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

Explanation

Let P (h,k) be the point which is equidistant from the points A(-5,4) and B(_1,6).
`:.` PA=PB [`:'` by distance formula , distance `=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`
`rArr(PA)^(2)=(PB)^(2)`
`rArr(-5=h)^(2)+(4-k)^(2)=(-1-h)^(2)+(6-k)^(2)`
`rArr25+h^(2)+10+16-8k=1+h^(2)+2h+36+k^(2)-12k`
`rArr25+10h+16=8k=1+2h+36-12k`
`rArr 8h+4k+41-37=0`
`rArr 8h+4k+4=0`
`rArr 2h +k+1=0` ... (i)
Mid- point of AB `=((-5-1)/(2),(4+6)/(2))=(-3,5)`
[`:'` mid point `=((x_(1)+x_(2))/(2),(y_(1)+y_(1))/(2))`
At point (-3,5), from Eq . (i) ,
`2h+k=2(-3)+5`
`=-6+5=-1`
`rArr2h+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 `2x+y+1=0`