**Coordinate Geometry Exercise 14.3 – Questions:**

**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)**

**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)**

**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)**

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

**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:**

**Find the distance between the following pairs of points**

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

** (vii) (a, b) and (c, b) **

Solution:

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

Solution:

**Find the distance between the origin and the point**

**(i) (-6, 8)**

Solution:

Distance between the origin and (x, y) = √(x^{2} + y^{2})

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) = √(x^{2} + y^{2})

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) = √(x^{2} + y^{2})

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,

5^{2} = x^{2} – 2x + 10

x^{2} – 2x + 10 – 25 = 0

x^{2} – 2x – 15 = 0

x^{2} + 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)

a^{2} – 4a + 40 = a^{2} – 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)

y^{2} – 4y + 29 = y^{2} – 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)

**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)

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)

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

**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)

Since AB = CA , the triangle is isosceles.

**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)

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

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

Solution:

A quadrilateral is a parallelogram if it’s opposite sides is equal.

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

Therefore, AB = CD and BC = DA

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

Solution:

A(4, 0), B(-2, -3), C(3, 2), D(-3, -1)

Therefore, AB = CD and BC = DA

**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)

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)

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)

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)

Since AB ≠ BC ≠ CA, the triangle isosceles.