Given :
P(x 3), Q(7, -1) and PQ= 5 .
To Find :
The possible value of x.
Solution :
We know, distance between two points in coordinate plane is given by :
[tex]PQ = \sqrt{(x-7)^2 + ( 3-(-1))^2}\\\\PQ^2 = (x-7)^2 + ( 3-(-1))^2\\\\(x-7)^2 + 4^2 = 5^2\\\\( x- 7)^2 = 3^2 \\\\x - 7 = \pm 3\\\\x = 10 \ and \ x = 4[/tex]
Therefore, the possible value of x are 10 and 4.
