Question

The distance between the points $P(x,-1)$ and $Q(3,2)$ is $5$ units. Find the value of $x$.

• A. $2,8$
• B. $-2,9$
• C. $1,8$
• D. $-1,7$

Answer 1

Answer: (D)

$-1,7$

Distance=$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2)}$

P$=(x_1,y_1)=(x,-1)$

Q$=(x_2,y_2)=(3,2)$

$25=(3-x{)}^2+(2+1{)}^2$

$25=(9+x^2-6x+9)$

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

$x^2-7x+x-7=0$

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

$x=-1,7$