Question

The value of $p$, so that the three lines $3x+y-2=0$, $px+2y-3=0$ and $2x-y-3=0$ may intersect at one point, is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (D)

Given, equation of lines are
$3x+y-2=\mathrm{0.....}$(i)
$px+2y-3=\mathrm{0.....}$(ii)
and $2x-y-3=\mathrm{0.....}$(iii)
Adding Eqs. (i) and (iii), we get
$5x-5=0$
$\Rightarrow x=1$
$\therefore$ From Eq. (i), $3\times 1+y-2=0$ $($ put $x=1)$
$\Rightarrow y=-1$
So, the point is $(1,-1)$.
$\because$ Lines are concurrent
Point $(1,-1)$ will satisfy Eq. (ii) i.e.,
$p\times 1+2(-1)-3=0(\text{ put }x=1,y=-1)$
$p=5$