Q84 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

If the lines $x+y-1=0, k x+2 y+1=0$ and $4 x+2 k y+7$ $=0$ are concurrent, then $k=$

  • A
    2
  • B
    $\frac{13}{2}$
  • Correct answer
    C
    $\frac{-13}{2}$
  • D
    $-2$

Solution

$\left|\begin{array}{ccc}1 & 1 & -1 \\ k & 2 & 1 \\ 4 & 2 k & 7\end{array}\right|=0$ $\begin{aligned} & \Rightarrow \quad(14-2 k)-(7 k-4)-\left(2 k^2-8\right)=0 \\ & \Rightarrow \quad 2 k^2+9 k-26=0 \\ & \Rightarrow \quad 2 k^2+13 k-4 k-26=0 \\ & \Rightarrow \quad k(2 k+13)-2(2 k+13)=0 \\ & \Rightarrow \quad(2 k+13)(k-2)=0 \\ & \Rightarrow \quad k=2, \frac{-13}{2}\end{aligned}$ But at $k=2$, lines will be coincident $\therefore \quad k=\frac{-13}{2}$.

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

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