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Question: The value of \(k\) such that the lines \(2x-3y+k=0\), \(3x-4y-13=0\) and \(8x-11y-33=0\) are concurr...

The value of kk such that the lines 2x−3y+k=02x-3y+k=0, 3x−4y−13=03x-4y-13=0 and 8x−11y−33=08x-11y-33=0 are concurrent, is
A. 20
B. −7-7
C. 7
D. −20-20

Explanation

Solution

We first assume the coordinates of the point in which the straight lines 2x−3y+k=02x-3y+k=0, 3x−4y−13=03x-4y-13=0 and 8x−11y−33=08x-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 kk.

Complete step by step answer:
We assume the point (h,k)\left( h,k \right) in which the straight lines 2x−3y+k=02x-3y+k=0, 3x−4y−13=03x-4y-13=0 and 8x−11y−33=08x-11y-33=0 are concurrent. It means that the point (h,k)\left( h,k \right) lies on every line of 2x−3y+k=02x-3y+k=0, 3x−4y−13=03x-4y-13=0 and 8x−11y−33=08x-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, ∣2−3k 3−4−13 8−11−33 ∣=0\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.
∣2−3k 3−4−13 8−11−33 ∣=2(132−143)+3(−99+104)+k(−33+32)=−k−7 \begin{aligned} & \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{aligned}
Now we simplify the equation −k−7=0-k-7=0.
−k−7=0 ∴k=−7 \begin{aligned} & -k-7=0 \\\ & \therefore k=-7 \\\ \end{aligned}

Hence, the correct option is B.

Note: We can also solve the lines 3x−4y−13=03x-4y-13=0 and 8x−11y−33=08x-11y-33=0 to find the intersecting point. We get
8(3x−4y−13)−3(8x−11y−33)=0 ⇒−32y−104+33y+99=0 ⇒y=58\left( 3x-4y-13 \right)-3\left( 8x-11y-33 \right)=0 \\\ \Rightarrow -32y-104+33y+99=0 \\\ \Rightarrow y=5
This gives x=11x=11. We put the value (x,y)=(11,5)\left( x,y \right)=\left( 11,5 \right) in the equation of 2x−3y+k=02x-3y+k=0 to find the value of kk.
2×11−3×5+k=0 ⇒k=15−22=−72\times 11-3\times 5+k=0 \\\ \Rightarrow k=15-22=-7