Question

Find the value of x, if the distance between the points (x, –1) and (3, 2) is 5. [2 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark

Let p (x , -1) and Q (3 ,2) be the given points. Then,

PQ = 5 (Given)
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\Rightarrow \sqrt{{(x-3)}^2+{(-1-2)}^2}=5$

$\Rightarrow {(x-3)}^2+9=5^2$ [Squaring both sides]

$\Rightarrow x^2-6x+18=25$

$\Rightarrow x^2-6x-7=0$

$\Rightarrow (x-7)(x+1)=0$

$\Rightarrow x=7orx=-1$