The general form of the equation of a straight line in the slope-intercept form is given to be:
[tex]y=mx+b[/tex]where m is the slope of the line and b is the intercept on the y-axis, that is the y-value when x = 0.
To calculate the slope, we can use the formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]From the question provided, we have the points:
[tex]\begin{gathered} 1\Rightarrow(x_1,y_1)=(-1,-2) \\ 2\Rightarrow(x_2,y_2)=(3,4) \end{gathered}[/tex]Using these values, we can calculate the slope to be:
[tex]m=\frac{4-(-2)}{3-(-1)}=\frac{4+2}{3+1}=\frac{6}{4}=\frac{3}{2}[/tex]Hence, we have the equation of the line to be:
[tex]y=\frac{3}{2}x+b[/tex]We currently do not have a value for b. We can find b, however, by substituting a point into the equation. Then we can solve for b. Using the second point, we have:
[tex]\begin{gathered} 2\Rightarrow(x,y)=(3,4) \\ \therefore \\ 4=\frac{3}{2}(3)+b \\ 4=\frac{9}{2}+b \\ b=4-\frac{9}{2} \\ b=-\frac{1}{2} \end{gathered}[/tex]Therefore, the equation of the line is:
[tex]\begin{gathered} y=\frac{3}{2}x-\frac{1}{2} \\ or \\ y=1.5x-0.5 \end{gathered}[/tex]