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 $5y-15x=13.$

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 $y-2x-3=0.$