# If the distance between the points (4, p) and (1, 0) is 5, then p =

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)

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