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(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 given

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$y-2x-3=0$

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(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 given line is $2x-y+9=0$,which can be written as $y=2x+9$. This is in the form $y=mx+c$,so the slope of the line is $m=2$. Since the tangent is parallel to the line $2x-y+9=0$,the slope of the tangent must be equal to the slope of the line. Therefore,$2x-2=2$,which gives $2x=4$,so $x=2$. Substituting $x=2$ into the curve equation,we get $y=(2)^{2}-2(2)+7=4-4+7=7$. The point of tangency is $(2, 7)$. The equation of the tangent line passing through $(2, 7)$ with slope $m=2$ is given by $y-y_{1}=m(x-x_{1})$. $y-7=2(x-2)$ $y-7=2x-4$ $y-2x-3=0$.

Find the equation of the tangent line to the curve $y=x^{2}-2x+7$ which is parallel to the line $2x-y+9=0$.

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