Question

If the lines \(x+y-2=0\), \(3x-4y+1=0\) and \(5x+ky-7=0\) are concurrent at \((\alpha, \beta)\), then equation of the line concurrent with the given lines and perpendicular to \(kx+y-k=0\) is

• A. \( x-3y=-2 \)
• B. \( x+4y=5 \)
• C. \( x+6y=7 \)
• D. \( x-2y=-1 \)

Hint:

1. Concurrent lines intersect at one point. Find this point using two equations. 2. Substitute the intersection point into the third equation to find unknowns like \(k\). 3. Slope of \(Ax+By+C=0\) is \(-A/B\). Perpendicular lines have slopes \(m_1, m_2\) such that \(m_1 m_2 = -1\). 4. Equation of line through \( (x_1,y_1) \) with slope \(m\) is \(y-y_1=m(x-x_1)\).

Answer 1

The Correct Option is D

Point of concurrence \((\alpha, \beta)\) from first two lines: \(L_A: x+y-2=0 \implies x=2-y\) \(L_B: 3x-4y+1=0\) Substitute \(x\) into \(L_B\): \(3(2-y)-4y+1=0 \implies 6-3y-4y+1=0 \implies 7-7y=0 \implies y=1\).
Then \(x=2-1=1\).
So \((\alpha, \beta) = (1,1)\).
Third line \(L_C: 5x+ky-7=0\) passes through \((1,1)\): \(5(1)+k(1)-7=0 \implies 5+k-7=0 \implies k-2=0 \implies k=2\).
The line to which the required line is perpendicular is \(L_P: kx+y-k=0\).
With \(k=2\), \(L_P: 2x+y-2=0\).
Slope of \(L_P\) is \(m_P = -2/1 = -2\).
The required line has slope \(m_R = -1/m_P = -1/(-2) = 1/2\).
The required line passes through the point of concurrence \((1,1)\) and has slope \(1/2\).
Equation: \(y-1 = \frac{1}{2}(x-1)\) \[ 2(y-1) = x-1 \implies 2y-2 = x-1 \] \[ x-2y+1=0 \implies x-2y=-1 \] This matches option (4).