Coordinate Geometry Solution of TS & AP Board Class 10 Mathematics ~ TGARIEA

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,y₁) = (2,3) and (x_2,y₂) = (4,1)

d =

d = units

Question 2.

Find the distance between the following pair of points

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

Answer:

(x_1,y₁) = (-5,7) and (x_2,y₂) = (-1,3)

d =

Question 3.

Find the distance between the following pair of points

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

Answer:

(x_1,y₁) = (-2,-3) and (x_2,y₂) = (3,2)

d =

Question 4.

Find the distance between the following pair of points

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

Answer:

(x_1,y₁) = (a,b) and (x_2,y₂) = (-a,-b)

d =

Question 5.

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

Answer:

let (x_1,y₁) = (0,0) and (x_2,y₂) = (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 =

⇒ For DB

DB =

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

Let us divide the parallelogram into 2 triangle ΔABD and ΔBCD

Area of both triangles

⇒ Area ΔABD =

= 77

⇒ Area ΔBCD =

= 77

Total area of parallelogram = Sum of Area of triangles

= ΔBCD + ΔABD

= 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 =

BD =

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

= =

= 72 units

⇒ Area of triangles

Area ΔABD =

= 36

⇒ Are

a ΔBCD =

= 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 =

BC =

CD =

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(x₁, y₁) and B(x₂, y₂) 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

PA² = PB²

(x-2)² + (-5)² = (x + 2)² + 9²

x²-4x + 4 + 25 = x² + 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

AB² = 10²

⇒ (1-x)² + 8² = 10²

⇒ x²-2x + 1 + 64 = 100

⇒ x²-2x-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

PQ² = 10²

⇒ (y + 3)² + 8² = 10²

y² + 6y + 9 + 64 = 100

y² + 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 =

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

PA² = PB²

(x + 2)² + (8-y)² = (x + 3)² + (y + 5)²

x² + 4x + 4 + 64 -16y + y² = x² + 6x + 9 + y² + 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 ΔABC =

Question 2.

Find the area of the triangle whose vertices are

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

Answer:

Area ΔABC =

= 32

Question 3.

Find the area of the triangle whose vertices are

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

Answer:

Area ΔABC =

= 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 ΔABC = 0

Area ΔABC =

∴ 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 ΔABC = 0

Area ΔABC =

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 ΔABC = 0

Area ΔABC =

⇒ K² -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 =

⇒ For BC

BC =

⇒ For AC

AC =

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

Area ΔXYZ =

= 1 sq cm

⇒ Let A(0,-1),B(2,1),C(0,3)

⇒ Area ΔABC =

= 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

ΔABD and ΔBDC

Area of both triangles

⇒ Area ΔBDC =

= 16.5 units

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

Area ΔABD =

= 11.5 units

⇒ Total area of quadrilateral = Sum of Area of triangles

= ΔBCD + ΔABD

= 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 =

BC =

AC =