Question

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

Answer 1

Let $A(3,a)$ and $B(4,1)$ be the points of a line $AB$

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

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

Squaring both sides, we get

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

$\Rightarrow 10=1+a^2-2a+1=0$

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

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

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

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

$\therefore a=-2$ or $a=4$