Question

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 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

• A. \( \frac{1}{\sqrt{2}} \)
• B. \( \sqrt{2} \)
• C. \( \sqrt{5} \)
• D. \( \frac{1}{\sqrt{5}} \)

Hint:

1. Solve systems of linear equations to find points of intersection. 2. The slope of a line through \( (x_1,y_1) \) and \( (x_2,y_2) \) is \( \frac{y_2-y_1}{x_2-x_1} \). Equation: \( y-y_1=m(x-x_1) \). 3. Parallel lines have equal slopes. 4. Distance between parallel lines \(Ax+By+C_1=0\) and \(Ax+By+C_2=0\) is \( \frac{|C_1-C_2|}{\sqrt{A^2+B^2}} \).

Answer 1

The Correct Option is A

Point of intersection for \(L_1\), \(P_1\): \(L_A: x-2y+3=0\) \(L_B: 2x-y=0 \implies y=2x\) Substitute \(y=2x\) into \(L_A\): \(x-2(2x)+3=0 \implies x-4x+3=0 \implies -3x=-3 \implies x=1\).
So \(y=2(1)=2\).
\(P_1 = (1,2)\).
Point of intersection for defining \(L_2\), \(P_2\): \(L_C: 3x-y+2=0 \implies y=3x+2\) \(L_D: x-3y-2=0\) Substitute \(y=3x+2\) into \(L_D\): \(x-3(3x+2)-2=0 \implies x-9x-6-2=0 \implies -8x=8 \implies x=-1\).
So \(y=3(-1)+2 = -1\).
\(P_2 = (-1,-1)\).
Line \(L_2\) passes through O(0,0) and \(P_2(-1,-1)\).
Slope of \(L_2\), \(m_2 = \frac{-1-0}{-1-0} = 1\).
Equation of \(L_2\): \(y-0 = 1(x-0) \implies y=x \implies x-y=0\).
Line \(L_1\) passes through \(P_1(1,2)\) and is parallel to \(L_2\), so slope \(m_1 = m_2 = 1\).
Equation of \(L_1\): \(y-2 = 1(x-1) \implies y-2=x-1 \implies x-y+1=0\).
Distance between parallel lines \(L_1: x-y+1=0\) and \(L_2: x-y=0\).
Using \( D = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}} \) for \(Ax+By+C_1=0\) and \(Ax+By+C_2=0\).
Here \(A=1, B=-1, C_1=1, C_2=0\).
\[ D = \frac{|1-0|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} \] This matches option (1).