Question

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

Answer 1

Let P(h,k) be the point which is equidistant from the points A(-5,4) and B(-1,6).
$\therefore PA=PB\Rightarrow (PA{)}^2=(PB{)}^2\Rightarrow (-5-h{)}^2+(4-k{)}^2=(-1-h{)}^2+(6-k{)}^2\text{by distance formula, distance}=\sqrt{(x_2-x_1{)}^2+(x_2-y_1{)}^2}\Rightarrow 25+h^2+10h+16+k^2-8k=1+h^2+2h+36+k^2-12k\Rightarrow 25+10h+16-8k=1+2h+36-12k\Rightarrow 8h+4k+41-37=0\Rightarrow 8h+4k+4=0\Rightarrow 2h+k+1=0\text{Mid-point of AB}=(\frac{-5-1}{2},\frac{4+6}{2})=(-3,5)\text{At point (-3,5)}2h+k=2(-3)+5=-6+5=-1\Rightarrow 2h+k+1=0$

So, the mid-point of AB satisfy the equation. 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 $2x+y+1=0$.