), have distance from the point (7, - 4).

To find the type of quadrilateral, let us find out the length of all four sides as well as two diagonals.

Now, using distance formula between two points.

Distance between the points (x₁, y₁) and (x₂, y₂);

d =

AB =

BC =

CD =

DA =

Calculate the distance of diagonals;

AC =

And BD =

Here, we see that the sides AB = CD and BC = DA

Also, diagonals are equal i.e., AC = BD

Which shows the quadrilateral is a rectangle.

According to the question,

Distance between AB = 9

Coordinates of A ( - 3, - 14)

Coordinates of B (a, - 5)

Distance between two points (x₁, y₁) and (x₂, y₂);

d =

So,

AB = = 9

AB = = 9

On squaring both the sides,

We get;

(a + 3)² + 81 = 81

(a + 3)² = 0

A = - 3

Hence, the value of a is - 3.

Let P (h, k) be the point which is equidistant from the points A( - 5, 4) and B( - 1, 6).

By distance formula;

Distance between two points (x₁, y₁) and (x₂, y₂);

d =

PA = PB

(PA)² = (PB)²

→ ( - 5 - h)² + (4 - k)² = ( - 1 - h)² + (6 - k)²

→25 + h² + 10h + 16 + k² - 8k = 1 + h² + 2h + 36 + k² – 12k ….

∵ [(a - b)² = a² + b² - 2ab]

→ 25 + 10h + 16 – 8k = 1 + 2h + 36 – 12k

→ 8h + 4k + 41 – 37 = 0

→ 8h + 4k + 4 = 0

→ 2h + k + 1 = 0 ……(i)

Now calculate the,

Mid - point =

Mid - point of AB =

At point ( - 3, 5), from Eq. (i);

2h + k = 2( - 3) + 5

= - 6 + 5 = - 1

→ 2h + k + 1 = 0

So, the mid - point of AB satisfies the Eq. (i).

Hence, all points which are solution of the equation 2h + k + 1 = 0 are equidistant from the points A and B.

Replacing h, k by x, y in above equation, we have 2h + k + 1 = 0

Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB passes through the mid - point of AB and divide the line in two equal parts.

Mid - point of a line segment passes through the points (x₁ + x₂) and (y₁ + y₂) is;

Mid - point of (x₁ + x₂) and (y₁ + y₂) =

So,

Mid - point of AB =

R =

Now, we draw a straight line on paper passes through the mid - point R.

We see that perpendicular bisector cuts the X - axis at the point Q

So, the coordinates of the point Q is on the X - axis which lies on the perpendicular bisector of the line segment joining the point AB.

To know the type of triangle formed by the points Q, A and B,

We have to find out the length of all three sides and see whatever condition of triangle is satisfied by these sides.

Now, using distance formula between two points (x₁, y₁) and (x₂, y₂);

d =

So,

AB = = 9

BQ =

QA =

We see that, BQ = QA ≠ AB

Which shows that the triangle formed by the points Q, A and B is an isosceles.

Since, the points

A = (5, 1)

B = ( - 2 – 3)

C = (8, 2m) are collinear

Area of ∆ABC = 0

1/2 [x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]=0

1/2 [5( - 3 - 2m) + ( - 2)(2m - 1) + 8(1 - ( - 3))]=0

1/2 ( - 15 – 10m – 4m + 2 + 32) = 0

1/2 ( - 14m + 19) = 0

m = 19/14

Hence, the value of m =

Given,

A (2, - 4) is equidistant from P (3, 8) = Q ( - 10, y) is equidistant from A (2, - 4)

By distance formula;

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

So, PA = QA

⇒√145= √(160 + y² + 8y)

On squaring both the sides,

We get;

145 – 160 + y² + 8y

y² + 8y + 160 - 145 = 0

y² + 8y + 15 = 0

y² + 5y + 3y + 15 = 0

y(y + 5) + 3(y + 5) = 0

(y + 5)(y + 3) = 0

If y + 5 = 0, then y = - 5

If y + 3 = 0, then y = - 3

∴ y = - 3, – 5

Now,

Distance P (3, 8) and Q ( - 10, y);

[Putting y = - 3]

→

Again, distance between P (3, 8) and Q (-10, y);

[Putting y = - 5]

→

Hence,

The values of y are - 3, - 5 and corresponding values of PQ are √290 and √338=13√2 respectively.

