Question

Find the distance of the line \(4x + 7y + 5 = 0\)  from the point (1, 2) along the line  \(2x - y = 0\)

Answer 1

The given lines are 
\(2x - y = 0 … (1) \)
\(4x + 7y + 5 = 0 … (2) \)
A (1, 2) is a point on line (1). 
Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain 
\(x =\frac{ -5}{18}\) and \(y = – \frac{5}{9}\)

∴Coordinates of point B are \(\left(\frac{-5}{18}, \frac{-5}{9}\right).\)

By using distance formula, the distance between points A and B can be obtained as
\(AB=\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2} \)units.

\(=\sqrt{\left(\frac{23}{18}\right)^2+\left(\frac{23}{9}\right)^2}\) uniits

\(= \sqrt{\left(\frac{23}{2\times9}\right)^2+\left(\frac{23}{9}\right)^2}\) units

\(= \sqrt{\left(\frac{23}{9}\right)^2\left(\frac{1}{2}\right)^2+\left(\frac{23}{9}\right)^2}\) units

\(= \sqrt{\left(\frac{23}{9}\right)^2+\left(\frac{1}{4}+1\right)}\) units

\(= \frac{23}{9}\sqrt{\frac{5}{4}}\) units

\(= \frac{23}{9} \times\frac{\sqrt5}{2} \)units

\(=\frac{23\sqrt{5}}{18}\) units

Thus, the required distance is \(\frac{23\sqrt{5}}{18}\) units.