91Ó°ÊÓ

The slope-intercept form of an equation of a straight line is $y=mx+c$, where m is the slope and c is the y-intercept.

If two lines are perpendicular, then the product of their slopes is $-1$.

The given equations are $ax+3y=9$ and $3x+y=-4$ such that they are perpendicular.

Convert the equation $ax+3y=9$ to slope-intercept form as follows:

$ax+3y=9$ (Given equation)

$ax+3y-ax=9-ax$ (Subtract $ax$ from both sides)

$3y=9-ax$ (Simplify)

$\frac{3y}{3}=\frac{9-ax}{3}$ (Divide both sides by 3)

$y=3-\frac{a}{3}x$ (Simplify)

$y=-\frac{a}{3}x+3$ (Rearrange)

Comparing with $y=mx+c$, $m=-\frac{a}{3}$. So, slope of $ax+3y=9$ is $-\frac{a}{3}$.

Similarly, convert the equation $3x+y=-4$ to slope-intercept form as follows:

$3x+y=-4$ (Given equation)

$3x+y-3x=-4-3x$ (Subtract $3x$ from both sides)

$y=-4-3x$ (Simplify)

$y=-3x-4$ (Rearrange)

Comparing with $y=mx+c$, $m=-3$. So, slope of $3x+y=-4$ is $-3$.

Since $ax+3y=9$and $3x+y=-4$are perpendicular, the product of their slopes must be $-1$.

Then, by the problem,

$\begin{array}{l}\left(−\frac{a}{3}\right)\left(−3\right)=−1\\ \left(\frac{a}{3}\right)\left(3\right)=−1\\ a=−1\end{array}$

So, $a=-1$.

This is the required solution.