Question

The value of k for which the graphs of (k - 1)x + y - 2 = 0 and (2 - k)x - 3y + 1 = 0 are parallel is:

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $2$
• D. $-2$
• E. None of the above

Answer 1

The correct option is A $\frac{1}{2}$

The graphs of (k - 1)x + y - 2 = 0 and (2 - k)x - 3y + 1 = 0 are parallel.
$\therefore \frac{k-1}{2-k}=\frac{1}{-3}\Rightarrow -3k+3=2-k$
$\Rightarrow -3k+k=2-3\Rightarrow -2k=-1$
$\Rightarrow k=\frac{1}{2}$
Two straight lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are parallel if
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
Alternative method
$(k-1)x+y-2=0$
$\Rightarrow y=(1-k)x+2\dots \dots \left(1\right)$
and, $(2-k)x-3y+1=0$
$\Rightarrow 3y=(2-k)x+1$
$\therefore y=\left(\frac{2-k}{3}\right)x+\frac{1}{3}\dots \dots \left(2\right)$
$\because m_1=m_2$
$\Rightarrow 1-k=\frac{2-k}{3}\Rightarrow 3-3k=2-k$
$\therefore k=\frac{1}{2}$