Question

Find the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0.

Answer 1

Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

$$\begin{gathered} d=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right| \\ \Rightarrow d=\left|\frac{3\times 4-5\times 5+7}{\sqrt{3^2+{\left(-5\right)}^2}}\right|=\frac{6}{\sqrt{34}} \end{gathered}$$

Hence, the required distance is $\frac{6}{\sqrt{34}}$.