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 [NCERT EXAMPLE]

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 $\mathrm{or}y=\frac{3}{5}x+\frac{11}{5}$ 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
$$\begin{gathered} m\times \frac{3}{5}=-1 \\ \Rightarrow m=-\frac{5}{3} \end{gathered}$$
Now, the equation of the required line passing through (− 1, − 1) and having slope $-\frac{5}{3}$ is given by
$$\begin{gathered} y+1=-\frac{5}{3}\left(x+1\right) \\ \Rightarrow 3y+3=-5x-5 \\ \Rightarrow 5x+3y+8=0 \end{gathered}$$