Let's begin by identifying key information given to us:
The equation passes through the point (-1, 2)
A line is perpendicular to x + 3y = 3
[tex]\begin{gathered} x+3y=3 \\ We\text{ will rearrange the terms to have:} \\ 3y=-x+3 \\ \text{Divide through the terms by the coefficient of y, 3. We have:} \\ \frac{3y}{3}=-\frac{1}{3}x+\frac{3}{3} \\ y=-\frac{1}{3}x+1 \\ But,y=mx+b \\ \Rightarrow m=-\frac{1}{3} \\ \therefore m_{perpendicular}=-\frac{1}{3} \end{gathered}[/tex]For a perpendicular line, the slope is given as the negative reciprocal of the slope of the original line. Mathematically represented thus:
[tex]\begin{gathered} m_{perpendicular}=-\frac{1}{m} \\ \Rightarrow m=-\frac{1}{m_{perpendicular}} \\ m=-\frac{1}{-\frac{1}{3}}=\frac{3}{1} \\ \Rightarrow m=3 \end{gathered}[/tex]We will proceed to solve for the equation of the original line using the point-slope equation. We have:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ (x_1,y_1)=(-1,2) \\ m=3 \\ y-2=3(x--1) \\ y-2=3x+3 \\ y=3x+3+2 \\ y=3x+5 \end{gathered}[/tex]Therefore, the equation of the line that passes through (-1, 2) and is perpendicular to x + 3y = 3 is y = 3x + 5
