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(A) The equation of the line is $5y-15x=13$. Rewriting in $y=mx+c$ form: $5y=15x+13 \Rightarrow y=3x+\frac{13}{5}$. The slope of this line is $m_1=3$.

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$36y+12x-227=0$

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(A) The equation of the line is $5y-15x=13$. Rewriting in $y=mx+c$ form: $5y=15x+13 \Rightarrow y=3x+\frac{13}{5}$. The slope of this line is $m_1=3$. Since the tangent is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$. Thus,$m_2 = -\frac{1}{3}$. The slope of the tangent to the curve $y=x^2-2x+7$ is given by $\frac{dy}{dx} = 2x-2$. Setting the slope equal to $-\frac{1}{3}$: $2x-2 = -\frac{1}{3} \Rightarrow 2x = 2 - \frac{1}{3} = \frac{5}{3} \Rightarrow x = \frac{5}{6}$. Now,find the $y$-coordinate: $y = (\frac{5}{6})^2 - 2(\frac{5}{6}) + 7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{25-60+252}{36} = \frac{217}{36}$. The point of tangency is $(\frac{5}{6}, \frac{217}{36})$. The equation of the tangent line is $y - \frac{217}{36} = -\frac{1}{3}(x - \frac{5}{6})$. Multiplying by $36$: $36y - 217 = -12(x - \frac{5}{6}) \Rightarrow 36y - 217 = -12x + 10$. Rearranging gives $36y + 12x - 227 = 0$.

Find the equation of the tangent line to the curve $y=x^{2}-2x+7$ which is perpendicular to the line $5y-15x=13$.

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