**Coordinate Geometry Solution of TS & AP Board Class 10 Mathematics**

###### Exercise 7.1

**Question 1.**

Find the distance between the following pair of points

(2, 3) and (4, 1)

**Answer:**

(x_{1,}y_{1}) = (2,3) and (x_{2,}y_{2}) = (4,1)

d =

=

=

=

d = units

**Question 2.**

Find the distance between the following pair of points

(-5, 7) and (-1, 3)

**Answer:**

(x_{1,}y_{1}) = (-5,7) and (x_{2,}y_{2}) = (-1,3)

d =

=

=

=

d =

**Question 3.**

Find the distance between the following pair of points

(-2, -3) and (3, 2)

**Answer:**

(x_{1,}y_{1}) = (-2,-3) and (x_{2,}y_{2}) = (3,2)

d =

=

=

d =

**Question 4.**

Find the distance between the following pair of points

(a, b) and (-a, -b)

**Answer:**

(x_{1,}y_{1}) = (a,b) and (x_{2,}y_{2}) = (-a,-b)

d =

=

=

d =

**Question 5.**

Find the distance between the points (0, 0) and (36, 15).

**Answer:**

let (x_{1,}y_{1}) = (0,0) and (x_{2,}y_{2}) = (36,15)

D =

=

=

=

d =

**Question 6.**

Verify whether the points (1, 5), (2, 3) and (-2, -1) are collinear or not.

**Answer:**

let A = (1,5) B = (2,3) and c = (-2,-1)

⇒ AB =

=

AB =

⇒ BC =

BC =

⇒ AC =

=

AC =

⇒ AB + BC is not equal to AC.

∴ Points are not collinear

**Question 7.**

Check whether (5, -2), (6, 4) and (7, 2) are the vertices of an isosceles triangle.

**Answer:**

let A = (5,-2) B = (6,4) and c = (7,2)

⇒ AB =

=

AB =

⇒ BC =

=

BC =

⇒ AC =

=

AC =

As all the sides are unequal the triangle is not isosceles.

**Question 8.**

In a class room, 4 friends are seated at the points A, B, C and D as shown in Figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani “Don’t you notice that ABCD is a square?” Phani disagrees.

Using distance formula, find which of them is correct. Why?

**Answer:**

let A = (3,4) B = (6,7) and C = (9,4) D = (6,1)

⇒ AB =

=

AB =

⇒ BC =

=

BC =

⇒ CD =

=

CD =

⇒ AD =

=

AD =

As all the sides are equal the ABCD is a square. Jarina is correct.

**Question 9.**

Show that the following points form an equilateral triangle A(a, 0), B(-a, 0),

**Answer:**

let A = (a,0) B = (-a,0) and

⇒ AB =

=

AB = 2a

⇒ BC =

=

=

BC = 2a

⇒ AC =

=

=

AC = 2a

As all the sides are equal the triangle is Equilateral.

**Question 10.**

Prove that the point (-7, -3), (5, 10), (15, 8) and (3,-5) taken in order are the corners of a parallelogram. And find its area.

**Answer:**

let A = (-7,-3) B = (5,10) and C = (15,8) D = (3,-5)

Let these points be a parallelogram.

So midpoints of AC and DB should be same.

⇒ To find midpoint of AC and DB

⇒ For AC =

AC =

AC =

⇒ For DB

DB =

DB =

DB =

As midpoints of AC and DB are same the points form a parallelogram.

Let us divide the parallelogram into 2 triangle Î”ABD and Î”BCD

Area of both triangles

⇒ Area Î”ABD =

=

=

=

=

= 77

⇒ Area Î”BCD =

=

=

=

= 77

Total area of parallelogram = Sum of Area of triangles

= Î”BCD + Î”ABD

= 154 units

**Question 11.**

Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus.

(Hint: Area of rhombus product of its diagonals)

**Answer:**

Let the points be

A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4)

Length of diagonals

AC =

AC =

AC =

BD =

BD =

BD =

⇒ Area of Rhombus = 1/2 × Product of diagonals

= =

= 72 units

⇒ Area of triangles

Area Î”ABD =

=

=

=

= 36

⇒ Are

a Î”BCD =

=

=

=

= 36

Sum of area of triangles = 36 + 36 = 72 units

Thus proved

**Question 12.**

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.

(-1, -2), (1, 0), (-1, 2), (-3, 0)

**Answer:**

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

AB =

AB =

AB =

AB =

BC =

BC =

BC =

BC =

CD =

CD =

CD =

AD =

AD =

AD =

As all sides are equal the quadrilateral is a square

**Question 13.**

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.

(-3, 5), (3, 1), (1, -3), (-1,-4)

**Answer:**

Let A(-3, 5), B(3, 1), C(1, -3) and D(-1,-4)

Let us see the points on coordinate axes.Let us first calculate the length of the sides,

We know that distance between two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) is given by,

Therefore,

Now calculating BC,

Calculating CD,

Calculating DA,

From the lengths we can see that, none of the sides are equal.

**Hence, the quadrilateral formed is of no specific type.**

**Question 14.**

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.

