Algebraically find the value of $a$.

Question

The graphs of $2y + x + 3 = 0\,$ and $3y + ax + 2 = 0\,$ meet at right angles. Algebraically find the value of $a$.

My instinct was making the two equations equal each other and solve like so:

1. $2y + x + 3 \, = 3y + ax + 2\,$
2. $-y + x + 1 \, = ax\,$
3. $\displaystyle \frac{-y + x + 1}{x} = \frac{ax}{x}\,$
4. $\displaystyle \frac{-y + x + 1}{x} \, = a\,$

I noticed my problem mentions the equations meeting at right angles. Would I need to change my equation for the correct answer; do I add the two equations and set them equal to 90?

Answer 1

The first line is

$$y=-\frac12 x -\frac32$$

The slope is $-\frac12$. To be perpendicular, the slope of the second line is $2$.

Using that, you should be able to recover $a$.

Answer 2

The slopes of perpendicular lines have product $-1$. Compute the slope of each line. The slope of the second will be an expression involving $a$. Multiply them and set the product to $-1$, then solve for $a$

Answer 3

For a line $ax+by+c=0$, the vector $(a,b)$ is orthogonal to it.

The two lines are orthogonal if

$$(1,2)\cdot(a,3)=0$$ or $$a+6=0.$$