Question

Please answer this question​

Answer 1

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

Let A(5,−2) and B(−3,2). Then,

PA=PB

$PA^2=PB^2$

$\small \bold {(5−0)^2+(−2−y)^2=(−3−0)^2+(2−y)^2}$

$\bold{25+4+y^2+4y=9+4+y^2−4y}$

8y=−16

y=−2

Therefore, the required point is (0,−2).

Answer 2

Answer:

P : (0, -2)

Step-by-step explanation:

Let the required point P be (0,y).

Given points are A(-3,2) & B(5,-2).

AP = $\sqrt{(0-(-3))^{2} + (y-2)^{2} }$= $\sqrt{9 + y^{2} + 4 - 4y} = \sqrt{y^{2} - 4y + 13}$

BP = $\sqrt{(0-5)^{2} + (y-(-2))^{2} } = \sqrt{25 + y^{2} + 4 + 4y} = \sqrt{ y^{2} + 4y + 29}$

Given that, AP = BP

=> $\sqrt{y^{2} - 4y + 13}$ = $\sqrt{ y^{2} + 4y + 29}$

=> 13 - 4y = 29 + 4y

=> 8y = 13 - 29 = -16

=> y = -2

P : (0, -2)