Question

Find the point on Y-axis which is
equidistant from the points (-5,2) and
(9, -2).​

Answer 1

The point (-0,-7) lies on y-axis which is equidistant from the points  (-5,2) and (9,-2).

Step-by-step explanation:

Let P(0,y) be the point on y-axis which is equidistant from the points A (-5,2) and B (9,-2).

Distance formula:$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AP=PB$

Squaring on both sides

$AP^2=PB^2$

Using the formula

$(0+5)^2+(y-2)^2=(9-0)^2+(-2-y)^2$

$25+y^2-4y+4=81+4+y^2+4y$

Using the identity

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

$y^2-4y+29=y^2+4y+85$

$y^2-4y-y^2-4y=85-29$

$-8y=56$

$x=\frac{56}{-8}=-7$

Hence, the point (-0,-7) lies on y-axis which is equidistant from the points  (-5,2) and (9,-2).

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