Find the equation of the tangent line to the curve $$y = x^2 - 2x +7$$ which is $(b)\; perpendicular\; to\; the\; line\; 5y-15x=13$

This is (b) part of the multi-part question q15

Toolbox:
• Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
• Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1) Step 1: Given :y=x^2-2x+7 Differentiating w.r.t x we get, \large\frac{dy}{dx}=$$(2x-2)$
$\quad\;\;=2(x-1)$
Step 2:
When the tangent is perpendicular to the line $5y-15x=13$
Since they are perpendicular,
slope of the tangent $\times$ slope of the line =$-1$
Slope of the line is $-\big(\large\frac{-15}{5}\big)$
$\Rightarrow 3$
Therefore $2(x-1)\times 3=-1$
$\Rightarrow 6(x-1)=-1$
$6x-6=-1$
$6x=5$
$x=\large\frac{5}{6}$
Step 3:
When $x=\large\frac{5}{6}$
$y=\big(\large\frac{5}{6}\big)$$-2\big(\large\frac{5}{6}\big)$$+7$
$y=\large\frac{25}{36}-\frac{10}{6}$$+7 \;\;=\large\frac{25-60+252}{36} \;\;=\large\frac{217}{36} Hence the points are (\large\frac{5}{6},\frac{217}{36}) Step 4: Equation of the tangent which is perpendicular to 5y-15x=13 at \big(\large\frac{5}{6},\frac{217}{36}\big) is y-\large\frac{217}{36}=\frac{-1}{3}$$(x-\large\frac{5}{6})$
On simplifying we get,
$36y-217=-12(x-\large\frac{5}{6})$
$36y-217=-12x+10$
$\Rightarrow 12x+36y-227=0$
This is the required equation of the tangent.