Find the value of k, if the point P (0, 2)

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

It is said that P(0,2) is equidistant from both A(3,k) and B(k,5).

So, using the distance formula for both these pairs of points we have

$A P=\sqrt{(3)^{2}+(k-2)^{2}}$

$B P=\sqrt{(k)^{2}+(3)^{2}}$

Now since both these distances are given to be the same, let us equate both.

$A P=B P$

$\sqrt{(3)^{2}+(k-2)^{2}}=\sqrt{(k)^{2}+(3)^{2}}$

Squaring on both sides we have,

$(3)^{2}+(k-2)^{2}=(k)^{2}+(3)^{2}$

$9+k^{2}+4-4 k=k^{2}+9$

$4 k=4$

$k=1$

Hence the value of ‘k’ for which the point ‘P’ is equidistant from the other two given points is.