Question

For what value(s) of $m$ will the distance between the points P(m, 2) and Q(3, -6) be 10 units?

• A. 3
• B. -3
• C. 9
• D. -9

Answer 1

Answer: (B), (C)

The correct options are
B -3
C 9

Given: P(m, 2) and Q(3, -6)
We know that the distance between two points $(x_1,y_1)and(x_2,y_2)$ is given by
$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2.}$

Thus, the distance between
$(m,2)and(3,-6)$ will be
$\sqrt{(3-m{)}^2+(-6-2{)}^2}$

Given:
$\sqrt{(3-m{)}^2+(-6-2{)}^2}=10$
$\Rightarrow \sqrt{(3-m{)}^2+(-8{)}^2}=10$
$\Rightarrow \sqrt{(3-m{)}^2+64}=10$
Squaring on both the sides,
$\Rightarrow (3-m{)}^2+64=100$
$\Rightarrow (3-m{)}^2=36$
Taking square root on both sides equation, we get
$\Rightarrow (3-m)=\pm 6$

If
$(3-m)=+6$
$\Rightarrow -m=6-3=3$
$\Rightarrow m=-3$
If
$(3-m)=-6$
$\Rightarrow -m=-6-3=-9$
$\Rightarrow m=9$
So,
$m$ can be either -3 or 9 and in both these cases, the distance between the points will be 10.