Solution
Given that a line passes through the point (-1, 2) and is perpendicular to line y = 3x + 1
The general form of the equation of a line is
[tex]\begin{gathered} y=mx+b \\ \text{Where m is the slope and b is the y-intercept} \end{gathered}[/tex]Let m₁ be the slope of the given line
The slope, m₁, of the given line y = 3x + 1 is
[tex]m_1=3[/tex]Since the line passing through the point is perpendicular to the given line, the formula to find the slope of perpendicular lines is
[tex]m_2=-\frac{1}{m_1}[/tex]Let m₂ be the slope of the line perpendicular to the given line.
The slope, m₂, of the line perpendicular to the given line is
[tex]\begin{gathered} m_2=-\frac{1}{m_1} \\ \text{Where } \\ m_1=3 \\ m_2=-\frac{1}{3}_{} \end{gathered}[/tex]The slope, m₂, of the line perpendicular to the given line is -1/3
To find the equation of a straight line, the formula is
[tex]\frac{y-y_1}{x-x_1}=m_{}[/tex]Where
[tex]\begin{gathered} m=m_2=-\frac{1}{3} \\ (x_1,y_1)\Rightarrow(-1,2) \end{gathered}[/tex]Substitute the values into the formula to find the equation of a straight line
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=m_2 \\ \frac{y-2}{x-(-1)}=-\frac{1}{3} \\ \frac{y-2}{x+1}=-\frac{1}{3} \end{gathered}[/tex]Crossmultiply
[tex]\begin{gathered} \frac{y-2}{x+1}=-\frac{1}{3} \\ 3(y-2)=-1(x+1) \\ \text{Open the brackets} \\ 3y-6=-x-1 \\ 3y=-x-1+6 \\ 3y=-x+5 \\ \text{Divide both sides by 3} \\ \frac{3y}{3}=\frac{-x+5}{3} \\ y=-\frac{1}{3}x+\frac{5}{3} \end{gathered}[/tex]Hence, the slope-intercept form of the line is
[tex]y=-\frac{1}{3}x+\frac{5}{3}[/tex]