Question

If the line $x-y-1=0,4x+3y=k$ and $2x-3y+1=0$ are concurrent, then $k$ is.

• A. $1$
• B. $-1$
• C. $25$
• D. $5$

Answer 1

Answer: (C)

$25$

Given, $x-y-1=0$ , $4x+3y=k$ & $2x-3y+1=0$

The given lines are concurrent i.e. all of them intersect at same point

Taking the above 2 equations:

$x-y-1=0$ -----------(i)

$2x-3y+1=0$ ----------(ii)

Multiplying 3 both side of eq. (i)

$3[x-y-1]=0$

$3x-3y-3=0$ -------------(iii)

Now, equating (ii) & (iii)

$\begin{array}{ccc}3x& -3y& -3=0\\ 2x& -3y& +1=0\end{array}$

$-+-$

$\overline{x-4=0}$

$\Rightarrow x=4$

$\therefore 2\left(4\right)-3y+1=0$

$\Rightarrow 8-3y+1=0$

$\Rightarrow 3y=9$

$\Rightarrow y=3$

$\therefore$ Point of intersection of 3 lines is (4,3)

This pt. lies on the line $4x+3y=k$

$\Rightarrow 4\left(4\right)+3\left(3\right)=k$

$k=25$