Find a point on the y-axis which is equidistant from (2, 2) and (9, 9).

Hint- If I have to find a point on y axis which is equidistant from the given points. Any point on y-axis is marked as a point having coordinate as (0, y), that is the coordinate of x axis is always 0 if we talk about any point on y axis.

Complete step-by-step answer:
Now the given points which are equidistant from the y axis is (2, 2) and (9, 9).
Any point on the y-axis is (0, y).
Using the distance formulae which is $D = \sqrt {{{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{x_2} - {x_1}} \right)}^2}} $………………… (1)
The distance between the points (0, y) and (2, 2) will be
${D_1} = \sqrt {{{\left( {2 - y} \right)}^2} + {{\left( {2 - 0} \right)}^2}} $
On solving we get
$ \Rightarrow {D_1} = \sqrt {{{\left( {2 - y} \right)}^2} + 4} $………………………… (2)
The distance between the points (0, y) and (9, 9) will be
${D_2} = \sqrt {{{\left( {9 - y} \right)}^2} + {{\left( {9 - 0} \right)}^2}} $
On solving we get
$ \Rightarrow {D_2} = \sqrt {{{\left( {9 - y} \right)}^2} + 81} $……………………….. (3)
Now since these point are equidistant thus equation (2) should be equal to equation (3)
 ${D_1} = {D_2}$
On putting the values we get
$\sqrt {{{\left( {2 - y} \right)}^2} + 4} = \sqrt {{{\left( {9 - y} \right)}^2} + 81} $
Squaring both the sides we get
${\left( {2 - y} \right)^2} + 4 = {\left( {9 - y} \right)^2} + 81$
Using the formula of ${(a - b)^2} = {a^2} + {b^2} - 2ab$
$
   \Rightarrow 4 + {y^2} - 4y + 4 = 81 + {y^2} - 18y + 81 \\
   \Rightarrow 8 - 4y = 162 - 18y \\
   \Rightarrow 14y = 154 \\
  y = \dfrac{{154}}{{14}} = 11 \\
$

Thus the point on the y-axis which is equidistant from (2, 2) and (9, 9) is$\left( {0,\;11} \right)$.
Note – Whenever we face such types of problems the key concept is simply to use distance formula. This will help us get the distance between 2 different points and then according to the conditions given in question we can find out the required quantity.