Question

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

• A. (-1, -3)
• B. (2, -3)
• C. (2, 1)
• D. (0, -2)

Answer 1

The correct option is D (0, -2)

Let the given points be A(3, 2) and B(-5, -2) and let the point P on the y-axis, which is equidistant from A and B, be (0, y). Then,
$AP=\sqrt{(0-3{)}^2+(y-2{)}^2}=\sqrt{9+(y-2{)}^2}$
and $BP=\sqrt{(0+5{)}^2+(y+2{)}^2}=\sqrt{25+(y+2{)}^2}$
Since, AP $=$ BP, we have
$9+(y-2{)}^2=25+(y+2{)}^2$
$\Rightarrow 9+y^2-4y+4=25+y^2+4y+4$
$\Rightarrow 8y=-16\Rightarrow y=-2$
Hence the required point is (0, -2).