Question

Find the point that equidistant from (3, 8) and (-2,5) on the y-axis

Answer 1

Given the points : ( 3 , 8 ) and ( -2 , 5 )

We need to find the coordinates of the point equidistant from the given point and should be on the y - axis

as the point will be on y - axis : the x - coordinates of the point will be 0

So, let the coordinates of the point is ( 0 , a )

The distance between any two points is given by the formula :

`d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`

So, the distance between ( 3 , 8 ) and ( 0 , a ) will be :

`d_1=\sqrt[]{(0-3)^2+(a-8)^2}=\sqrt[]{9+(a-8)^2}`

And the distance between ( -2 , 5 ) and ( 0 , a ) will be :

`d_2=\sqrt[]{(0+2)^2+(a-5)^2}=\sqrt[]{4+(a-5)^2}`

the two distances are equal

`\begin{gathered} d_1=d_2 \\ \sqrt[]{9+(a-8)^2}=\sqrt[]{4+(a-5)^2} \end{gathered}`

squaring both sides then solve for a:

`\begin{gathered} 9+(a-8)^2=4+(a-5)^2 \\ 9+a^2-16a+64=4+a^2-10a+25 \\ -16a+73=-10a+29 \\ -16a+10a=29-73 \\ -6a=-44 \\ \\ a=(-44)/(-6)=(22)/(3)=7(1)/(3) \end{gathered}`

so, the answer is the coordinates of the point is ( 0 , 7 1/3 )