(4, 5), (7, 6), (4, 3), (1, 2)

**Answer:**

Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2)

If ABCD is parallelogram

Midpoint of diagonals AC and BD

⇒ For AC =

X(4,4)

⇒ For BD =

X(4,4)

As the midpoints are same the diagonals bisect each other

Thus, the points form a parallelogram

**Question 15.**

Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

**Answer:**

Let P(x,0) be the point

A(2,-5) and B(-2,9)

⇒ PA =

⇒ PB =

PA = PB

PA^{2} = PB^{2}

(x-2)^{2} + (-5)^{2} = (x + 2)^{2} + 9^{2}

x^{2}-4x + 4 + 25 = x^{2} + 4x + 4 + 81

8x = -56

X = 7

Point is (-7,0)

**Question 16.**

If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.

**Answer:**

7 or -5

Let A(x,7) and B(1,15)

be the point

⇒ AB =

AB = 10

AB^{2} = 10^{2}

⇒ (1-x)^{2} + 8^{2} = 10^{2}

⇒ x^{2}-2x + 1 + 64 = 100

⇒ x^{2}-2x-35 = 0

⇒ (x-7)(x + 5) = 0

⇒ X = 7 or x = -5

**Question 17.**

Find the value of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

**Answer:**

3 or -9

Let P(2,-3) and Q(10,y)

be the point

⇒ PQ =

⇒ PQ = 10

PQ^{2} = 10^{2}

⇒ (y + 3)^{2} + 8^{2} = 10^{2}

y^{2} + 6y + 9 + 64 = 100

y^{2} + 6y-27 = 0

⇒ (y + 9)(y-3) = 0

⇒ y = 9 or y = -3

**Question 18.**

Find the radius of the circle whose centre is (3, 2) and passes through (-5, 6).

**Answer:**

Let P be center such that P(3,2) and Q be point on circumference Q(-5,6)

PQ =

PQ =

PQ =

PQ =

PQ =

**Question 19.**

Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.

**Answer:**

Let A(1, 5), B(5, 8) and C(13, 14)

⇒ AB =

AB =

AB = 5

⇒ BC =

BC =

BC = 10

⇒ AC =

AC =

AC = 15

Largest side is AC = 15 which is equal to Sum of other two sides AB and BC i.e. (10 + 5 = 15)

∴ triangle cannot be drawn infact the points are collinear

**Question 20.**

Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5)

**Answer:**

Let P(x,y) be the point

A(-2,8) and B(-3,-5)

⇒ PA =

⇒ PB =

PA = PB

PA^{2} = PB^{2}

(x + 2)^{2} + (8-y)^{2} = (x + 3)^{2} + (y + 5)^{2}

x^{2} + 4x + 4 + 64 -16y + y^{2} = x^{2} + 6x + 9 + y^{2} + 10y + 25

-2x-26y = -34

⇒ X + 13y = 17

###### Exercise 7.2

**Question 1.**

Find the coordinates of the point which divides the line segment joining the points (-1,7) and (4, -3) in the ratio 2 : 3.

**Answer:**

Let P(x,y) be the point

P(x,y) = (1,3)

**Question 2.**

Find the coordinates of the points of trisection of the line segment joining (4,-1) and (-2, -3).

**Answer:**

points A(4, -1) and B(-2, -3) are shown in the graph above

Now points of trisection means the points which divides the line in three parts. From the figure it is clear that, there will be 2 points which will do that. Let us call them P and Q.

Now clearly point P divides the BA in the ratio 1:2 and point Q divides the line in the ratio 2:1

⇒ Let P and Q be points of trisection

∴ P divides BA internally in ratio 1:2

Such that AP = PQ = QB

Let us apply section formula to the points A and B such that P divides BA in the ratio 1:2⇒ Q divides BA internally in ratio 2:1

Hence, Points of trisection are and

**Question 3.**

Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

**Answer:**

Let P and Q be line

∴ A divides PQ internally in ratio a:b

⇒ Given that A(-1,6)

6a-3b = -a-b , 10b-8a = 6a + 6b

7a = 2b, 14a = 4b

∴ a:b = 2:7

**Question 4.**

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

**Answer:**

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)

If ABCD is a parallelogram

AC and BD bisect each other

⇒ Midpoint of AC

⇒ For AB =

=

⇒ Midpoint of BD

For BD =

=

∴

1 + x = 7 , 8 = 5 + y

x = 6 y = 3

**Question 5.**

Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

**Answer:**

Let O be the center O(2,-3)

⇒ A(x,y) B(1,4)

∴ O divides AB internally in ratio 1:1

x + 1 = 4

x = 3

y + 4 = -6

y = -10

**Question 6.**

If A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that and P AB lies on the segment AB.

**Answer:**

AP = 3/7AB

∴ P divides AB in ratio 3:4

**Question 7.**

Find the coordinates of points which divide the line segment joining A(-4, 0) and B(0, 6) into four equal parts.

