Question

Find the point on y-axis which is equidistant from the points 5 - 2 and - 32

Answer 1

Answer:

$\textbf{The required point is (0,-2)}$

Step-by-step explanation:

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

$\text{Formula used:}$

$\text{The distance between two points (x_1,y_1) and (x_2,y_2) is}$

$\boxed{\bf{d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}}$

$\text{Let the given points be A(5,-2) and B(-3,2)}$

$\text{Let the required point on y-axis be P(0,y)}$

$\text{Given:PA=PB}$

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

$\implies\:\sqrt{25+(y+2)^2}=\sqrt{9+(y-2)^2}$

squaring on both sides, we get

$\implies\:25+y^2+4+4y=9+y^2+4-4y$

$\implies\:25+4y=9-4y$

$\implies\:8y=-16$

$\implies\:y=-2$

$\therefore\textbf{The required point is (0,-2)}$