Question

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

Answer 1

Answer :-

P = (0, -2)

Given :-

The point on y- axis is equidistant from the point A (-3 , 2) and point B (5, -2)

To find :-

The point

Solution :-

Lets say that the point P on y-axis is equidistant from point A(-3,2) and B(5, -2)

So, as it is on the y-axis Since , x -axis is absent So, lets consider

P = ( 0, y )

A =( -3,2)

B = (5, -2)

As they are equidistant So,

PA = PB

• PA = Distance between P and A
• PB = Distance between P and B

By using distance formula

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

So,

PA = PB

$\sqrt{[0-(-3)]^2+(y-2)^2 }= \sqrt{(0-5)^2+[y-(-2)]^2}$

$\sqrt{(0+3)^2+(y-2)^2 }= \sqrt{(0-5)^2+(y+2)^2}$

Squaring on both sides

$(0+3)^2+(y-2)^2 =(0-5)^2+(y+2)^2$

$(3)^2+(y-2)^2 =(-5)^2+(y+2)^2$

$9+(y-2)^2 =25+(y+2)^2$

Transposing R.H.S terms to L.H.S

$9+(y-2)^2 -25-(y+2)^2=0$

$-16 +(y-2)^2-(y+2)^2 =0$

As we know that ,

(a+b)² = a²+2ab +b²

(a-b)² = a²-2ab+b²

$-16 +(y^2-4y+4)-(y^2+4y+4) =0$

$-16 +y^2-4y+4-y^2-4y-4=0$

$-16 \not+y^2-4y\not+4\not -y^2-4y\not-4=0$

$-16-8y=0$

$-8y=16$

$y = -2$

So, y is -2 The required point P (0, y) = (0, -2)

So, the required point is (0, -2)