Question

Find a point on Y-axis which is equidistant from P(-6,4) and Q(2,-8).

Answer 1

Given that the point lies on y-axis, Therefore, its x-coordinate will be 0.

Let the point A(0,y) be equidistant from points.

Given points are P(-6,4) and Q(2,-8).

Given that there distances are equal.

= > PA = QA

$= > \sqrt{(0 + 6)^2 + (y - 4)^2} = \sqrt{(0 - 2)^2 + (y + 8)^2}$

$= > \sqrt{36 + y^2 + 16 - 8y} = \sqrt{4 + y^2 + 64 + 16y}$

$= > \sqrt{y^2 - 8y + 52} = \sqrt{y^2 + 16y + 68}$

On squaring both sides, we get

= > y^2 - 8y + 52 = y^2 + 16y + 68

= > -24y = 16

= > y = -2/3.

Therefore, the required point is (0, -2/3).

Hope this helps!

Answer 2

HELLO DEAR,

given vertices are p(-6 , 4) and Q(2 , -8).

let point on y-axis be B(0 , y).

BP² = QB²

(0 + 6)² + (y - 4)² = (0 - 2)² + (y + 8)²

36 + y² + 16 - 8y = 4 + y² + 64 + 16y

52 + y² - 8y = y² + 68 + 16y

-24y = 16

y = -2/3

therefore, point on y-axis is (0 , -2/3)

I HOPE ITS HELP YOU DEAR,
THANKS