By definition, when the line cuts the x-axis, the value of "y" is:
[tex]y=0[/tex]Therefore, to find the x-intercept of the line given in the exercise, you must substitute that value of "y" into the equation and solve for "x":
[tex]\begin{gathered} 0=3x+12 \\ -12=3x \\ \frac{-12}{3}=x \\ x=-4 \end{gathered}[/tex]The equation of a line in Slope-Intercept form is:
[tex]y=mx+b[/tex]Where "m" is the slope and "b" is the y-intercept.
Kwnowing that the line is:
[tex]y=3x+12[/tex]You can identify that the y-intercept is:
[tex]b=12[/tex]So, you have these points:
[tex](-4,0);(0,12)[/tex]The formula used to find the distance between two points, is:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]In this case:
[tex]\begin{gathered} x_2=0 \\ x_1=-4 \\ y_2=12 \\ y_1=0 \end{gathered}[/tex]Substituting values and evaluating, you get that the distance between the y-intercept and x-intercept of the line, is:
[tex]d=\sqrt[]{(0-(-4))^2+(12-0)^2}=\sqrt[]{16+144}=\sqrt[]{160}[/tex]The answer is the option (2).
