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) = √(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,
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)
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)
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)
- 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.