Q52. - Algebra 2

Question

If the graph of the equation $\mathbf{ax}+\mathbf{3}\mathbf{y}=\mathbf{9}$ is perpendicular to the graph of the equation $\mathbf{3}\mathbf{x}+\mathbf{y}=-\mathbf{4}$, find the value of $\mathbf{a}$.

Step-by-Step Solution

Answer

The required value of $a$ is -1.

Step 1 – State the concept

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 $-1$.

Step 2 – List the given data

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

Step 3 – Calculate the slopes

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 $-3$.

Step 4 – Compute the required value

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.