Question

Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.

Answer 1

Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0 

Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

∴ \(p = \begin{vmatrix}\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\end{vmatrix} = \) \(\begin{vmatrix}\frac{12(-2)-5(3)-13}{\sqrt{12^2+(-5)^2}}\end{vmatrix} \)

= \(\begin{vmatrix}\frac{-24-15-13}{\sqrt{144+25}}\end{vmatrix} \)=\(\begin{vmatrix}\frac{-52}{\sqrt{13}}\end{vmatrix} \)= 4 units

Answer 2

Here's the required solution