Question:
The family of lines, forming an isosceles triangle with the lines \(3x-4y-2=0\) and \(12x-5y+6=0\), is
The family of lines, forming an isosceles triangle with the lines \(3x-4y-2=0\) and \(12x-5y+6=0\), is
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When three lines form an isosceles triangle, the third line is usually an angle bisector of the other two lines. Use the angle bisector formula directly for faster calculations.
Updated On: Jun 25, 2026
- \(9x+7y+c=0\)
- \(7x-9y+c=0\)
- \(9x-7y+c=0\)
- \(x\pm y+c=0\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the slopes of the given lines.
The first line is \[ 3x-4y-2=0 \] Rewriting in slope-intercept form: \[ 4y=3x-2 \] \[ y=\frac{3}{4}x-\frac{1}{2} \] Hence, slope of the first line is \[ m_1=\frac{3}{4} \] Now consider the second line: \[ 12x-5y+6=0 \] Rewriting: \[ 5y=12x+6 \] \[ y=\frac{12}{5}x+\frac{6}{5} \] Hence, slope of the second line is \[ m_2=\frac{12}{5} \]
Step 2: Use the condition for an isosceles triangle.
If three lines form an isosceles triangle, then one line must make equal angles with the other two lines.
Therefore, the required family of lines represents the angle bisectors of the two given lines.
Equation of angle bisectors of \[ a_1x+b_1y+c_1=0 \] and \[ a_2x+b_2y+c_2=0 \] is given by \[ \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}} \] Substituting: \[ \frac{3x-4y-2}{\sqrt{3^2+(-4)^2}} = \pm \frac{12x-5y+6}{\sqrt{12^2+(-5)^2}} \] Since \[ \sqrt{3^2+(-4)^2}=5 \] and \[ \sqrt{12^2+(-5)^2}=13, \] we get \[ \frac{3x-4y-2}{5} = \pm \frac{12x-5y+6}{13} \]
Step 3: Solve for the angle bisectors.
Taking positive sign: \[ 13(3x-4y-2)=5(12x-5y+6) \] \[ 39x-52y-26=60x-25y+30 \] \[ 21x+27y+56=0 \] This simplifies to another family not present in the options.
Now taking negative sign: \[ 13(3x-4y-2)=-5(12x-5y+6) \] \[ 39x-52y-26=-60x+25y-30 \] \[ 99x-77y+4=0 \] Dividing by \(11\): \[ 9x-7y+\frac{4}{11}=0 \] Hence, the family of lines is \[ 9x-7y+c=0 \]
Step 4: Final conclusion.
Therefore, the correct answer is \[ \boxed{9x-7y+c=0} \]
The first line is \[ 3x-4y-2=0 \] Rewriting in slope-intercept form: \[ 4y=3x-2 \] \[ y=\frac{3}{4}x-\frac{1}{2} \] Hence, slope of the first line is \[ m_1=\frac{3}{4} \] Now consider the second line: \[ 12x-5y+6=0 \] Rewriting: \[ 5y=12x+6 \] \[ y=\frac{12}{5}x+\frac{6}{5} \] Hence, slope of the second line is \[ m_2=\frac{12}{5} \]
Step 2: Use the condition for an isosceles triangle.
If three lines form an isosceles triangle, then one line must make equal angles with the other two lines.
Therefore, the required family of lines represents the angle bisectors of the two given lines.
Equation of angle bisectors of \[ a_1x+b_1y+c_1=0 \] and \[ a_2x+b_2y+c_2=0 \] is given by \[ \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}} \] Substituting: \[ \frac{3x-4y-2}{\sqrt{3^2+(-4)^2}} = \pm \frac{12x-5y+6}{\sqrt{12^2+(-5)^2}} \] Since \[ \sqrt{3^2+(-4)^2}=5 \] and \[ \sqrt{12^2+(-5)^2}=13, \] we get \[ \frac{3x-4y-2}{5} = \pm \frac{12x-5y+6}{13} \]
Step 3: Solve for the angle bisectors.
Taking positive sign: \[ 13(3x-4y-2)=5(12x-5y+6) \] \[ 39x-52y-26=60x-25y+30 \] \[ 21x+27y+56=0 \] This simplifies to another family not present in the options.
Now taking negative sign: \[ 13(3x-4y-2)=-5(12x-5y+6) \] \[ 39x-52y-26=-60x+25y-30 \] \[ 99x-77y+4=0 \] Dividing by \(11\): \[ 9x-7y+\frac{4}{11}=0 \] Hence, the family of lines is \[ 9x-7y+c=0 \]
Step 4: Final conclusion.
Therefore, the correct answer is \[ \boxed{9x-7y+c=0} \]
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