Answer
Slope of $-1$ is perpendicular to slope of $1$.
Therefore, the tangents are perpendicular at their points of intersection.
Work Step by Step
$y_1=x$
$y_2=\frac{1}{x}$
Intersects when $y_1 = y_2$.
$x=\frac{1}{x}$
$x^2=1$
$x=1$ or $-1$
Tangent line of $y_1=x$
$\frac{dy_1}{dx}=1$
Tangent line of $y_2=\frac{1}{x}$
$\frac{dy_2}{dx}=-\frac{1}{(x)^2}$
Tangent line of $y_1=x$
$\frac{dy_1}{dx}=1$
Any $x$ value will have a slope of $1$.
Tangent line of $y_2=\frac{1}{x}$:
When $x=1$,
$-\frac{1}{(1)^2}=-1$
When $x=-1$,
$-\frac{1}{(-1)^2}=-1$
Slope of $-1$ is perpendicular to slope of $1$.
Therefore, the tangents are perpendicular at their points of intersection.
