Q58 / 160 · TS EAMCET 2003 · Mathematics

If the lines $4 x+3 y-1=0, x-y+5=0$ and $k x+5 y-3=0$ are concurrent, then $k$ is equal to :

  • A
    4
  • B
    5
  • Correct answer
    C
    6
  • D
    7

Solution

Since, the given three lines are concurrent, then $\left|\begin{array}{rrr}4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3\end{array}\right|=0$ $\Rightarrow 4(3-25)-3(-3-5 k)-1(5+k)=0$ $\Rightarrow-88+9+15 k-5-k=0$ $\Rightarrow \quad 14 k=84$ $\therefore \quad k=6$

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

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