**Answer:**

⇒ Let x,y and z divide the line into 4 equal parts such that

AX = XY = YZ = ZB

⇒ X divides AB in ration 1:3

⇒ Y divides AB in ratio 1:1

Y(-2,3)

⇒ Z divides AB in ratio 3:1

**Question 8.**

Find the coordinates of the points which divides the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

**Answer:**

⇒ Let x,y and z divide the line into 4 equal parts such that

AX = XY = YZ = ZB

⇒ X divides AB in ration 1:3

⇒ Y divides AB in ratio 1:1

Y(0,5)

⇒ Z divides AB in ratio 3:1

**Question 9.**

Find the coordinates of the point which divides the line segment joining the points (a+b, a-b) and (a-b, a+b) in the ratio 3 : 2 internally.

**Answer:**

**Question 10.**

Find the coordinates of centroid of the triangle with vertices following:

(-1, 3), (6, -3) and (-3, 6)

**Answer:**

The coordinates of centroid are

**Question 11.**

Find the coordinates of centroid of the triangle with vertices following:

(6, 2), (0, 0) and (4, -7)

**Answer:**

**Question 12.**

Find the coordinates of centroid of the triangle with vertices following:

(1, -1), (0, 6) and (-3, 0)

**Answer:**

###### Exercise 7.3

**Question 1.**

Find the area of the triangle whose vertices are

(2, 3) (-1, 0), (2, -4)

**Answer:**

Area Î”ABC =

=

=

=

**Question 2.**

Find the area of the triangle whose vertices are

(-5, -1), (3, -5), (5, 2)

**Answer:**

Area Î”ABC =

=

=

=

= 32

**Question 3.**

Find the area of the triangle whose vertices are

(0, 0) (3, 0) and (0, 2)

**Answer:**

Area Î”ABC =

=

=

=

= 3

**Question 4.**

Find the value of ‘K’ for which the points are collinear.

(7, -2) (5, 1) (3, K)

**Answer:**

For points to be collinear area of Î”ABC = 0

Area Î”ABC =

=

=

=

=

∴ 8 - 2k = 0

K = 4

**Question 5.**

Find the value of ‘K’ for which the points are collinear.

(8, 1), (k, -4), (2, -5)

**Answer:**

For points to be collinear area of Î”ABC = 0

Area Î”ABC =

=

=

=

18-6k = 0

K = 3

**Question 6.**

Find the value of ‘K’ for which the points are collinear.

(K, K) (2, 3) and (4, -1).

**Answer:**

For points to be collinear area of Î”ABC = 0

Area Î”ABC =

=

=

=

=

⇒ K^{2} -11k-12 = 0

⇒ K(k-12) + (k-12) = 0

⇒ (K-12)(k + 1) = 0

⇒ K = 12 or k = -1

**Question 7.**

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area of the area of the given triangle.

**Answer:**

1 sq. unit; 1 : 4

⇒ Let A(0,-1),B(2,1),C(0,3)

So midpoints of AB and BC and CD

⇒ Midpoint formula

⇒ For AB =

AB =

AB =

⇒ For BC

BC =

BC =

BC =

⇒ For AC

AC =

AC =

AC =

(1,0)(1,2)(0,1)

Area Î”XYZ =

=

=

=

= 1 sq cm

⇒ Let A(0,-1),B(2,1),C(0,3)

⇒ Area Î”ABC =

=

=

=

= 4 sq cm

∴ Ratio of Area of triangles is 1:4

**Question 8.**

Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)

**Answer:**

28 sq. units

⇒ Let A = (-4, -2), B = (-3, -5), C = (3, -2) and D = (2, 3)

We draw Line BD and divide the quadrilateral into 2 triangles

Î”ABD and Î”BDC

Area of both triangles

⇒ Area Î”BDC =

=

=

=

=

=

= 16.5 units

A = (-4, -2), B = (-3, -5), D = (2, 3)

Area Î”ABD =

=

=

=

=

= 11.5 units

⇒ Total area of quadrilateral = Sum of Area of triangles

= Î”BCD + Î”ABD

= 16.5 + 11.5 units

= 28 sq units.

**Question 9.**

Find the area of the triangle formed by the points (8, -5), (-2, -7) and (5, 1) by using Heron’s formula.

**Answer:**

Not possible

Let A = (8,-5) B = (-2,-7) and C = (5,1)

AB =

AB =

AB =

BC =

BC =

BC =

AC =

AC =

AC =

###### Exercise 7.4

**Question 1.**

Find the slope of the line passing the two given points

(4, -8) and (5, -2)

**Answer:**

Slope

**Question 2.**

Find the slope of the line passing the two given points

(0, 0) and

**Answer:**

**Question 3.**

Find the slope of the line passing the two given points

(2a, 3b) and (a, -b)

**Answer:**

**Question 4.**

Find the slope of the line passing the two given points

(a, 0) and (0, b)

**Answer:**

**Question 5.**

Find the slope of the line passing the two given points

A(-1.4, -3.7), B(-2.4, 1.3)

**Answer:**

**Question 6.**

Find the slope of the line passing the two given points

A(3, -2), B(-6, -2)

**Answer:**

**Question 7.**

Find the slope of the line passing the two given points

**Answer:**

**Question 8.**

Find the slope of the line passing the two given points

A(0, 4), B(4, 0)

**Answer:**