Question

The lines $x+y-1=0,(m-1)x+(m^2-7)y-5=0$ and $(m-2)x+(2m-5)y=0$ are

• A. concurrent for three values of $m$
• B. concurrent for one value of $m$
• C. concurrent for no value of $m$
• D. are parallel for m $=3$

Answer 1

Answer: (C), (D)

The correct options are
C concurrent for no value of $m$
D are parallel for m $=3$

$x+y-1=0$

$(m-1)x+(m^2-7)y-5=0$

$(m-2)x+(2m-5)y=0$

Lines are concurrent if the following determinant is zero

$\left|\begin{array}{ccc}1& 1& -1\\ m-1& m^2-7& -5\\ m-2& 2m-5& 0\end{array}\right|$

$=5(2m-5)-1(0+(5(m-2))-1((2m-5)(m-1)-(m^2-7)(m-2))$

$=5m-15-4m^2-19+m^3$

$m^3-4m^2+5m-34=0$

No real values of $m$

For parallel condition

$\frac{1}{m-1}=\frac{1}{m^2-7}$

$m^2-7-m-1=0$

$m^2-m-6=0$

$m^2-3m+2m-6=0$

$m=3,-2$

negative values are not possible,hence $m=3$