Question

The lines $x-2y+1=0$, $2x-3y-1=0$ and $3x-y+k=0$ are concurrent. The angle between the lines $3x-y+k=0$ and $mx-3y+6=0$ is $45^\circ$. If m is an integer, then $m-k=$

• A. -6
• B. 18
• C. 6
• D. -18

Hint:

To check for concurrency of three lines $L_1=0, L_2=0, L_3=0$, you can either find the intersection of two and check if it lies on the third, or you can check if the determinant of the coefficients is zero. For the angle formula involving absolute value, remember to solve for both the positive and negative cases.

Answer 1

The Correct Option is C

Step 1: Find Intersection of First Two Lines: Solve: 1) \( x - 2y = -1 \) 2) \( 2x - 3y = 1 \) Multiply (1) by 2: \( 2x - 4y = -2 \). Subtract from (2): \( (2x-3y) - (2x-4y) = 1 - (-2) \implies y = 3 \). Substitute \( y=3 \) into (1): \( x - 6 = -1 \implies x = 5 \). Intersection Point: \( (5,3) \).
Step 2: Find k using Concurrency: Since lines are concurrent, the third line passes through \( (5,3) \). \( 3(5) - 3 + k = 0 \implies 15 - 3 + k = 0 \implies k = -12 \).
Step 3: Find m using Angle Condition: Line 3: \( 3x - y - 12 = 0 \). Slope \( m_1 = 3 \). Line 4: \( mx - 3y + 6 = 0 \). Slope \( m_2 = m/3 \). Angle is \( 45^\circ \), so \( \tan 45^\circ = 1 \). \( |\frac{m_1 - m_2}{1 + m_1 m_2}| = 1 \) \( |\frac{3 - m/3}{1 + 3(m/3)}| = 1 \) \( |\frac{(9-m)/3}{1+m}| = 1 \implies |9-m| = 3|1+m| \). Case 1: \( 9-m = 3(1+m) \implies 9-m = 3+3m \implies 4m=6 \implies m=1.5 \) (Not integer). Case 2: \( 9-m = -3(1+m) \implies 9-m = -3-3m \implies 2m = -12 \implies m = -6 \) (Integer).
Step 4: Calculate \( m-k \): \( m - k = (-6) - (-12) = -6 + 12 = 6 \).