How do you find the perpendicular of a line?

Question

I need to reflect a point on a random line, I know that in order to do that said point and its homologous need to form a perpendicular line to the reflection line. So let's say that the line is L, the point is A and its homologous is A', the line formed between A and A' and line L need to form a right angle when they intersect. How do I go about this?

Answer 1

Let $L = mx+b$ be the line we are reflecting over

Now we want to find the line perpendicular $L'$ to the reflection line, which will have a slope of $-\frac1m$ and will pass through the point $A = (a,b)$. Plugging in the point $A$ into $L'$ we have:

$b = -\frac1m(a) + C$

Then all that is needed is to solve for the intercept of the line, represented by $C$.

Once that is done, if you want to find the coordinates of the new point after the reflection, $A'$, a nice formula mentioned by Harish Chandra Rajpoot can be found here.

Answer 2

Start by finding the slope of the line $L$, say $m_l$.
Then, you can find the perpendicular line $L'$ by point-slope formula, with (A, $\frac{-1}{m_l}$).
Call the intersection point of $L$ and $L'$ by $B$. Then we have $AB$ is a line segment on $L'$.

You can find the answer by finding a point $A'$ on $L'$ satisfying that (x-coordinate of $A$ - x-coordinate of $B$) = (x-coordinate of $B$ - x-coordinate of $A'$)

Answer 3

Unless $\theta$ is a multiple of $\frac\pi2$, we have the identity $\tan\left(\theta+\frac\pi2\right) = -\cot\theta$, therefore if a line has a slope of $m\ne0$, lines perpendicular to it will have a slope of $-\frac1m$. This will work for “most” lines, but doesn’t quite help you with horizontal or vertical lines.

More generally, if you have a line given by the equation $ax+by+c=0$, you probably know that its direction vector is $\langle b,-a\rangle$ (or, if your prefer $\langle -b,a \rangle$). Two nonzero vectors are perpendicular if their dot product is equal to zero, so we can see that $\langle a,b \rangle$ is perpendicular to the line’s direction vector, which means that lines with the equation of the form $bx-ay+d=0$ are perpendicular to the first line.