How to do this line rotation (matrices) question? Question: The line $y=10-2x$ is rotated anticlockwise about the origin such that its image has an $x$-intercept of $(p,0)$ and $y$-intercept of $(0,q)$. Determine the angle of rotation θ $(0 ≤ θ ≤ 90)$ such that $p=q$.

I know that i probably need to do this and possibly letting $x' = p$ and $y'=0$. But i dont know its original point $(x,y)$ so what should i do?
I can find the x/y  intercepts as well as the gradient of the line but does that help?
 A: The points $(5,0)$ and $(0,10)$ are the intercepts of the given line $y=10-2x$. 
Let's find their images after being rotated by $t$ degrees counterclockwise around the origin:
$$\begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}\begin{pmatrix}5\\ 0\end{pmatrix}=\begin{pmatrix}5\cos t\\ 5\sin t\end{pmatrix}=\begin{pmatrix}x_1\\ y_1\end{pmatrix}; \\
\begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}\begin{pmatrix}0\\ 10\end{pmatrix}=\begin{pmatrix}-10\sin t\\ 10\cos t\end{pmatrix}=\begin{pmatrix}x_2\\ y_2\end{pmatrix}$$
The slope of the line passing through these two points must be $-1$ (Why? If not sure, see appendix.):
$$\frac{y_2-y_1}{x_2-x_1}=-1 \Rightarrow \frac{10\cos t-5\sin t}{-10\sin t-5\cos t}=-1\Rightarrow \tan t=\frac13 \Rightarrow t=\arctan \frac13=18.43^\circ. $$
Appendix:

Because its intercepts are $(p,0)$ and $(0,p)$.

A: Hint
The line of cartesian equation $2x+y-10 = 0$ is having for direction the vector $u = (-1,2)$ and passes through the point $P=(5,0)$. Do you see that?
The rotated line has for direction the rotated vector and passes through the rotated point.
Compute those to find the equation of the rotated line based on the angle $\theta$.
Then find $\theta$ based on the condition $p=q$.
A: The image $(x',y')$ by this rotation of the generic point $(x,10-x)$ of the first line is : $$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta)\end{pmatrix}\begin{pmatrix}x\\10-2x\end{pmatrix}$$
This image must belong to a line with this equation $x'+y'=p$ (for a certain constant $p$), whatever $x$.
Condition $x'+y'=p$ gives an expression $Ax+B=p$, which must be independent of $x$ ; thus we must have $A=0$.
This gives constraint $-\cos(\theta)+3\sin(\theta)=0$
From there, we deduce $\theta=$atan$(1/3)=0.3218 rad\approx 18.5 $ degrees.
A: You can do some geometry: if the line is rotated by an angle $\theta$, its normal vector is rotated likewise. Now, let $\varphi$ the polar angle of the given line – with cartesian equation $\;2x+y=10$. A normal vector is $(2,1), so $\;\varphi=\arctan\frac12$.
On the other hand, the polar angle of a normal angle to the rotated line, which has equation $x+y=p,\:$ is $\:\frac \pi4=\arctan 1$, so the angle of rotation is
$$\theta=\arctan 1-\arctan\tfrac12.$$
Can you end the computation?
