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(C) Let the point be $P(3, -1)$ and the line be $L: 2x + y - 1 = 0$. The image $P'(x', y')$ of point $P$ with respect to line $L$ is given by the form

Vedclass · Accepted Answer

$\sqrt{\frac{89}{5}}$

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(C) Let the point be $P(3, -1)$ and the line be $L: 2x + y - 1 = 0$. The image $P'(x', y')$ of point $P$ with respect to line $L$ is given by the formula: $\frac{x' - 3}{2} = \frac{y' - (-1)}{1} = -2 \frac{2(3) + 1(-1) - 1}{2^2 + 1^2}$ $\frac{x' - 3}{2} = \frac{y' + 1}{1} = -2 \frac{6 - 1 - 1}{5} = -2 \frac{4}{5} = -\frac{8}{5}$ Thus,$x' = 3 + 2(-\frac{8}{5}) = 3 - \frac{16}{5} = -\frac{1}{5}$ And $y' = -1 + 1(-\frac{8}{5}) = -1 - \frac{8}{5} = -\frac{13}{5}$ The image is $P'(-\frac{1}{5}, -\frac{13}{5})$. The distance between $A(2, 1)$ and $P'(-\frac{1}{5}, -\frac{13}{5})$ is: $D = \sqrt{(2 - (-\frac{1}{5}))^2 + (1 - (-\frac{13}{5}))^2}$ $D = \sqrt{(\frac{11}{5})^2 + (\frac{18}{5})^2} = \sqrt{\frac{121 + 324}{25}} = \sqrt{\frac{445}{25}} = \sqrt{\frac{89}{5}}$

The distance between the point $(2, 1)$ and the image of the point $(3, -1)$ with respect to the line $2x + y - 1 = 0$ is

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