(C) Let the point be $(x, y)$ on the line $x+y+3=0$. So,$y = -x-3$. The point is $(x, -x-3)$.The distance of this point from the line $x+2y+2=0$ is gi

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(C) Let the point be $(x, y)$ on the line $x+y+3=0$. So,$y = -x-3$. The point is $(x, -x-3)$.The distance of this point from the line $x+2y+2=0$ is given by $\frac{|x+2(-x-3)+2|}{\sqrt{1^2+2^2}} = \sqrt{5}$.$\frac{|x-2x-6+2|}{\sqrt{5}} = \sqrt{5} \implies |-x-4| = 5$.$|x+4| = 5 \implies x+4 = 5$ or $x+4 = -5$.So,$x_1 = 1$ and $x_2 = -9$.The value of $|x_1-x_2| = |1 - (-9)| = |10| = 10$.

Class 11 Mathematics Straight Lines Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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If $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line $x+y+3=0$ such that each of them is at a distance of $\sqrt{5}$ units from the line $x+2y+2=0$,then the value of $|x_1-x_2|$ is:

AP EAMCET 2018 All AP EAMCET

A
$6$
B
$20$
C
$10$
D
$2$

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Straight Lines

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Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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If $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line $x+y+3=0$ such that each of them is at a distance of $\sqrt{5}$ units from the line $x+2y+2=0$,then the value of $|x_1-x_2|$ is:

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