Question

Prove that the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6.

Answer 1

Let $d_1$ be the distance between lines 2x + 3y = 19 and 2x + 3y = 6,
while $d_2$ is the distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6

$$\begin{gathered} \therefore d_1=\left|\frac{-19-\left(-6\right)}{\sqrt{2^2+3^2}}\right|\mathrm{and}{d}_2=\left|\frac{7-\left(-6\right)}{\sqrt{2^2+3^2}}\right| \\ \Rightarrow d_1=\left|\frac{-13}{\sqrt{13}}\right|=\sqrt{13}\mathrm{and}{d}_2=\left|\frac{13}{\sqrt{13}}\right|=\sqrt{13} \end{gathered}$$

Hence, the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6