Question:
Suppose, in triangle \(\triangle ABC\), \(x-y+5=0\), \(x+2y=0\) are respectively the equations of the perpendicular bisectors of the sides \(AB\) and \(AC\). If \(A=(1,-2)\), the equation of the line joining \(B\) and \(C\) is
Suppose, in triangle \(\triangle ABC\), \(x-y+5=0\), \(x+2y=0\) are respectively the equations of the perpendicular bisectors of the sides \(AB\) and \(AC\). If \(A=(1,-2)\), the equation of the line joining \(B\) and \(C\) is
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If a line is the perpendicular bisector of a segment, then the endpoints of the segment are reflections of each other across that line.
Updated On: Jun 22, 2026
- \(6x+7y=0\)
- \(14x+23y-40=0\)
- \(2x-11y=0\)
- \(2x+y=0\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the concept of perpendicular bisector.
Since \(x-y+5=0\) is the perpendicular bisector of \(AB\), point \(B\) is the reflection of point \(A\) in the line \(x-y+5=0\).
Similarly, since \(x+2y=0\) is the perpendicular bisector of \(AC\), point \(C\) is the reflection of point \(A\) in the line \(x+2y=0\).
Step 2: Find point \(B\).
Point \(A=(1,-2)\).
For the line
\[ x-y+5=0 \] we have \(a=1,\;b=-1,\;c=5\).
Using reflection formula,
\[ x'=x-\frac{2a(ax+by+c)}{a^2+b^2} \] \[ y'=y-\frac{2b(ax+by+c)}{a^2+b^2} \] Now,
\[ ax+by+c=1(1)+(-1)(-2)+5=8 \] and
\[ a^2+b^2=1^2+(-1)^2=2 \] Therefore,
\[ x'=1-\frac{2(1)(8)}{2}=1-8=-7 \] \[ y'=-2-\frac{2(-1)(8)}{2}=-2+8=6 \] So,
\[ B=(-7,6) \]
Step 3: Find point \(C\).
For the line
\[ x+2y=0 \] we have \(a=1,\;b=2,\;c=0\).
Now,
\[ ax+by+c=1(1)+2(-2)+0=-3 \] and
\[ a^2+b^2=1^2+2^2=5 \] Therefore,
\[ x'=1-\frac{2(1)(-3)}{5} \] \[ x'=1+\frac{6}{5}=\frac{11}{5} \] Also,
\[ y'=-2-\frac{2(2)(-3)}{5} \] \[ y'=-2+\frac{12}{5}=\frac{2}{5} \] So,
\[ C=\left(\frac{11}{5},\frac{2}{5}\right) \]
Step 4: Find the equation of line \(BC\).
The line passes through
\[ B=(-7,6) \] and
\[ C=\left(\frac{11}{5},\frac{2}{5}\right) \] Using two-point form,
\[ \frac{y-6}{x+7} = \frac{\frac{2}{5}-6}{\frac{11}{5}+7} \] \[ \frac{y-6}{x+7} = \frac{\frac{2-30}{5}}{\frac{11+35}{5}} \] \[ \frac{y-6}{x+7} = \frac{-28}{46} = -\frac{14}{23} \] Thus,
\[ 23(y-6)=-14(x+7) \] \[ 23y-138=-14x-98 \] \[ 14x+23y-40=0 \]
Step 5: Final conclusion.
Hence, the equation of the line joining \(B\) and \(C\) is
\[ \boxed{14x+23y-40=0} \]
Since \(x-y+5=0\) is the perpendicular bisector of \(AB\), point \(B\) is the reflection of point \(A\) in the line \(x-y+5=0\).
Similarly, since \(x+2y=0\) is the perpendicular bisector of \(AC\), point \(C\) is the reflection of point \(A\) in the line \(x+2y=0\).
Step 2: Find point \(B\).
Point \(A=(1,-2)\).
For the line
\[ x-y+5=0 \] we have \(a=1,\;b=-1,\;c=5\).
Using reflection formula,
\[ x'=x-\frac{2a(ax+by+c)}{a^2+b^2} \] \[ y'=y-\frac{2b(ax+by+c)}{a^2+b^2} \] Now,
\[ ax+by+c=1(1)+(-1)(-2)+5=8 \] and
\[ a^2+b^2=1^2+(-1)^2=2 \] Therefore,
\[ x'=1-\frac{2(1)(8)}{2}=1-8=-7 \] \[ y'=-2-\frac{2(-1)(8)}{2}=-2+8=6 \] So,
\[ B=(-7,6) \]
Step 3: Find point \(C\).
For the line
\[ x+2y=0 \] we have \(a=1,\;b=2,\;c=0\).
Now,
\[ ax+by+c=1(1)+2(-2)+0=-3 \] and
\[ a^2+b^2=1^2+2^2=5 \] Therefore,
\[ x'=1-\frac{2(1)(-3)}{5} \] \[ x'=1+\frac{6}{5}=\frac{11}{5} \] Also,
\[ y'=-2-\frac{2(2)(-3)}{5} \] \[ y'=-2+\frac{12}{5}=\frac{2}{5} \] So,
\[ C=\left(\frac{11}{5},\frac{2}{5}\right) \]
Step 4: Find the equation of line \(BC\).
The line passes through
\[ B=(-7,6) \] and
\[ C=\left(\frac{11}{5},\frac{2}{5}\right) \] Using two-point form,
\[ \frac{y-6}{x+7} = \frac{\frac{2}{5}-6}{\frac{11}{5}+7} \] \[ \frac{y-6}{x+7} = \frac{\frac{2-30}{5}}{\frac{11+35}{5}} \] \[ \frac{y-6}{x+7} = \frac{-28}{46} = -\frac{14}{23} \] Thus,
\[ 23(y-6)=-14(x+7) \] \[ 23y-138=-14x-98 \] \[ 14x+23y-40=0 \]
Step 5: Final conclusion.
Hence, the equation of the line joining \(B\) and \(C\) is
\[ \boxed{14x+23y-40=0} \]
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