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Q.
The value of $p$, so that the three lines $3 x+y-2=0$, $p x+2 y-3=0$ and $2 x-y-3=0$ may intersect at one point, is
Straight Lines
Solution:
Given, equation of lines are
$3 x+y-2 =0 .....$(i)
$p x+2 y-3 =0 .....$(ii)
and $2 x-y-3 =0 .....$(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.,
$p \times 1+2(-1)-3 =0 (\text { put } x=1, y=-1)$
$p =5$