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# Find the equation of the tangent line to the curve $y = x^2 - 2x +7$ which is $(a)\; parallel\; to\; the\; line\; 2x - y + 9 = 0$

This is (a) 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:
It is given that the tangent is parallel to the line $2x-y+9=0$
The slope of the line is $-\bigg(\large\frac{coeff\;of\;x}{coeff\;of\;y}\bigg)$
Slope of the line =$-\big(\large\frac{2}{-1}\big)$
$\qquad\qquad\quad\;\;\;=2$
Step 3:
Since they are parallel,their slopes are equal,
Therefore $2(x-1)=2$
$\Rightarrow x=2$
When $x=2$,
$y=(2)^2-2(2)+7$
$\;\;\;=4-4+7$
$\;\;\;=7$
Step 4:
Therefore tangent parallel to $2x-y+9=0$ at $(2,7)$ is
$(y-7)=2(x-2)$
$\Rightarrow y-7=2x-4$
$\Rightarrow 2x-y+3=0$

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