Question

Find a point on the y-axis that is equidistant from the points (8, −8) and (1, 1).

Answer 1

$\text{We know that any point on the y-axis is of the form (0, y),}\\ \text{because on y-axis we have x=0.}\\ \\ \text{So let (0, y) be the point on the y-axis that is equidistance from}\\ \text{the points }(8,-8) \text{ and }(1,1)\\ \\ \text{so using the distance formula, we have}\\ \\ \text{Distance between (0, y) and }(8,-8)=\text{Distance between (0,y) and }(1,1)\\ \\ \Rightarrow \sqrt{(0-8)^2+(y-(-8))^2}=\sqrt{(0-1)^2+(y-1)^2}$

$\Rightarrow \sqrt{64+(y+8)^2}=\sqrt{1+(y-1)^2}\\ \\ \text{Squaring both sides, we get}\\ \\ (\sqrt{64+(y+8)^2})^2=(\sqrt{1+(y-1)^2})^2\\ \\ \Rightarrow 64+(y+8)^2=1+(y-1)^2\\ \\ \Rightarrow 64+(y^2+16y+64)=1+(y^2-2y+1)\\ \\ \Rightarrow y^2+16y+128=y^2-2y+2\\ \\ \Rightarrow y^2+16y-y^2+2y=2-128\\ \\ \Rightarrow 18y=-126\\ \\ \Rightarrow y=\frac{-126}{18}\\ \\ \Rightarrow y=-7$

So the point on the y-axis that is equidistant from the points (8,-8) and (1,1) is: (0,-7)