Find a point on line which builds angle of 45 degrees with other two points (not on line)

Question

I have two points $P(0,8)$ and $Q(7,9)$ and line $p: x - 3y + 2 = 0$. Find point $M$ which is on line $p$ and it builds $45°$ angle with points $P$ and $Q$ ($PMQ = 45°$).

I can't solve this problem no matter what I do.

Answer 1

Think about it this way, if you have a circle intersecting line $p$, such that $\overline{PQ}$ is a chord of the circle, you want that the chord $\overline{PQ}$ makes $\angle{45}$ with the circle. (As it will make $\angle{45}$ with all points in the circle)

Now, as a chord makes an angle at the center which is twice of angle made at the circle, the angle made by the chord at the center is $2 \cdot \angle45 = 90$

So you have a right isosceles triangle $\triangle OPQ$, ($\overline{OP} = \overline{OQ}$ as $O$ is the center of the circle), so $\overline{PQ}$ = $r\sqrt{2}$ (Pythogoras in $\triangle OPQ$), $\overline{PQ}$ = $5\sqrt{2}$, so $r$ = $5$ You need a circle with radius $5$ whose center makes a right angle with its chord $\overline{PQ}$, can you figure out the rest?

No? That’s alright!

Let O $(p,q)$ be the center of the circle with radius 5. As $P(0,8), Q(7, 9)$ lie on the circle: $$p^2 + (8 - q)^2 = (7 - p)^2 + (9 - q)^2 = 25 \implies p = 4, q = 5$$ Note: The idea behind $3^2 + 4^2 = 5^2$ is pythagorean triplets, ambiguous and it perhaps works when tested!

Thus, there is a circle at center $O(4, 5)$ with radius 5. We have to find it’s intersection point with the line $x - 3y + 2 = 0$ $$(x - 4)^2 + (y - 5)^2 = 25, x - 3y + 2 = 0 \implies (x,y) \in \{(1,1), \Bigg(\frac{44}{5}, \frac{18}{5}\Bigg)\} \tag*{$\blacksquare$}$$