Equation of the tangent to the curve $\mathrm{y}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)-\mathrm{x}+7 = 0$ where it crosses the x-axis is

$$\begin{gathered} \mathrm{y}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)-\mathrm{x}+7 = 0 \\ \therefore \mathrm{y} = \frac{\left(\mathrm{x}-7\right)}{\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)} = \frac{\mathrm{x}-7}{{\mathrm{x}}^2 - 5\mathrm{x} + 6} \\ \mathrm{The} \mathrm{curve} \mathrm{crosses} \mathrm{x}-\mathrm{axis} \Rightarrow \mathrm{y} = 0 \end{gathered}$$$\therefore \mathrm{x}= 7 \therefore \mathrm{m} = {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(7,0\right)} = \frac{\left({\mathrm{x}}^2 - 5\mathrm{x} + 6\right) - \left(\mathrm{x}-7\right)\left(2\mathrm{x}-5\right)}{{\left({\mathrm{x}}^2 - 5\mathrm{x} + 6\right)}^2 } = \frac{20 - 0}{{20}^2} = \frac{1}{20}$Equation tangent at $$\begin{gathered} \left(7,0\right) \mathrm{is} \left(\mathrm{y} - 0\right) = \frac{1}{20}\left(\mathrm{x}-7\right) \\ \mathrm{x}-20\mathrm{y} = 7 \end{gathered}$$