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: Understanding the Concept
The problem involves two main geometric concepts: concurrency of lines and the angle between two lines. Concurrency means the three lines intersect at a single common point.
Step 2: Key Formula or Approach
1. Find the point of intersection of the first two lines by solving them as a system of linear equations. 2. Since the three lines are concurrent, this point of intersection must also lie on the third line. Substitute the coordinates of the point into the third line's equation to find the value of $k$. 3. Use the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$: $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$. 4. Solve the resulting equation for $m$, keeping in mind that $m$ must be an integer. 5. Calculate the final expression $m-k$.
Step 3: Detailed Explanation
1. Find the point of concurrency: We solve the system: \[ x-2y+1=0 \quad \text{(1)} \] \[ 2x-3y-1=0 \quad \text{(2)} \] From (1), $x = 2y-1$. Substitute this into (2): \[ 2(2y-1) - 3y - 1 = 0 \] \[ 4y - 2 - 3y - 1 = 0 \implies y - 3 = 0 \implies y = 3 \] Substitute $y=3$ back into $x=2y-1$: \[ x = 2(3) - 1 = 5 \] The point of concurrency is $(5,3)$. 2. Find k: The point $(5,3)$ lies on the third line, $3x-y+k=0$. \[ 3(5) - (3) + k = 0 \implies 15 - 3 + k = 0 \implies 12+k=0 \implies k=-12 \] 3. Find m: The two lines are $L_3: 3x-y-12=0$ and $L_4: mx-3y+6=0$. The angle between them is $45^\circ$. Slope of $L_3$: $m_3 = -(\text{coeff } x)/(\text{coeff } y) = -3/(-1) = 3$. Slope of $L_4$: $m_4 = -m/(-3) = m/3$. Using the angle formula with $\tan 45^\circ = 1$: \[ 1 = \left|\frac{m_3-m_4}{1+m_3m_4}\right| = \left|\frac{3 - m/3}{1 + 3(m/3)}\right| = \left|\frac{(9-m)/3}{1+m}\right| = \left|\frac{9-m}{3(1+m)}\right| \] This leads to two possibilities: Case (a): $\frac{9-m}{3(1+m)} = 1 \implies 9-m = 3+3m \implies 6 = 4m \implies m = 3/2$. This is not an integer. Case (b): $\frac{9-m}{3(1+m)} = -1 \implies 9-m = -3(1+m) \implies 9-m = -3-3m \implies 2m = -12 \implies m = -6$. This is an integer. So, we must have $m=-6$. 4. Calculate m-k: \[ m-k = (-6) - (-12) = -6 + 12 = 6 \] Step 4: Final Answer
The values are $k=-12$ and $m=-6$. The required value is $m-k=6$.