The image of a point (2, -1) with respect to | vedclass

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(B) Let the point be $P(x_1, y_1) = (2, -1)$ and the line be $ax + by + c = 0$,where $x - y + 1 = 0$. Let the image of the point be $P'(x', y')$. The

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$(-2, 3)$

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(B) Let the point be $P(x_1, y_1) = (2, -1)$ and the line be $ax + by + c = 0$,where $x - y + 1 = 0$. Let the image of the point be $P'(x', y')$. The formula for the image of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is given by: $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$ Substituting the values: $\frac{x' - 2}{1} = \frac{y' - (-1)}{-1} = -2 \frac{1(2) - 1(-1) + 1}{1^2 + (-1)^2}$ $\frac{x' - 2}{1} = \frac{y' + 1}{-1} = -2 \frac{2 + 1 + 1}{1 + 1}$ $\frac{x' - 2}{1} = \frac{y' + 1}{-1} = -2 \frac{4}{2} = -4$ Now,solving for $x'$ and $y'$: $x' - 2 = -4 \implies x' = -2$ $y' + 1 = (-1)(-4) = 4 \implies y' = 3$ Thus,the image is $(-2, 3)$.

The image of a point $(2, -1)$ with respect to the line $x - y + 1 = 0$ is

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