Find a point which is equidistant from

Let P (h, k) be the point which is equidistant from the points A (- 5, 4) and B (-1, 6).

$\therefore \quad P A=P B \quad\left[\because\right.$ by distance formula, distance $=\sqrt{\left.\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}$

$\Rightarrow \quad(P A)^{2}=(P B)^{2}$

$\Rightarrow \quad(-5-h)^{2}+(4-k)^{2}=(-1-h)^{2}+(6-k)^{2}$

$\Rightarrow \quad 25+h^{2}+10 h+16+k^{2}-8 k=1+h^{2}+2 h+36+k^{2}-12 k$

$\Rightarrow \quad 25+10 h+16-8 k=1+2 h+36-12 k$

$\Rightarrow \quad 8 h+4 k+41-37=0$

$\Rightarrow \quad 8 h+4 k+4=0$

$\Rightarrow \quad 2 h+k+1=0$ $\ldots$ (i)

Mid-point of $A B=\left(\frac{-5-1}{2}, \frac{4+6}{2}\right)=(-3,5)$

$\left[\because\right.$ mid-point $\left.=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\right]$

At point $(-3,5)$, from Eq. (i), 

$2 h+k=2(-3)+5$

$=-6+5=-1$

$\Rightarrow \quad 2 h+k+1=0$

So, the mid-point of AB satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation 2h + k +1 = 0, are

equidistant from the points A and B.

Replacing h, k by x, y in above equation, we have 2 x +y+1= 0