Question

If the straight lines $3x-5y=7and4x+ay+9=0$are perpendicular to one another , find the value of $a$.

Answer 1

Step1- Finding the slope of equations:

Given lines are,

$$\begin{gathered} 3x-5y=1 \\ \Rightarrow y=\frac{3}{5}x-\frac{1}{5} \end{gathered}$$

and

$$\begin{gathered} 4x+ay+9=0 \\ \Rightarrow y=\frac{-4}{a}x-\frac{9}{a} \end{gathered}$$

On comparing both equations with the general equation of a line $y=mx+c$ we get

$$\begin{gathered} \text{From},y=\frac{3}{5}x-\frac{1}{5} \\ {m}_1=\frac{3}{5} \end{gathered}$$

$$\begin{gathered} \text{From},y=\frac{-4}{a}x-\frac{9}{a} \\ {m}_2=\frac{-4}{a} \end{gathered}$$

Step2- Finding the value of $a$:

Also it is given that the lines are perpendicular to each other

When two lines are perpendicular to each other then , $m_1\times m_2=-1$

$$\begin{gathered} \Rightarrow m_1.m_2=-1 \\ \Rightarrow \left(\frac{3}{5}\right)\left(\frac{-4}{a}\right)=-1 \\ \Rightarrow \frac{12}{5a}=1 \\ \Rightarrow 12=5a \\ \Rightarrow a=\frac{12}{5} \end{gathered}$$

Hence , the value of $a$ is $\frac{12}{5}$.