A line L_1 passing through the point of inter | vedclass

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(A) First,find the intersection point of $L_2$ lines: $3x-y+2=0$ and $x-3y-2=0$. Solving these,we get $x=-1, y=-1$. Since $L_2$ passes through $(0,0)$

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$\frac{1}{\sqrt{2}}$

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(A) First,find the intersection point of $L_2$ lines: $3x-y+2=0$ and $x-3y-2=0$. Solving these,we get $x=-1, y=-1$. Since $L_2$ passes through $(0,0)$ and $(-1,-1)$,its equation is $y=x$,or $x-y=0$.Next,find the intersection point of $L_1$ lines: $x-2y+3=0$ and $2x-y=0$. Solving these,we get $x=-1, y=-2$. Since $L_1$ is parallel to $L_2$ $(x-y=0)$,its equation is $x-y+k=0$. Substituting $(-1,-2)$,we get $-1-(-2)+k=0$,so $k=-1$. Thus,$L_1$ is $x-y-1=0$.The distance between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.Here,$d = \frac{|-1-0|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$.

$A$ line $L_1$ passing through the point of intersection of the lines $x-2y+3=0$ and $2x-y=0$ is parallel to the line $L_2$. If $L_2$ passes through the origin and also through the point of intersection of the lines $3x-y+2=0$ and $x-3y-2=0$,then the distance between the lines $L_1$ and $L_2$ is

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