Question

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

Answer 1

$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{is} \\ \mathrm{y}={\mathrm{x}}^2-5\mathrm{x}+6 \\ \mathrm{differentiating}\mathrm{y}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x},\mathrm{we}\mathrm{get} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}-5 \\ \mathrm{Now},\mathrm{Tangent}\mathrm{at}(2,0)\mathrm{is} \\ (\frac{\mathrm{dy}}{\mathrm{dx}}{)}_{2,0}=2\times 2-5=-1\Rightarrow {\mathrm{m}}_1=-1 \\ \mathrm{Similarly}, \\ (\frac{\mathrm{dy}}{\mathrm{dx}}{)}_{3,0}=2\times 3-5=1\Rightarrow {\mathrm{m}}_2=1 \\ {\mathrm{m}}_1\times {\mathrm{m}}_2=-1 \\ \mathrm{Thus},\mathrm{tangents}\mathrm{are}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}. \end{gathered}$$