Given,

Let the vertices of triangle;

(x₁, y₁) → ( - 8, 4)

(x₂, y₂) = ( - 6, 6)

(x₃, y₃) = ( - 3, 9)

We know that, the area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃)

Area of triangle = 1/2 [x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]=0

= 1/2 [ - 8(6 – 9) – 6(9 – 4) + ( - 3)(4 – 6)]

= 1/2 [ - 8( - 3) – 6(5) – 3( - 2)]

= 1/2 24 – 30 + 6)

= 1/2 (30 – 30)

= 1/2 (0) = 0

Hence, the area of triangle is 0.

Let the required ratio be λ:1

So, the coordinates of the point M of division A ( - 4, - 6) and B ( - 1, 7) are;

Here we have;

x₁ = - 4,

x₂ = - 1 and

y₁ = - 6

y₂ = 7

=

According to the question,

Line segment joining A ( - 4, - 6) and B ( - 1, 7) is divided by the X - axis.

So y - coordinate must be zero.

So, the required ratio is 6:7 and the point of division M is;

=

=

=

Hence, the point of division is

Let P divide AB internally in the ratio m:n.

Using the section formula,

Internal section formula, the coordinates of point P divides the line segment joining the point (x₁, y₁), (x₂, y₂) in the ratio m₁:m₂ internally is

We get;

On equating,

We get;

And

3m + 3n = 8m – 2n

5n – 5m = 0

n = m

And, 5m + 5n = - 60 + 18n

65m – 13n = 0

13(5m – n) = 0

5m – n = 0

Since,

m = n does not satisfy.

5m - n = 0

5m = n

Hence, the required ratio is 1:5.

Let P (9a – 2, - b) divides AB internally in the ratio 3:1.

By the section formula;

Internal section formula, the coordinates of point P divides the line segment joining the point (x₁, y₁), (x₂, y₂) in the ratio m₁:m₂ internally is

For 9a – 2, we have;

9a – 2 =

→ 36a – 8 = 27a + 1

And for – b we have;

- b =

Hence, the values of a and b are 1 and - 3.

Given,

(a, b) is the mid - point of line segment AB

Since, mid - point of a line segment having points (x₁, y₁) and (x₂, y₂) =

So,

(a, b) =

Now, equating coordinates on both sides, we get

a = …. (i)

And

b = - 1 …(ii)

Given,

a – 2b = 18

Using equation (ii);

a – 2( - 1) = 18

a + 2 = 18

a = 16

Using equation (i);

16 =

32 = 10 + k

k = 22

Hence, the required value of k is 22

→ k = 22

∴ A = (10, - 6),

B = (22, 4)

Now,

By using the distance formula;

Distance between two points (x₁, y₁) and (x₂, y₂);

So,

Distance between A (10, - 6) and B (22, 4)

AB =

=

Hence, the distance of AB is

Using the given condition;

As distance between two points (x₁, y₁) and (x₂, y₂);

d =

So,

Distance between the centre C (2a, a – 7) and the point P (11, - 9), which lie on the circle = Radius of circle.

…….(i)

Given that,

Length of diameter =

By putting this value in Eq. (i),

We get;

By squaring both sides,

We get;

50 = (11 – 2a)² + (2 + a)²

→ 50 = 121 + 4a² – 44a + 4 + a² + 4a

→ 5a² – 40a + 75 = 0

→ a² – 8a + 15 = 0

→ a² – 5a – 3a + 15 = 0

By factorization method;

a(a – 5) – 3(a – 5) = 0

(a – 5)(a – 3) = 0

a = 3, 5

Hence, the values of a are 5 and 3.

