Question:
The straight line passing through \((-1,1)\) and remaining parallel to the line common to the pairs of lines provided by \(6x^2-xy-12y^2=0\) and \(15x^2+14xy-8y^2=0\), is
The straight line passing through \((-1,1)\) and remaining parallel to the line common to the pairs of lines provided by \(6x^2-xy-12y^2=0\) and \(15x^2+14xy-8y^2=0\), is
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To find a common line between two homogeneous second-degree equations, first factorize both equations completely and identify the repeated linear factor.
Updated On: Jun 25, 2026
- \(5x-2y+7=0\)
- \(3x+4y-1=0\)
- \(3x-4y+7=0\)
- \(2x-3y+5=0\)
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The Correct Option is B
Solution and Explanation
Step 1: Factorize the first pair of lines.
Given: \[ 6x^2-xy-12y^2=0 \] We factorize: \[ 6x^2-9xy+8xy-12y^2=0 \] \[ 3x(2x-3y)+4y(2x-3y)=0 \] \[ (3x+4y)(2x-3y)=0 \] Thus, the two lines are \[ 3x+4y=0 \] and \[ 2x-3y=0 \]
Step 2: Factorize the second pair of lines.
Given: \[ 15x^2+14xy-8y^2=0 \] We factorize: \[ 15x^2+20xy-6xy-8y^2=0 \] \[ 5x(3x+4y)-2y(3x+4y)=0 \] \[ (5x-2y)(3x+4y)=0 \] Thus, the two lines are \[ 5x-2y=0 \] and \[ 3x+4y=0 \]
Step 3: Identify the common line.
The common line in both pairs is \[ 3x+4y=0 \] Therefore, the required line must be parallel to \[ 3x+4y=0 \] Hence, its equation is of the form \[ 3x+4y+c=0 \]
Step 4: Use the given point \((-1,1)\).
Substitute \((-1,1)\) into \[ 3x+4y+c=0 \] \[ 3(-1)+4(1)+c=0 \] \[ -3+4+c=0 \] \[ 1+c=0 \] \[ c=-1 \] Thus, the required line is \[ 3x+4y-1=0 \]
Step 5: Final conclusion.
Therefore, the correct answer is \[ \boxed{3x+4y-1=0} \]
Given: \[ 6x^2-xy-12y^2=0 \] We factorize: \[ 6x^2-9xy+8xy-12y^2=0 \] \[ 3x(2x-3y)+4y(2x-3y)=0 \] \[ (3x+4y)(2x-3y)=0 \] Thus, the two lines are \[ 3x+4y=0 \] and \[ 2x-3y=0 \]
Step 2: Factorize the second pair of lines.
Given: \[ 15x^2+14xy-8y^2=0 \] We factorize: \[ 15x^2+20xy-6xy-8y^2=0 \] \[ 5x(3x+4y)-2y(3x+4y)=0 \] \[ (5x-2y)(3x+4y)=0 \] Thus, the two lines are \[ 5x-2y=0 \] and \[ 3x+4y=0 \]
Step 3: Identify the common line.
The common line in both pairs is \[ 3x+4y=0 \] Therefore, the required line must be parallel to \[ 3x+4y=0 \] Hence, its equation is of the form \[ 3x+4y+c=0 \]
Step 4: Use the given point \((-1,1)\).
Substitute \((-1,1)\) into \[ 3x+4y+c=0 \] \[ 3(-1)+4(1)+c=0 \] \[ -3+4+c=0 \] \[ 1+c=0 \] \[ c=-1 \] Thus, the required line is \[ 3x+4y-1=0 \]
Step 5: Final conclusion.
Therefore, the correct answer is \[ \boxed{3x+4y-1=0} \]
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