# If the distance between the points $(4, p)$ and $(1,0)$ is 5 , then the value of $p$ is(A) 4 only(B) $\pm 4$(C) $-4$ only(D) 0

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Given:

The distance between the points $(4,\ p)$ and $(1,\ 0)$ is $5$.

To do:

We have to find the value of $p$.

Solution:

As given, $x_1=4,\ y_1=p,\ x_2=1,\ y_2=0$

Using the distance formula,

$5=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$

$\Rightarrow 5=\sqrt{ (1-4)^2+( 0-p)^2}$

$\Rightarrow 5^2=( -3)^2+( -p)^2$

$\Rightarrow 25=9+p^2$

$\Rightarrow p^2=25-9$

$\Rightarrow p^2=16$

$\Rightarrow p=\sqrt{16}$

$\Rightarrow p=\pm4$

Therefore, the value of $p$ is $\pm4$.

Updated on 10-Oct-2022 13:28:28