Given that, the line segment joining the points A (3, 2) and B (5, 1) is divided at the point p in the ratio 1:2

By using the section formula for internal ratio;

∴ Coordinates of point P =

As given in the question;

The point P lies on the line 3x – 18y + 5 = 0

∴

→11 - 30 + k = 0

k = 19 = 0

k = 19

Hence, the value of k is 19

Let the vertices of ∆ABC are;

A = (x₁, y₁)

B = (x₂, y₂)

C = (x₃, y₃)

Given,

D, E and F are the mid - points of the sides BC, CA and AB respectively.

Where Coordinates are;

D =

E = (7, 3)

F =

By mid - point formula of a line segment having points (x₁, y₁) and (x₂, y₂) =

So,

D is the mid - point of BC

∴

And

x₂ + x₃ = - 1 …..(i)

And

y₂ + y₃ = 5 … (ii)

As E (7, 3) is the mid - point of CA

∴

And

∴ x₃ + x₁ = 14 .. (iii)

And

Y₃ + y₁ = 6 …. (iv)

Also,

F is the mid - point of AB

∴

And

∴ x₁ + x₂ = 7 … (v)

And

y₁ + y₂ = 7 …(vi)

On adding equations (i), (iii) and (v),

We get,

2(x₁ + x₂ + x₃) = 20

x₁ + x₂ + x₃ = 10………(vii)

On subtracting Equations (i), (iii) and (v) from Eq. (vii) respectively,

We get;

x₁ = 11,

x₂ = - 4

x₃ = 3

On adding Equations (ii), (iv) and (vi),

We get;

2(y₁ + y₂ + y₃) = 18

y₁ + y₂ + y₃ = 9…..(viii)

On subtracting Equations (ii), (iv) and (vi) from Eq. (viii) respectively,

We get;

y₁ = 4

y₂ = 3

y₃ = 2

Hence, the vertices of ΔABC are A (11, 4), B ( - 4, 3) and C (3, 2)

So,

∵ Area of ∆ABC = 1/2 [x₁ (y₂ – y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

∆ = [11(3 – 2) + ( - 4)(2 – 4) + (4 – 3)]

= [11 × 1 + ( - 4)( - 2) + 3(1)]

= [11 + 8 + 3] = 22/2 = 11

∴ Required area of ΔABC = 11

Given,

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ΔABC right angled at B.

By using Pythagoras theorem,

AC² = AB² + BC² ……(i)

Now by using distance formula;

As distance between two points (x₁, y₁) and (x₂, y₂);

So,

AB =

BC =

Put the values of AB, BC and AC in Eq. (i),

We get;

25 = a² – 4a + 20 + 25 + a² – 10a

2a² – 14a + 20 = 0

a² – 7a + 10 = 0

a² – 2a – 5a + 10 = 0

a(a – 2) – 5(a – 2) = 0

(a – 2)(a – 5) = 0

∴ a = 2, 5

Here, a ≠ 5, since at a = 5, the length of BC = 0.

It is not possible because the sides AB, BC and CA form a right-angled triangle.

So,

a = 2

Now, the coordinate of A, B and C becomes (2, 9), (2, 5) and (5, 5), respectively.

∵ Area of ∆ABC = 1/2 [x₁ (y₂ – y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

[2(5 - 5) + 2(5 - 9) + 5(9 - 5)]

= [2 × 0 + 2( - 4) + 5(4)]

= (0 – 8 + 20) = × 12 = 6

Hence, the required area of ΔABC is 6 sq. units.

According to the question,

Given,

PR = PQ

→

∴ RQ:PR = 2:3

Or

PR:RQ = 3:2

Let’s take,

R (x, y) be the point which divides the line segment joining the points P ( - 1, 3) and Q (2, 5) in the ratio 3:2

∴ (x, y) =

∵ By internal section formula;

=

Hence, the required coordinates of the point R is.

We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k - 1, 5k) are collinear.

