I'm stuck on problem 8 from chapter 4 in Spivaks Calculus.
Prove the graphs of the functions
$f(x)=mx+b$
$g(x)=nx+c$
are perpendicular if $mn=-1$ by computing the squares of the lengths of the sides of the triangle.
What I tried:
- In right $\Delta ACQ$ we find: $AC^2 = CQ^2 + AQ^2 = 1 + m^2$
- In right $\Delta BCQ$ we find: $BC^2 = CQ^2 + BQ^2 = 1 + n^2$
If we can show that $\angle C$ is a right angle, then we are done.
So assuming $\Delta ABC$ is a right triangle, we would find:
$$AC^2 + BC^2 = AB^2$$
Since $AB=m+n$:
$$AC^2 + BC^2 = (m + n)^2$$
If we plug in the values from above:
$$(1 + m^2) + (1+n^2) = (m + n)^2$$
But that doesn't seem like it's correct or leading anywhere.
Did I miss something? I'm also not sure how to make use of the fact that $mn=-1$, meaning that $m$ is the negative reciprocal of $n$ or vice versa.
You're almost there. Consider that the distance from $(1,m)$ to $(1,n)$ is $(m-n)^2$ ($n$ is a negative number), your final line will now read
$m^2 + n^2 + 2 = (m-n)^2$.
Expand this out and simplify.