Straight Lines and Pair of Straight Lines

First, we need to find the point where the two lines $x-2y+3=0$ and $2x-y=0$ meet.Step 1: Find Intersection PointTo solve the two equations:$\;x-2y+3=0$$\;2x-y=0$Use the second equation: $2x-y=0 \Rightarrow y=2x$.Put $y=2x$ into the first equation:$\;x-2(2x)+3=0 \Rightarrow x-4x+3=0$$\;-3x+3=0$$\;x=1$Substitute $x=1$ back: $y=2x=2 \times 1=2$So, the intersection is at the point $(1,2)$.Step 2: Find Another Point for Line $L_2$$L_2$ passes through the origin $(0,0)$ and the intersection of $3x-y+2=0$ and$x-3y-2=0 .$We need the intersection point of $3x-y+2=0$ and $x-3y-2=0$.Multiply the second equation by 3 :$3x-9y-6=0$Subtract $3x-y+2=0$ from $3x-9y-6=0$ :$\;{[3x-9y-6]-[3x-y+2]}=0$$\;(3x-9y-6)-3x+y-2=0$$\;-8y-8=0$$y=-1$Substitute $y=-1$ into $x-3y-2=0$ :$\;x-3(-1)-2=0$$\;x+3-2=0$$\;x=-1$So, the intersection point is $(-1,-1)$.Step 3: Find Equation of $L_2$$L_2$ passes through $(0,0)$ and $(-1,-1)$.Slope $m=\;\frac{-1-0}{-1-0}=1$.Using slope-intercept form: $y=x$.In general form: $-x+y=0$Step 4: Find Equation of $L_1$$L_1$ is parallel to $L_2$ (so same slope $m=1$ ) and passes through ( 1,2 ).Write $y=x+c$.Plug in $(1,2)$ :$2=1+c \Rightarrow c=1 .$So Equation of $L_1$ is $y=x+1$, or $-x+y=1$Step 5: Find Distance Between $L_1$ and $L_2$The distance between two parallel lines $-x+y=1$ and $-x+y=0$ is:Distance $=\;\frac{|1-0|}{\sqrt{1^2+1^2}} = \;\frac{1}{\sqrt{2}}$