Question:
If the distance between the points (4, p) and (1, 0) is 5, then the value of pis
(a) 4 only
(b) ±4
(c) – 4 only
(d) 0
Solution:
(b) According to the question, the distance between the points (4, p) and (1, 0) = 5
i.e., $\quad \sqrt{(1-4)^{2}+(0-p)^{2}}=5$
$\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$
$\Rightarrow \quad \sqrt{(-3)^{2}+p^{2}}=5$
$\Rightarrow \quad \sqrt{9+p^{2}}=5$
On squaring both the sides, we get
$9+p^{2}=25$
$\Rightarrow$ $p^{2}=16 \Rightarrow p=\pm 4$
Hence, the required value of p is ± 4,