How do I approach this geometrical problem?

Question

For a point $P=(x,y)$ write $f(P)=ax+by$.

Let

• $f(A)=f(B)=10$.
• $C$ be a point not lying on the line joining $A$ and $B$.
• $C^{'}$ be the reflection of $C$ w.r.t. this line.

If $f(C)=15$, find $f(C^{'})$.

How do I approach this kind of problem?

Answer 1

Let $u(P) = f(P) - 10$. Then $u$ is zero on the line $AB$, and $u(C) = 5$.

Observation 1: Let $g(P)$ denote "how far to the left of line $AB$ the point $P$ is. ("To the left" means "if you stand on your graph paper at $A$, looking at $B$, is $P$ to the left or to the right?" This assumes that looking down at the origin, the $x$-axis points to the right, and the y-axis points in the direction of the top of your head.) To the left of the line $AB$, the function $g$ is positive, and to the right, it's negative. In fact the graph of $z = g(P)$ is a plane that intersects the $xy$-plane exactly along the line $AB$. You can prove this using similar triangles, for instance, but I'm not going to go through that here.

Observation 2: The function $P \mapsto u(P)$ has much the same property: the graph of $z = u(P)$ is a plane intersecting the $xy$-plane along the line $AB$. The only difference is that $u(C) = 5$, while $g(C) = k = $ "however far $C$ is to the left of the line $AB$" (where $g(C) < 0$ means that $C$ is to the right of the line).

Now suppose that we multiply $g$ by $5/k$ to get a new function $h$. The graph of $h$ will again be a plane passing through $AB$. And $h(C) = (5/k) g(C) = (5/k)k = 5$, so $h(C) = u(C)$. The graphs of both $h$ and $u$ are planes containing the line $AB$ and the point $(C_x, C_y, 5)$. That means that they're the same plane!

So I've now shown you that $u$ is a multiple of $g$, namely $u(P) = \frac{u(C)}{g(C)} g(P)$ for any point $P$.

Obviously for $g$, we have $g(C') = -g(C)$. From this, we can conclude that $u(C') = -u(C)$. So $u(C') = -5$. But $u(C) = f(C) -10$, so $-5 = f(C) -10$, so $10-5 = f(C)$ and we get $f(C) = 5$.