Q83 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

The equation of the straight line whose slope is $\frac{-2}{3}$ and which divides the line segment joining $(1,2),(-3,5)$ in the ratio $4: 3$ externally is

  • Correct answer
    A
    $2 x+3 y-12=0$
  • B
    $3 x+2 y+27=0$
  • C
    $2 x+3 y-9=0$
  • D
    $2 x+3 y+12=0$

Solution

Let $P\left(x_1, y_2\right)$ be the point of which divide line segment in $4: 3$ extermally $\left(x_1, y_1\right)=\left(\frac{-12-3}{1}, \frac{20-6}{1}\right)=(-15,14)$ Equation of line having slope $\frac{-2}{3}$ and posses through $\begin{aligned} & (-15,14) \text { is ; } y-14=\frac{-2}{3}(x+15) \Rightarrow 3 y-42=-2 x-30 \\ & 2 x+3 y-12=0\end{aligned}$

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

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