Question

Find the equation of the straight line passing through the point of intersection of the lines 5x − 6y − 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x − 5y + 11 = 0 .

Answer 1

The point of intersection of lines 5x − 6y − 1 = 0 and 3x + 2y + 5 = 0 is given by (− 1, − 1)
Now, the slope of the line 3x − 5y + 11 = 0 \[\text { or } y = \frac{3}{5}x + \frac{11}{5}\] is  \[\frac{3}{5}\]

Now, we know that the product of the slopes of two perpendicular lines is − 1.
Let the slope of the required line be m

\[m \times \frac{3}{5} = - 1\]

\[ \Rightarrow m = - \frac{5}{3}\]

Now, the equation of the required line passing through (− 1, − 1) and having slope \[- \frac{5}{3}\] is given by

\[y + 1 = - \frac{5}{3}\left( x + 1 \right)\]

\[ \Rightarrow 3y + 3 = - 5x - 5\]

\[ \Rightarrow 5x + 3y + 8 = 0\]