Find the value of $a$ when the distance between the points $(3, a)$ and $(4, 1)$ is $\sqrt{10}$.

Given:

The distance between the points $(3, a)$ and $(4, 1)$ is $\sqrt{10}$.

To do:

We have to find the value of $a$.

Solution:

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

 The distance between the points $(3, a)$ and $(4, 1)$ is,

$\sqrt{10}=\sqrt{(4-3)^2+(1-a)^2}$

Squaring on both sides, we get,

$(\sqrt{10})^2=(\sqrt{(1)^2+(1-a)^2})^2$

$10=1+1+a^2-2a$

$a^2-2a+2-10=0$

$a^2-2a-8=0$

$a^2-4a+2a-8=0$

$a(a-4)+2(a-4)=0$

$(a-4)(a+2)=0$

$a-4=0$ or $a+2=0$

$a=4$ or $a=-2$

The value of $a$ is $4$ or $-2$.