If the distance between the points

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,

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