#### Find a point which is equidistant from the points A (–5, 4) and B (–1, 6)? How many such points are there?

Solution.   Let P(x, y) is a point which is equidistant from point A(–5, 4) and B(–1, 6) i.e.   PA = PB
Squaring both sides we get
$PA^{2}= PB^{2}$
$\left ( -5-x \right )^{2}+\left ( 4-y \right )^{2}= \left ( -1-x \right )^{2}+\left ( 6-y \right )^{2}$
25 + x2 + 10x + 16 + y2 – 8y = 1 + x2 + 2x + 36 + y2 – 12y
{Using : (a + b)2 = a2 + b2 + 2ab; (a – b)2 = a2 + b2 – 2ab}
25 + 10x + 16 – 8y = 1 + 2x + 36 – 12y
10x – 8y + 41 – 2x + 12y – 37 = 0
8x + 4y + 4 = 0
Dividing by 4 se get
2x + y + 1 = 0 ……..(1)
$Mid-point\, of\, AB= \left ( \frac{-5-1}{2},\frac{4+6}{2} \right )$
= (-3,5)
$\because Mid-point= \left \{ \left ( \frac{x_{1}+x_{2}}{2} \right ) ,\left ( \frac{y_{1}+y_{2}}{2} \right )\right \}$

Put point (–3, 5) in eqn. (1)
2(–3) + 5 + 1 = 0
– 6 + 6 = 0
0 = 0
Mid-point of AB satisfy equation (1)
Hence, infinite numbers of points are there.