If the point P(x_1, y_1) lying on the curve y | vedclass

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(C) The slope of the line $y = x - 3$ is $m = 1$. Since $P(x_1, y_1)$ is the closest point on the curve $y = x^2 - x + 1$ to the line, the tangent at

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$1$

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(C) The slope of the line $y = x - 3$ is $m = 1$. Since $P(x_1, y_1)$ is the closest point on the curve $y = x^2 - x + 1$ to the line, the tangent at $P$ must be parallel to the given line. Thus, the derivative $\frac{dy}{dx} = 2x - 1$ must equal $1$. $2x_1 - 1 = 1 \implies 2x_1 = 2 \implies x_1 = 1$. Substituting $x_1 = 1$ into the curve equation: $y_1 = (1)^2 - 1 + 1 = 1$. So, the point $P$ is $(1, 1)$. The perpendicular distance from $P(1, 1)$ to the line $3x + 4y - 2 = 0$ is given by the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$. $d = \frac{|3(1) + 4(1) - 2|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 4 - 2|}{\sqrt{9 + 16}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1$.

If the point $P(x_1, y_1)$ lying on the curve $y = x^2 - x + 1$ is the closest point to the line $y = x - 3$, then the perpendicular distance from $P$ to the line $3x + 4y - 2 = 0$ is

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