Find the equation of the tangent line to the curve $y=x^2-2x+7$ which is.
(a) parallel to the line $2x-y+9=0$.
(b) perpendicular to the line $5y-15x=13.$

(a)

The equation of the given curve is  $y=x^2-2x+7$.
On differentiating with respect to $x$, we get: 
$\frac{dy}{dx}=2x-2$
The equation of the line is $2x-y+9=0.\Rightarrow y=2x+9$
This is of the form $y=mx+c.$
Slope of the line $=2$
If a tangent is parallel to the line $2x-y+9=0$, then the slope of the tangent is equal to the slope of the line.
Therefore, we have: $2=2x-2$
$\Rightarrow 2x=4\Rightarrow x=2$
Now, at $x=2$
$\Rightarrow y=2^2-2\times 2+7=7$
Thus, the equation of the tangent passing through $(2,7)$ is given by, 
$y-7=2(x-2)$
$\Rightarrow y-2x-3=0$
Hence, the equation of the tangent line to the given curve (which is parallel to line $(2x-y+9=0)$ is $y-2x-3=0.$

(b)

The equation of the line is $5y-15x=13.$
Slope of the line $=3$
If a tangent is perpendicular to the line $5y-15x=13$, 
then the slope of the tangent is $\frac{-1}{\text{Slope of the line}}=\frac{-1}{3}$.
$\Rightarrow \frac{dy}{dx}=2x-2=\frac{-1}{3}$
$\Rightarrow 2x=\frac{-1}{3}+2$
$\Rightarrow 2x=\frac{5}{3}$
$\Rightarrow x=\frac{5}{6}$
Now, at  $x=\frac{5}{6}$
$\Rightarrow y=\frac{25}{36}-\frac{10}{6}+7=\frac{25-60+252}{36}=\frac{217}{36}$
Thus, the equation of the tangent passing through $\left(\frac{5}{6},\frac{217}{36}\right)$ is given by, 
$\left(y-\frac{217}{36}\right)=-\frac{1}{3}\left(x-\frac{5}{6}\right)$
$\Rightarrow =\frac{36y-217}{36}-\frac{1}{18}(6x-5)$   
$\Rightarrow 36y-217=-2(6x-5)$
$\Rightarrow 36y-217=-12x+10$
$\Rightarrow 36y+12x-227=0$  
Hence, the equation of the tangent line to the given curve 
(which is perpendicular to line $5y-15x=13)$ is $36y+12x-227=0$