Q7 of 169 Page 644

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

Let the point on the y-axis be P(0, y)

Given: P is equidistant from A(-4, 3) and B(5, 2).

i.e., PA = PB

Squaring both sides, we get

⇒ (-4 – 0)² + (3 – y)² = (5 – 0)² + (2 – y)²

⇒ 16 + 9 – 6y + y² = 25 + 4 – 4y + y²

⇒ 25 – 6y = 29 – 4y

⇒ 2y = -4

⇒ y = -2

Therefore, the required point on the y-axis is (0, -2).

Find the distance between the points A(x₁, y₁) and B(x₂, y₂), when

(i) AB is parallel to the x-axis

(ii) AB is parallel to the y-axis.

A is a point on the x-axis with abscissa -8 and B is a point on the y-axis with ordinate 15. Find the distance AB.

Using the distance formula, show that the points A(3, -2), B(5, 2) and C(8,8) are collinear.

Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right-angled triangle.