Question

Find a point on Y-axis which is equidistant from P(-6,4) and Q(2,-8).

Answer 1

Hint – In order to solve this problem assume a point on the y-axis then find the distance from both the points with the help of distance formula and equate. This will give you the right answer.

Complete step-by-step answer:
Let the point on the y-axis be (0,y).
The distance formula to find the distance between the point $({x_1},{y_1}),({x_2},{y_2})$ is $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} $
Then the distance of P form the point (0,y) is
$
  {d_1} = \sqrt {{{(0 - ( - 6))}^2} + {{(y - 4)}^2}} \\
  {d_1} = \sqrt {36 + {y^2} + 16 - 8y} = \sqrt {{y^2} + 52 - 8y} \\
$
And the distance of the point Q from (0,y) is
$
  {d_2} = \sqrt {{{(0 - (2))}^2} + {{(y - ( - 8))}^2}} \\
  {d_2} = \sqrt {4 + {y^2} + 64 + 16y} = \sqrt {{y^2} + 68 + 16y} \\
$
As the points are equidistant so we can equate ${d_1} = {d_2} = d$
So we do,
$\sqrt {{y^2} + 68 + 16y} $= $\sqrt {{y^2} + 52 - 8y} $
On squaring both sides we get,
${y^2} + 68 + 16y$= ${y^2} + 52 - 8y$
$
  68 + 16y = 52 - 8y \\
  16 = - 24y \\
  y = - \dfrac{{16}}{{24}} = - \dfrac{2}{3} \\
$
Hence the point on the y-axis is $\left( {0, - \dfrac{2}{3}} \right)$.

Note – In these types of problems you first have to assume the point knowing that if it's on the y-axis its x coordinate will be zero and if it's on the x-axis its y coordinate is zero. Then we can find the distance with the help of distance formula form the given points then apply the condition provided in the problem to get the answer of the question.