Tag: Coordinate Geometry Exercise 14.3 class 10

Coordinate Geometry Exercise 14.3 – Class 10

Coordinate Geometry Exercise 14.3 – Questions:

  1. Find the distance between the following pairs of points

(i) (8, 3) and (8,-7)

(ii) (1,-3) and (-4, 7)

(iii) (-4, 5) and (-12, 3)

(iv) 6, 5) and (4, 4)

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

(vi) (2, 8) and (6, 8)

 (vii) (a, b) and (c, b)

(viii) (cosθ, -sinθ) and (sinθ, -cosθ)

  1. Find the distance between the origin and the point

(i) (-6, 8)

(ii) (5, 12)

(iii) (-8, 15)

(i)The distance between the points (3, 1) and (0, x) is 5 units. Find x.

(ii) A point P(2, 1) is equidistant from the points (a, 7) and (-3, a). Find a.

(iii) Find a point on y-axis which is equidistant from the points (5, 2) and (-4, 3)

  1. Find the perimeter of the triangles whose vertices have the following coordinates

(i)(-2, 1), (4, 6) and (6, -3)

(ii) (3, 10), (5, 2), (14, 12)

  1. Prove that the points A(1, -3), B(-3, 0) and C(4, 1) are the vertices of a right isosceles triangle.
  2. Find the radius of the circle whose centre is (-5, 4) and which passes through the point (-7, 1).
  3. Prove that each of the set of coordinates are the vertices of parallelogram.

(i) (-5, -3), (1, -11), (7, -6), (1, 2)

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

  1. The coordinates of vertices of triangles are given. Identify the types of the triangle.

(i) (2, 1), (10, 1), (6, 9)

(ii) (1, 6), (3, 2), (10,8)

(iii) (3, 5), (-1, 1), (6, 2)

(iv) (3, -3), (3, 5), (11, -3)

 

Coordinate Geometry Exercise 14.3 – Solutions:

  1. Find the distance between the following pairs of points

(i) (8, 3) and (8,-7)

Solution:

Coordinate Geometry Exercise 14.3


 (ii) (1,-3) and (-4, 7)

Solution:

Coordinate Geometry Exercise 14.3

 

 (iii) (-4, 5) and (-12, 3)

Solution:

Coordinate Geometry Exercise 14.3

 

 (iv) 6, 5) and (4, 4)

Solution:

Coordinate Geometry Exercise 14.3

 

 (v) (2,0) and (0, 3)

Solution:

Coordinate Geometry Exercise 14.3Coordinate Geometry Exercise 14.3

 (vi) (2, 8) and (6, 8)

Solution:

Coordinate Geometry Exercise 14.3

 (vii) (a, b) and (c, b)

Solution:

Coordinate Geometry Exercise 14.3

 

 (viii) (cos θ, -sinθ) and (sinθ, -cosθ)

Solution:

Coordinate Geometry Exercise 14.3

 

  1. Find the distance between the origin and the point

(i) (-6, 8)

Solution:

Distance between the origin and (x, y) = √(x2 + y2)

Here, (x, y) = (-6, 8)

d = √[(-6)2 + (8)2]

d = √[36+64]

d = √100

d = 10 units

 

 (ii) (5, 12)

Solution:

Distance between the origin and (x, y) = √(x2 + y2)

Here, (x, y) = (5, 12)

d = √[(5)2 + (12)2]

d = √[25+144]

d = √169

d = 13 units

 

 (iii) (-8, 15)

Solution:

Distance between the origin and (x, y) = √(x2 + y2)

Here, (x, y) = (-8, 15)

d = √[(-8)2 + (15)2]

d = √[64+225]

d = √289

d =  17 units

 

3. (i)The distance between the points (3, 1) and (0, x) is 5 units. Find x.

Solution:

We know,

Coordinate Geometry Exercise 14.3

52 = x2 – 2x +  10

x2 – 2x + 10  – 25 = 0

x2 – 2x – 15  = 0

x2 + 3x – 5x – 15 = 0

x(x + 3) – 5(x + 3) = 0

(x + 3)(x – 5) = 0

x + 3 = 0 or x – 5 = 0

x = -3 or x = 5

 

(ii) A point P(2, 1) is equidistant from the points (a, 7) and (-3, a). Find a.

