Question

If the straight lines $3x-5y=7$ and $4x+ay+9=0$are perpendicular to one another, find the value of a.

Answer 1

Step 1: Finding the slope of the line $3x-5y=7$:
The equation of the line is given as:

$3x-5y=7$

$\Rightarrow 5y=3x-7$

$\Rightarrow y=\frac{3}{5}x-\frac{7}{5}$

Comparing with $y=mx+c$

$m_1=\frac{3}{5}$ (where $m_1$ is the slope of the line)

Step 2: Finding the slope of the line $4x+ay+9=0$:
The equation of the line is given as:

$4x+ay+9=0$

$\Rightarrow ay=-4x-9$

$\Rightarrow y=\frac{-4}{a}x-\frac{9}{a}$

Comparing with $y=mx+c$

$m_2=\frac{-4}{a}$ (where $m_2$ is the slope of the line)

Step 3: Finding the value of ‘a’ using the condition for perpendicular lines:

We know that: $m_1=\frac{3}{5}$and $m_2=\frac{-4}{a}$

Also, the two given lines are perpendicular to each other.

$\Rightarrow m_1.m_2=-1$

Substituting known values in the above expression, we get;

$\frac{3}{5}\times \frac{-4}{a}=-1$

$\Rightarrow -12=-5a$

$\Rightarrow a=\frac{12}{5}$

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