Q206 / 270 · NDA 2017 (Phase 2) · Mathematics

The equation of the line passing through the point $(2,3)$ and the point of intersection of lines $2 x-3 y+7=0$ and $7 x+4 y+2=0$ is

  • A
    $21 \mathrm{x}+46 \mathrm{y}-180=0$
  • Correct answer
    B
    $21 x-46 y+96=0$
  • C
    $46 x+21 y+155=0$
  • D
    $46 x-21 y-29=0$

Solution

Required line is $(2 x-3 y+7)+k(7 x+4 y+2)=0$ Given that this line passes through $(2,3)$ $\begin{aligned} & \therefore(2(2)-3(3)+7)+\mathrm{k}(7(2)+4(3)+2)=0 \\ & \Rightarrow 4-9+7+\mathrm{k}(14+12+2)=0 \\ & \Rightarrow 2+\mathrm{k}(28)=0\end{aligned}$ $\Rightarrow 28 \mathrm{k}=-2 \Rightarrow \mathrm{k}=\frac{-1}{14}$ $\therefore$ Required line $\begin{aligned} & \Rightarrow(2 x-3 y+7)+\left(\frac{-1}{14}\right)(7 x+4 y+2)=0 \\ & \Rightarrow 28 x-42 y+98-7 x-4 y-2=0 \\ & \Rightarrow 21 x-46 y+96=0\end{aligned}$

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

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