Solution:

Let P(2, 1), Q(a , 7) and R(-3, a)

Coordinate Geometry Exercise 14.3Coordinate Geometry Exercise 14.3

a2 – 4a + 40 = a2 – 2a + 26

-4a + 2a + 40 – 26 = 0

-2a – 14 = 0

2a = 14

a = 7 units

 

 (iii) Find a point on y-axis which is equidistant from the points (5, 2) and (-4, 3)

Solution:

Let A(5, 2),  B(4, 3) and the point on y-axis be Y(0, y)

 

Coordinate Geometry Exercise 14.3

y2 – 4y + 29 = y2 – 6y + 25

-4y + 6y + 29 – 25 = 0

2y + 4 = 0

2y = -4

y = -2 units

Therefore, a point on y-axis which is equidistant from the points (5, 2) and (-4, 3) is (0, -2)

 

  1. Find the perimeter of the triangles whose vertices have the following coordinates

(i)(-2, 1), (4, 6) and (6, -3)

Solution:

Perimeter of the triangle = sum of 3 sides of the triangle

P(∆ABC) = AB + BC + CA

Let A(-2, 1), B(4, 6) and C(6, -3)

 

Coordinate Geometry Exercise 14.3

Therefore, Perimeter of triangle ABC = AB + BC + CA = √61+√85+√80

 

(ii) (3, 10), (5, 2), (14, 12)

Solution:

Perimeter of the triangle = sum of 3 sides of the triangle

P(∆ABC) = AB + BC + CA

Let A(3, 10), B(5, 2) and C(14, 12)

 

Coordinate Geometry Exercise 14.3

Therefore, Perimeter of triangle ABC = AB + BC + CA = √68+√181+√125

 

  1. Prove that the points A(1, -3), B(-3, 0) and C(4, 1) are the vertices of a right isosceles triangle.

Solution:

To show that the triangle is isosceles, we must show that the length of 2 sides of the triangle are equal.

Let A(1, -3), B(-3, 0) and C(4, 1)

 

Coordinate Geometry Exercise 14.3

Since AB = CA , the triangle is isosceles.

 

  1. Find the radius of the circle whose centre is (-5, 4) and which passes through the point (-7, 1).

Solution:

Let O(-5, 4) and P(-7, 1)

Coordinate Geometry Exercise 14.3

 

  1. Prove that each of the set of coordinates is the vertices of parallelogram.

(i) (-5, -3), (1, -11), (7, -6), (1, 2)

Solution:

Coordinate Geometry Exercise 14.3A quadrilateral is a parallelogram if it’s opposite sides is equal.

Let A(-5, -3), B(1, -11), C(7, -6), D(1, 2)

Coordinate Geometry Exercise 14.3

Therefore, AB = CD and BC = DA

 

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

Solution:

Coordinate Geometry Exercise 14.3A(4, 0), B(-2, -3), C(3, 2), D(-3, -1)

Coordinate Geometry Exercise 14.3

Therefore, AB = CD and BC = DA

 

  1. The coordinates of vertices of triangles are given. Identify the types of the triangle.

(i) (2, 1), (10, 1), (6, 9)

Solution:

Let A(2, 1), B(10, 1) and  C(6, 9)

Coordinate Geometry Exercise 14.3

Since BC = CA, the triangle is isosceles.

 

 (ii) (1, 6), (3, 2), (10,8)

Solution:

Let A(1, 6), B(3, 2), C(10,8)

Coordinate Geometry Exercise 14.3

Since AB ≠ BC ≠ CA, the triangle isosceles.

 

 (iii) (3, 5), (-1, 1), (6, 2)

Solution:

Let A(3, 5), B(-1, 1), and C(6, 2)

Coordinate Geometry Exercise 14.3

Since AB ≠ BC ≠ CA, the triangle scalene.

 

 (iv) (3, -3), (3, 5), (11, -3)

Solution:

Let A(3, -3), B(3, 5), C(11, -3)

Coordinate Geometry Exercise 14.3

Since AB ≠ BC ≠ CA, the triangle isosceles.