Q86 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

The equation of the perpendicular bisectors of the sides \(A B\) and \(A C\) of a \(\triangle A B C\) are \(x-y+5=0\) and \(x+2 y+5=0\), respectively. If \(A\) is \((1,-2)\), then the equation of the straight line \(B C\) is

  • Correct answer
    A
    \(14 x+23 y-40=0\)
  • B
    \(12 x+17 y-28=0\)
  • C
    \(14 x-29 y-30=0\)
  • D
    \(7 x-12 y+15=0\)

Solution

The point \(B\left(x_1, y_1\right)\) is the reflection of point \(A\) w.r.t. line \(x-y+5=0\), so \(\begin{aligned} \Rightarrow \quad \frac{x_1-1}{1} & =\frac{y_1+2}{-1} \\ & =-2 \frac{1+2+5}{2}=-8 \\ \Rightarrow \quad x_1 & =-7, y_1=6 \end{aligned}\) so, \(B(-7,6)\) Similarly point \(C\left(x_2, y_2\right)\) is the reflection of point \(A\) w.r.t., line \(x+2 y=0\), so \(\Rightarrow \quad \frac{x_2-1}{1}=\frac{y_2+2}{2}=-2 \frac{1-4}{5}=\frac{6}{5}\) \(\Rightarrow \quad x_2=\frac{11}{5}\) and \(y_2=\frac{2}{5}\) So, \(C\left(\frac{11}{5}, \frac{2}{5}\right)\). Now, equation of line \(B C\) is \(\begin{array}{cc} & y-6=\frac{6-\frac{2}{5}}{-7-\frac{11}{5}}(x+7) \\ \Rightarrow & y-6=\frac{28}{-46}(x+7) \\ \Rightarrow & y-6=\frac{-14}{23}(x+7) \\ \Rightarrow & 14 x+23 y=138-98 \\ \Rightarrow & 14 x+23 y-40=0 \end{array}\) Hence, option (1) is correct.

Hence, the correct answer is option: 'A'.

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