Find the value of a when the distance between the points (3, a)

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The distance between two points $(3, a)$ and $(4,1)$ is given as $\sqrt{10}$. Substituting these values in the formula for distance between two points we have,

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

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

Now, squaring the above equation on both sides of the equals sign

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

 

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

$8=a^{2}-2 a$

Thus we arrive at a quadratic equation. Let us solve this now,

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

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

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

 

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

The roots of the above quadratic equation are thus 4 and $-2$.

 

Thus the value of ' $a$ ' could either be 4 or $-2$.