One of the following points belong to the lines $3x+y=4,6x+2y=8$  is

Given equation :

$3x+y=4\Rightarrow 3x+y-4=0$

$6x+2y=8\Rightarrow 6x+2y-8=0$

Here $a_1=3;b_1=1;c_1=-4$

          $a_2=6;b_2=2;c_2=-8$

$\frac{a_1}{a_2}=\frac{3}{6};\frac{b_1}{b_2}=\frac{1}{2};\frac{c_1}{c_2}=\frac{-4}{-8}=\frac{1}{2}$

$=\frac{1}{2}$

$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\therefore$    The pair of lines have in finite solutions (coincident lines) in $3x+y=4$

$a\to ifx=0$

$3\left(0\right)+y=4$

$0+y=4$

$y=4$

$\therefore \left(0,4\right)$  belongs to the lines

$b\to ifx=3$

$3\left(3\right)+y=4$

$9+y=4$

$y=4-9$

$y=-5$

$\therefore \left(3,-5\right)$  belongs to the lines

$c\to x=-1$

$3\left(-1\right)+y=4$

$-3+y=4$

$y=4+3$

$y=7$

$\therefore \left(-1,7\right)$  belongs to the lines

$d\to ifx=2$

$3\left(2\right)+y=4$

$6+y=4$

$y=4-6$

$y=-2$

$\therefore \left(2,-2\right)$  belongs to the lines