Find all possible values of x for which the distance between the points
A (x, -1) and B (5, 3) is 5 units.
We know that the distance between two points (x1,y1) and (x2,y2) ,
d=√(x2−x1)2+(y2−y1)2
Given d = 5 units and points are A (x, -1) and B (5, 3)
∴√(5−x)2+(3−−1)2=5
(5−x)2+(3+1)2=25
(5−x)2+16=25
(5−x)2=25−16
(5−x)2=9
(5−x)=√9
(5−x)=±3
x=5±3
x=5+3 or 5−3
x=8 or 2