Then, area of ΔABC = 0

1/2 [x₁ (y₂ – y₃) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )] = 0

Here,

x₁ = k + 1,

x₂ = 3k

x₃ = 5k – 1

And

y₁ = 2k

y₂ = 2k + 3

y₃ = 5k

[(k + 1)(2k + 3 - 5k) + 3k(5k – 2k) + (5k – 1)(2k + 3)]

[(k + 1)( - 3k + 3) + 3k(3k) + (5k - 1)(2k – 2k – 3)] = 0

[ - 3k² + 3k - 3k + 3 + 9k² - 15k + 3] = 0

(6k² – 15k + 6) = 0 (multiply by 2)

→ 6k² – 15k + 2 = 0

→ 2k² – 5k + 2 = 0

[Divide by 3]

→ 2k² – 4k – k + 2 = 0

→ 2k (k – 2) – 1(k – 2) = 0

→ (k – 2)(2k – 1) = 0

If k – 2 = 0, then k = 2

If 2k – 1 = 0, then k= 1/2

∴ k = 2, 1/2

Hence, the required values of k are 2 and 1/2.

Let the line 2x + 3y – 5 = 0 divides the line segment AB in the ratio of λ:1 at point P.

∴ Coordinates of P =

∵ Internal Division

=

But P lies on 2x + 3y – 5 = 0

→ 2(2λ + 8) + 3(λ – 9) – 5(λ + 1) = 0

→ 4λ + 16 + 3λ - 27 - 5λ – 5 = 0

→ 2λ – 16 = 0

→ λ = 8

→ λ:1 = 8:1

So, the point P divides the line in the ratio 8:1

∴ Point of division P =

Hence, the required point of division is

Let three vertices of the triangle be A, B and C

And the coordinates of third vertex be (x, y)

We have,

A ( - 4, 3),

B (4, 3) and

C (x, y)

We know that, in equilateral triangle the angle between two adjacent sides is 60 and all three sides are equal.

∴ AB = BC = CA

→ AB² = BC² = CA²

Now, taking first two parts

AB² = BC²

(4 + 4)² + (3 - 3)² = (x – 4)² + (y – 3)²

64 + 0 = x² + 16 – 8x + y² + 9 – 6y

x² + y² – 8x – 6y = 39……. (ii)

Now, taking first and third parts;

AB² = CA²

(4 + 4)² + (3 - 3)² = ( - 4 - x)² + (3 – y)²

64 + 0 = 16 + x² + 8x + 9 + y² – 6y

x² + y² + 8x – 6y = 39……. (iii)

On subtracting Eq. (ii) from Eq. (iii),

We get:

→ x = 0

Now, by putting the value of x in Eq. (ii),

We get;

0 + y² – 0 – 6y = 39

→ y² – 6y – 39 = 0

∴ y =

∵ Solution of ax² + bx + c = 0 is x =

y =

So, the points of third vertex are (0,3 + 4√3) or (3 - 4√3)

But given that, the origin lies in the interior of the ΔABC and the x - coordinate of third vertex is zero.

Then, y - coordinate of third vertex should be negative.

Hence, the required coordinate of third vertex,

C = [∵ c ≠ ]

Given,

A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD;

Let the fourth vertex of parallelogram be (x, y),

We know that, the diagonals of a parallelogram bisect each other

∴ Mid - point of BD = Mid - point of AC

→

∵ Mid - point of a line segment joining the points (x₁, y₁) and (x₂, y₂) =

→

→ 8 + x = 15 →x = 7

And

→ 2 + y = 5 → y = 3

So, fourth vertex of a parallelogram is D (7, 3)

Now,

Mid - point of side

DC =

E =

∵ Area of ΔABC with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃);

= [x₁(y₂ - y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

∴ Area of ΔADE with vertices A (6, 1), D (7, 3) and E

∆ =

=( - 3)/4 but area can’t be negative

Hence, the required area of ΔADE is sq. units

Given,

The vertices of ΔABC = A, B and C

Coordinates of A, B and C;

A(x₁, y₁)

B(x₂, y₂)

C(x₃, y₃)

(i) As per given information D is the mid - point of BC and it bisect the line into two equal parts.

Coordinates of the mid - point of BC;

BC –

(ii) Let the coordinates of a point P be (x, y)

Given,

The point P(x, y), divide the line joining A(x₁, y₁) and D in the ratio 2:1

Then,

Coordinates of P =

By using internal section formula;

=

(iii) ∴ Let the coordinates of a point Q be (p, q)

Given,

The point Q (p, q),

Divide the line joining B(x₂, y₂) and E in the ratio 2:1,

Then,

Coordinates of Q =

=

Since, BE is the median of side CA, So BE divides AC in to two equal parts.

