Question

Find value of x for which the distance between the points P(x,4) and Q(9,10) is 10 units.

Answer 1

The given points are `P (x,4) and Q (9,10)

`∴ PQ = sqrt((x-9)^2 +(4-10)^2)`

`=sqrt((x-9)^2 +(-6)^2)`

`=sqrt(x^2 -18x+81+36)`

`=sqrt(x^2-18x+117)`

∵ PQ = 10

`∴ sqrt(x^2 -18x +117)`

`⇒x^2 -18x +117=100`                    (Squaring both sides)

` ⇒ x^2 -18x +17 `

`x^2 -17x -x +17=0`

⇒ ( x - 17 ) ( x - 1) = 0

⇒ ( x - 17 ) (x - 1 ) = 0 

⇒ x- 17 = 0 or x - 1 = 0

⇒ x = 17 or x = 1

Hence, the values of x are 1 and 17.