hello
the points given were
[tex]3,1[/tex]and the equation is parallel to the line
[tex]3x-y-4=0[/tex]let's rearange this equation
[tex]y=3x-4[/tex]first we find the slope of the line using slope-intercept form
from the equation above, the slope and intercept are
[tex]\begin{gathered} \text{slope(M)=3}_{} \\ \text{ intercept(c)=-4} \end{gathered}[/tex]now the equation of the line passing through a point (x1, y1) is given as
[tex]y-y_1=m(x-x_1)[/tex]now the points of the equation are (3, 1)
[tex]\begin{gathered} x_1=3 \\ y_1=1 \\ \text{slope(m)}=3 \end{gathered}[/tex][tex]\begin{gathered} y-1=3(x-3) \\ y-1=3x-9 \\ y=3x-9+1 \\ y=3x-8 \end{gathered}[/tex]the equation of the line is given as y = 3x - 8