Question:
If the distance between the points (4, p) and (1, 0) is 5, then p =
(a) ± 4
(b) 4
(c) −4
(d) 0
Solution:
It is given that distance between $\mathrm{P}(4, p)$ and $\mathrm{Q}(1,0)$ is 5 .
In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,
$\mathrm{AB}^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$
So,
$5^{2}=(4-1)^{2}+(p-0)^{2}$
On further simplification,
$p^{2}=16$
$p=\pm 4$
So,
$p=\pm 4$
So the answer is (a)