∴ mid - point of AC = Coordinate of E;

E =

So, the required coordinate of point Q;

Q =

Now,

Let the coordinates of a point E be (⍺, β)

Given,

Point R (⍺, β) divide the line joining C(x3, y3) and F in the ratio 2:1,

Then the coordinates of R;

=

=

Since, CF is the median of side AB.

So, CF divides AB in to two equal parts.

∴ mid - point of AB = Coordinate of F;

F =

So, the required coordinate of point R;

=

(iv) Coordinate of the centroid of the ΔABC;

In parallelogram, we know that, diagonals are bisects each other i.e., mid - point of AC = mid - point of BD

= - 1

Since, mid - point of a line segment having points (x₁, y₁) and (x₂, y₂) is

1 + a = - 2

a = - 3

So, the value of a is – 3

Given,

AB as base of a parallelogram and drawn a perpendicular from D to AB which meets AB at P.

So, DP is a height of a parallelogram.

Now, equation of base AB, passing through the points (1, - 2) and (2, 3) is;

(y – y₁) =

(y + 2) =

(y + 2) = 5(x - 1)

5x – y = 7

Slop of AB = m₁ = ….. (i)

Let the slope of DP be m₂.

Since, DP is perpendicular to AB.

By condition of perpendicularity,

m₁.m₂ = - 1 5.m₂ = - 1

m₂ = -

Now,

Equation of DP, having slope - and passing the point ( - 4, - 3) is;

(y– y₁) = m₂(x – x₁)

→ (y + 3) = - (x + 4)

→ 5y + 15 = - x – 4

→ x + 5y = - 19 ……(ii)

On adding Equations (i) and (ii), then we get the intersection point P.

By putting the value of y from Eq. (i) to Eq. (ii),

We get;

X + 5(5x – 7) = - 19

X + 25x – 35 = - 19

26x = 16

x =

Put the value of x in Eq. (i);

We get;

y = 5

∴ Coordinates of point P =

By distance formula,

Distance between two points (x₁, y₁) and (x₂, y₂) is;

D =

So, length of the height of a parallelogram,

DP =

DP =

Hence, the required length of height of a parallelogram is

By observing the positions of the students as given in the figure we can say that, A, B, C and D are forming a quadrilateral. So the vertices of this quadrilateral will be A (3, 5), B (7, 9), C (11, 5) and D (7, 1).

Now,

To find out the type of this quadrilateral we have to find all its sides;

By distance formula;

D =

So,

AB =

AB =

AB

BC =

CD =

And DA =

We see that, AB = BC = CD = DA i.e., all sides are equal.

Now, we find length of both diagonals;

AC =

= 8

And BD =

= 8

Here, AC = BD

Since AB = BC = CD = DA and AC = BD

So we can say that ABCD is a square. As we also know that diagonals of a square bisect each other. So, P be position of Jaspal in which he is equidistant from each of the four students A, B, C and D

Calculate the coordinates of point P;

Coordinates of P ₌ Mid - point of AC

= = (7, 5)

Since, mid - point of a line segment having points (x₁, y₁) and (x₂, y₂) is

Hence, the required position of Jaspal is (7, 5).

With the help of above information lets draw a figure in which every place are indicated with his coordinates and direction also.

By using the distance formula, we know the distance between two points (x₁, y₁) and (x₂, y₂);

D =

Now calculate the;

Distance between house and bank;

=

=

Distance between bank and daughter’s school;

=

= 10

Distance between daughter’s school and office;

=

= 12

Total distance (House + Bank + School + Office) travelled = 5 + 10 + 12 = 27 units

Distance between house to offices =

= 24.59 = 24.6 km

So,

Extra distance travelled by Ayush in reaching his office = 27 – 24.6 = 2.4 km

Hence, the extra distance travelled by Ayush is 2.4 km.