The distance from the origin to the image of | vedclass

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(C) The formula for the image $(x', y')$ 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}

Vedclass · Accepted Answer

$6$

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(C) The formula for the image $(x', y')$ of a point $(x_1, y_1)$ with respect to the line $ax+by+c=0$ is given by $\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}$.For the point $(1,1)$ and line $x+y+5=0$, we have $a=1, b=1, c=5$.$\frac{x'-1}{1} = \frac{y'-1}{1} = -2 \frac{1(1)+1(1)+5}{1^2+1^2} = -2 \frac{7}{2} = -7$.Thus, $x'-1 = -7 \Rightarrow x' = -6$ and $y'-1 = -7 \Rightarrow y' = -6$.The image point is $(-6, -6)$.The distance $D$ from the origin $(0,0)$ to the point $(-6, -6)$ is calculated as:$D = \sqrt{(-6-0)^2 + (-6-0)^2} = \sqrt{36+36} = \sqrt{72} = 6 \sqrt{2}$.

The distance from the origin to the image of $(1,1)$ with respect to the line $x+y+5=0$ is (in $\sqrt{2}$)

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