If the line x-y+1=0 cuts the lines 2x+2y+3=0 | vedclass

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(A) The given lines are $L_1: 2x+2y+3=0$ and $L_2: 3x+3y+2=0$. Dividing $L_1$ by $2$,we get $x+y+\frac{3}{2}=0$. Dividing $L_2$ by $3$,we get $x+y+\fr

Vedclass · Accepted Answer

$\frac{5}{6\sqrt{2}}$

Vedclass · Answer

(A) The given lines are $L_1: 2x+2y+3=0$ and $L_2: 3x+3y+2=0$. Dividing $L_1$ by $2$,we get $x+y+\frac{3}{2}=0$. Dividing $L_2$ by $3$,we get $x+y+\frac{2}{3}=0$. Since the slopes of both lines are $-1$,they are parallel. The line $L_3: x-y+1=0$ has a slope of $1$. Since the product of the slopes of $L_1$ (or $L_2$) and $L_3$ is $(-1) 	imes (1) = -1$,the line $L_3$ is perpendicular to both parallel lines. The distance $AB$ between the points of intersection is the perpendicular distance between the two parallel lines $x+y+\frac{3}{2}=0$ and $x+y+\frac{2}{3}=0$. The distance $d$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$. Here,$a=1, b=1, c_1=\frac{3}{2}, c_2=\frac{2}{3}$. $AB = \frac{|\frac{3}{2}-\frac{2}{3}|}{\sqrt{1^2+1^2}} = \frac{|\frac{9-4}{6}|}{\sqrt{2}} = \frac{5}{6\sqrt{2}}$.

If the line $x-y+1=0$ cuts the lines $2x+2y+3=0$ and $3x+3y+2=0$ at the points $A$ and $B$ respectively,then $AB=$

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