IPMAT Indore 2022 (MCQ) - The value of k for which the following lines are concurrent is x-y-1=0 2x+3y-12=0 2x-3y+k=0 | PYQs + Solutions

IPMAT Indore 2022

Geometry

Straight Lines

Medium

The value of $k$ for which the following lines are concurrent is
$x-y-1=0 \newline 2x+3y-12=0 \newline 2x-3y+k=0$

-1

✅ Correct Option: 3

To find the value of

k

for which the three lines are concurrent (pass through a common point).

For lines

a_1x + b_1y + c_1 = 0

a_2x + b_2y + c_2 = 0

, and

a_3x + b_3y + c_3 = 0

to be concurrent, their determinant must equal zero:

\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0

From our three lines:

\newline

- Line 1:

x-y-1=0

with coefficients

(1, -1, -1)

\newline

- Line 2:

2x+3y-12=0

with coefficients

(2, 3, -12)

\newline

- Line 3:

2x-3y+k=0

with coefficients

(2, -3, k)

Setting up the determinant:

\begin{vmatrix} 1 & -1 & -1 \\ 2 & 3 & -12 \\ 2 & -3 & k \end{vmatrix} = 0

Expanding along the first row:

1 \cdot \begin{vmatrix} 3 & -12 \\ -3 & k \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & -12 \\ 2 & k \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} = 0

1 \cdot (3k - (-12)(-3)) + 1 \cdot (2k - (-12)(2)) - 1 \cdot (2(-3) - 3(2)) = 0

1 \cdot (3k - 36) + 1 \cdot (2k + 24) - 1 \cdot (-6 - 6) = 0

3k - 36 + 2k + 24 + 12 = 0

5k = 0

k = 0