Question

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

Answer 1

Let the point be P(0,y)

Since the point is on y-axis, the x coordinate will be 0 .

$\boxed{ \bf{distance \: = \sqrt{(x {\tiny2 } - x {\tiny1}) {}^{2} + (y {\tiny2} - y {\tiny1})}{}^{2} }}$

$\boxed{ \bf{Refer \: to \: \: attachment }}$

Since P is equidistant ;

AP = BP

Hence

AP^2 = BP^2

y=-2

Hence, point P(0,-2)

$\boxed{\textbf{Identity used}}$

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

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

Answer 2

☆Since the point is on y-axis it's coordinates are (0,y).

☆ If P(0,y) be the point equidistant from A(5,-2) and B(-3,2),

→ AP=BP Given

→ AP^2=BP^2

(squaring as to remove roots.)

》(0-5)2+(y+2)2 = (-3-0)2+(2-y)2

》25+y2+4y+4 = 9+4-4y+y2

》16= -8y

》So,y= -2.

$BE BRAINLY$