In math lesson, we were solving the problem shown below.

What is the real number a when lines don't have any common point? $$ p: ax+3y-1=0, q: x+2y-4=0 $$

I told my teacher correct answer. When she asked how did I solve it, I responded that I have assumed If lines are parallel, their normal vectors have to be parallel too, therefore, the slope of the vectors must be the same. From this assumption, I have derived the equation below to solve the problem.

$$ {3 \over a}={2 \over 1} $$

She said it is speculative and my solution is wrong. Since it worked and the equation can be derived from $\vec{n}_p = k\vec{n}_q$ which she used on the whiteboard, from my point of view it is fully logical to solve the problem in this way. I have asked her why it is speculative. She practically repeated the same thing without giving any logical counter-argument.

I would like to ask: is my solution correct? If so, is there any way to prove it mathematically and If not, why not?


1 Answer 1


I think it's fine. To be more precise, to have no intersection in $2D$, they are parallel and not identical.

$p$ can be rewritten as $$\frac{a}3x+y=\frac13$$

$q$ can be rewritten as $$\frac12 x+y=2$$

Clearly, we have $\frac13 \ne 2$ and hence we only require $$\frac{a}3=\frac12.$$

which is your condition.


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