Recall that the slopes of two perpendicular lines, satisfy that:
[tex]m_1\times m_2=-1.[/tex]
Therefore, the slope of the line perpendicular to -3x-2 must-have slope
[tex]m=\frac{1}{3}\text{.}[/tex]
Now, to determine the equation of the line, we will use the following formula for the equation of a line with slope m, that passes through the point (x₁,y₁):
[tex]y-y_1=m(x-x_1)\text{.}[/tex]
Substituting (x₁,y₁)=(-7,-1/3), and m=1/3 in the above formula, we get:
[tex]y-(-\frac{1}{3})=\frac{1}{3}(x-(-7))\text{.}[/tex]
Simplifying the above result, we get:
[tex]y+\frac{1}{3}=\frac{1}{3}x+\frac{7}{3}\text{.}[/tex]
Recall that the slope-intercept form of the equation of a line is:
[tex]y=mx+b,[/tex]
where b is the y-intercept and m is the slope.
Taking the equation of the line to its slope-intercept form we get:
[tex]\begin{gathered} y=\frac{1}{3}x+\frac{7}{3}-\frac{1}{3}, \\ y=\frac{1}{3}x+\frac{6}{3}, \\ y=\frac{1}{3}x+2. \end{gathered}[/tex]
Answer:
[tex]y=\frac{1}{3}x+2.[/tex]
