Question:
Find the value of a, if the distance between the points A(- 3, – 14) and B (a, – 5) is 9 units.
Solution:
According to the question,
Distance between A (- 3, -14) and 8 (a, – 5), AB = 9
$\left[\because\right.$ distance between two points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}}\right]$
$\Rightarrow \quad \sqrt{(a+3)^{2}+(-5+14)^{2}}=9$
$\Rightarrow \quad \sqrt{(a+3)^{2}+(9)^{2}}=9$
On squaring both the sides, we get
$(a+3)^{2}+81=81$
$\Rightarrow \quad(a+3)^{2}=0 \Rightarrow a=-3$
Hence, the required value of a is – 3.