Question

The value of $k$ such that the lines $2x-3y+k=0$, $3x-4y-13=0$ and $8x-11y-33=0$ are concurrent, is
A. 20
B. $-7$
C. 7
D. $-20$

Answer 1

Hint: We first assume the coordinates of the point in which the straight lines $2x-3y+k=0$, $3x-4y-13=0$ and $8x-11y-33=0$ are concurrent. The area created by the lines at that point is 0 and using the condition of the coefficient matrix of the equations, we find the determinant value 0. We solve the equation and find the value of $k$.

Complete step by step answer:
We assume the point $\left( h,k \right)$ in which the straight lines $2x-3y+k=0$, $3x-4y-13=0$ and $8x-11y-33=0$ are concurrent. It means that the point $\left( h,k \right)$ lies on every line of $2x-3y+k=0$, $3x-4y-13=0$ and $8x-11y-33=0$. The point satisfies the equations.The three lines represent the same point with area being 0 when the coefficient matrix of the equations has determinant value 0 which means the matrix is singular matrix.Therefore, $\left| \begin{matrix}
   2 & -3 & k \\
   3 & -4 & -13 \\
   8 & -11 & -33 \\
\end{matrix} \right|=0$.

We now need to simplify the determinant by expanding through the first column.
$\begin{align}
  & \left| \begin{matrix}
   2 & -3 & k \\
   3 & -4 & -13 \\
   8 & -11 & -33 \\
\end{matrix} \right| =2\left( 132-143 \right)+3\left( -99+104 \right)+k\left( -33+32 \right)=-k-7 \\
\end{align}$
Now we simplify the equation $-k-7=0$.
$\begin{align}
  & -k-7=0 \\
 & \therefore k=-7 \\
\end{align}$

Hence, the correct option is B.

Note: We can also solve the lines $3x-4y-13=0$ and $8x-11y-33=0$ to find the intersecting point. We get
$8\left( 3x-4y-13 \right)-3\left( 8x-11y-33 \right)=0 \\
\Rightarrow -32y-104+33y+99=0 \\
\Rightarrow y=5 $
This gives $x=11$. We put the value $\left( x,y \right)=\left( 11,5 \right)$ in the equation of $2x-3y+k=0$ to find the value of $k$.
$2\times 11-3\times 5+k=0 \\
\Rightarrow k=15-22=-7 $