Let us find the slope of each line
slope of line l
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{5-1}{3-(-3)}=\frac{4}{6}=\frac{2}{3} \end{gathered}[/tex]slope of line m
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{2-(-4)}{6-(-3)}=\frac{6}{9}=\frac{2}{3} \end{gathered}[/tex]The slopes are the same therefore no one is greater than each other. let us find the y -intercept and x - intercept of each line if required.
y-intercept of line l is
[tex]\begin{gathered} y=mx+b \\ 1=\frac{2}{3}(-3)+b \\ b=y-\text{intercept} \\ 1=-2+b \\ 1+2=b \\ b=3 \end{gathered}[/tex]y-intercept of line m is
[tex]\begin{gathered} y=mx+b \\ -4=\frac{2}{3}(-3)+b \\ -4=-2+b \\ -4+2=b \\ b=-2 \end{gathered}[/tex]The y-intercept of line l is greater than the y-intercept of line m.
The answer is B. The x-intercept of line m is greater than the x-intercept of line l.
we can prove it is option B by making y = 0 below
[tex]\begin{gathered} \text{ line l} \\ y=mx+b \\ 0=\frac{2}{3}x+3 \\ -3=\frac{2}{3}x \\ -9=2x \\ x=-\frac{9}{2} \\ \\ \text{ line m} \\ y=mx+b \\ 0=\frac{2}{3}x-2 \\ 2=\frac{2}{3}x \\ 6=2x \\ x=\frac{6}{2} \\ x=3 \end{gathered}[/tex]