Rotating a Linear Equation around a point other than the origin

Question

I've been trying to research this question but continually landing on explanations that don't make sense. I would appreciate if you could give me some direction in solving it. It's for a test that's already been submitted, but I'm still curious lol.

A line has the equation $3x - 5y + 40 = 0$. The line is rotated $45$ degrees counterclockwise around the point $(20,20)$ to obtain a second line. What is the x intercept of the second line?

Answer 1

The blue line is the graph of $3x-5y+40=0$. We note first that $(20,20)$ lies on this graph, so we know that point will also go through the rotated red line as well. The other major challenge will be determining the slope of the red line.

Manipulating the blue equation into slope-intercept form, we get $y=\frac35x+8$, so the slope of the blue line is $\frac35$. Therefore, the angle $\theta$ that the blue line makes with the x-axis satisfies $\tan\theta=\frac35$. Now we want to want to rotate the line by a $45^\circ$ rotation $\rho$, which satisfies $\tan\rho=1$. Using the tangent sum formula: $$\tan(\theta+\rho)=\frac{\tan\theta+\tan\rho}{1-\tan\theta\tan\rho}=\frac{\frac35+1}{1-\frac35}=\frac{8/5}{2/5}=4$$ Therefore, the slope of the red line is $4$. Remembering that it goes through $(20,20)$, its equation in point-slope form is $y-20=4(x-20)$. However, we are asked to find its x-intercept. This is easily found by setting $y=0$ into its equation: $$0-20=4x-80\\60=4x\\x=15$$

Answer 2

Here is another (more tedious but using elementary formulas) way for this particular problem.

Let L : 3x – 5y + 40 = 0 and P = (20, 20).

L cut the x-axis at R whose co-ordinates can then be found. Therefore, PR is known.

Let N be the normal to L at P. Then, equation of N can be found. If N cuts the x-axis at S, using the same method, PS is known.

Let the target line cut the x-axis at T. Since PT bisects $\angle RPS$, T divides RS in the ratio $\dfrac {PR}{PS}$ by the angle bisector theorem. This means the co-ordinates of T can be found by section formula.

Finally, the equation of PT can be obtained by using two-point form.

This works only because the rotation is $45^0$.