Question

Find the distance of the point (1, 2) from the straight line with slope 5 and passing through the point of intersection of x + 2y = 5 and x − 3y = 7.

Answer 1

To find the point intersection of the lines x + 2y = 5 and x − 3y = 7, let us solve them.

\[\frac{x}{- 14 - 15} = \frac{y}{- 5 + 7} = \frac{1}{- 3 - 2}\]

\[ \Rightarrow x = \frac{29}{5}, y = - \frac{2}{5}\]

So, the equation of the line passing through

\[\left( \frac{29}{5}, - \frac{2}{5} \right)\] with slope 5 is

\[y + \frac{2}{5} = 5\left( x - \frac{29}{5} \right)\]

\[ \Rightarrow 5y + 2 = 25x - 145\]

\[ \Rightarrow 25x - 5y - 147 = 0\]

Let d be the perpendicular distance from the point (1, 2) to the line

\[25x - 5y - 147 = 0\]

\[\therefore d = \left| \frac{25 - 10 - 147}{\sqrt{{25}^2 + 5^2}} \right| = \frac{132}{5\sqrt{26}}\]

Hence, the required perpendicular distance is \[\frac{132}{5\sqrt{26}}\]