Question

Prove that the tangents to the curve $y=x^2-5x+6$ at the points $(2,0)$ and $(3,0)$ are at right angles.

Answer 1

Equation of tangent to any curve $y=f\left(x\right)$ at $(x_1,y_1)$ is
$(y-y_1)=\frac{dy}{dx}{|}_{(x_1,y_1)}(y-y_1)$.
Clearly, the slope of the tangent to the curve has slope $\frac{dy}{dx}{|}_{(x_1,y_1)}$ at $(x_1,y_1)$
Now, for the given curve $y=x^2-5x+6$, the tangents at $(2,0)$ and $(3,0)$ have the slope $2\times 2-5=-1$ and $2\times 3-5=1$.
It is clear product of slopes $=-1$.
Hence, the problem.