How to find coordinates of reflected point?

The formula for finding the foot of the perpendicular from a point $(x_1,y_1)$ to the line $ax+by+c=0$ is given by: $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}$$

For finding the image of the point in the same line, we just multiply the rightmost term by 2.

So, the image of the point $(x_1,y_1)$ in the line $ax_1+by_1+c=0$ is given by: $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}$$

The image of the point is at the same distance from the line as the point itself is from the line. So, we have to multiply it by 2. That's what I think.

Here's the proof from my book:

Please excuse the image size.. :P

Find line perpendicular to $ 2y=x+1$ which passes through $(3, -3)$

$$2x+y=3$$

Find intersection point of those two lines ( Foot of perpendicular )

$$(1,1)$$

Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)

$$(-1,5)$$

In general, the reflection of the point $(p, q)$ in the line $ay + bx + c = 0$ is

$$\left(\frac{p(a^{2}-b^{2})-2b(aq+c)}{a^{2}+b^{2}},\frac{q(b^{2}-a^{2})-2a(bp+c)}{a^{2}+b^{2}}\right)$$

Proof here (no complex numbers needed!).

$$\\$$

So in this case, your point is $(p,q)=(3,-3)$ and your line is $2y-x-1=0$ (rewritten in the form $ay+bx+c=0$). So just plug in the numbers to get $(-1,5)$ as desired.

Or you can use the spreadsheet titled "Reflection of Point in Line" here.

(I was looking for a general solution but couldn't seem to find any online, so I thought I might as well share my work here.)