How to find the fixed number of two vertical lines to each other?

Question

I would like to ask for an explanation of this exercise, please.

When two straight lines $3x - (a - 3)y - 6 = 0$ and $(a + 1)x + y - 1 = 0$ are vertical to each other, then the fixed number $a$ is?

I've been struggling to solve this.

Thank you for your time.

Answer 1

First we need to rewrite these in the form $y=mx+c$

\begin{align}3x-(a-3)y-6&=0\\ (a-3)y&=3x-6 \\ y &= \frac{3x-6}{a-3} \\ y &= \frac{3}{a-3}x - \frac{6}{a-3} \end{align}

and

\begin{align} (a+1)x+y-1&= 0 \\ y &= -(a+1)x + 1 \end{align}

These lines are parallel when their gradient is equal (the $m$ in each of the equations), therefore

\begin{align} \frac{3}{a-3} &= -(a+1) \\ 3&=-(a+1)(a-3) \\ 3 &= -(a^2+a-3a-3) \\ 3 &= -a^2 +2a+3 \\ a^2-2a&=0 \\ a(a-2) &= 0 \end{align}

We can therefore say that $a=0$ or $a=2$

Answer 2

When two lines $y=m_1x+c$ and $y=m_2x+c'$ are perpendicular to each other, $$m_1m_2=-1$$

Now, rearrange the equations in the form above:

$$y = \frac{3}{a-3}x - \frac{6}{a-3}$$ and $$y = -(a+1)x + 1$$

Now, use the product-of-slopes condition.