Answer:
[tex]y=\frac{1}{3} x+7[/tex]
Step-by-step explanation:
We need to find the equation of a line perpendicular to [tex]y=-3x-3[/tex], which passes through the point (-3, 6).
Recall that a line perpendicular to a line of the form: [tex]y=mx+b[/tex], must have a slope which is the opposite of the reciprocal of the slope of the original line. that is, a slope of the form;
[tex]slope=-\frac{1}{m}[/tex]
Then, in our case, since the original line has slope "-3", a perpendicular line to it should have a slope given by:
[tex]slope=-\frac{1}{-3} =\frac{1}{3}[/tex]
We now know the slope, and also a point for this new line, so we use the point-slope form of a line:
[tex]y-y_0=m_\perp\,(x-x_0)\\y-6=\frac{1}{3} (x-(-3))\\y-6=\frac{1}{3} x+\frac{3}{3} \\y-6=\frac{1}{3} x+1\\y=\frac{1}{3} x+7[/tex]
