Question:
The equation of straight line through the intersection of the lines \(2x+y=1\) and \(3x+2y=5\) and passing through the origin is
The equation of straight line through the intersection of the lines \(2x+y=1\) and \(3x+2y=5\) and passing through the origin is
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Line through intersection of \(L_1=0\) and \(L_2=0\) is \(L_1+\lambda L_2=0\). Use the extra condition (point passing) to find \(\lambda\).
Updated On: Jan 3, 2026
- \(7x+3y=0\)
- \(7x-y=0\)
- \(3x+2y=0\)
- \(x+y=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Family of lines through intersection point.
Line through intersection of:
\[ 2x+y-1=0 \quad \text{and}\quad 3x+2y-5=0 \] is:
\[ (2x+y-1)+\lambda(3x+2y-5)=0 \] Step 2: Since line passes through origin \((0,0)\).
Substitute \(x=0,y=0\):
\[ (-1)+\lambda(-5)=0 \Rightarrow -1-5\lambda=0 \Rightarrow \lambda=-\frac{1}{5} \] Step 3: Substitute \(\lambda\) back.
\[ (2x+y-1)-\frac{1}{5}(3x+2y-5)=0 \] Multiply by 5:
\[ 5(2x+y-1)-(3x+2y-5)=0 \] \[ 10x+5y-5-3x-2y+5=0 \] \[ 7x+3y=0 \] Final Answer: \[ \boxed{7x+3y=0} \]
Line through intersection of:
\[ 2x+y-1=0 \quad \text{and}\quad 3x+2y-5=0 \] is:
\[ (2x+y-1)+\lambda(3x+2y-5)=0 \] Step 2: Since line passes through origin \((0,0)\).
Substitute \(x=0,y=0\):
\[ (-1)+\lambda(-5)=0 \Rightarrow -1-5\lambda=0 \Rightarrow \lambda=-\frac{1}{5} \] Step 3: Substitute \(\lambda\) back.
\[ (2x+y-1)-\frac{1}{5}(3x+2y-5)=0 \] Multiply by 5:
\[ 5(2x+y-1)-(3x+2y-5)=0 \] \[ 10x+5y-5-3x-2y+5=0 \] \[ 7x+3y=0 \] Final Answer: \[ \boxed{7x+3y=0} \]
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