P is a point on x+y+5=0,whose perpendicular d | vedclass

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(B) Let the coordinates of point $P$ be $(x, y)$. Since $P$ lies on $x+y+5=0$,we have $y = -x-5$. So,$P = (x, -x-5)$.The perpendicular distance $d$ of

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$(1,-6)$

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(B) Let the coordinates of point $P$ be $(x, y)$. Since $P$ lies on $x+y+5=0$,we have $y = -x-5$. So,$P = (x, -x-5)$.The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $Ax+By+C=0$ is given by $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.Given the distance from $2x+3y+3=0$ is $\sqrt{13}$,we have:$\frac{|2x+3(-x-5)+3|}{\sqrt{2^2+3^2}} = \sqrt{13}$$\frac{|2x-3x-15+3|}{\sqrt{13}} = \sqrt{13}$$|-x-12| = 13$$|x+12| = 13$This gives two cases:$1) x+12 = 13 \Rightarrow x = 1$. Then $y = -1-5 = -6$. So,$P = (1, -6)$.$2) x+12 = -13 \Rightarrow x = -25$. Then $y = -(-25)-5 = 20$. So,$P = (-25, 20)$.Comparing with the given options,the correct point is $(1, -6)$.

$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

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