Can you work out problem #11 of section four in Official SAT #5?

Sure. The math concept you need to know here is that perpendicular lines have negative reciprocal slopes. For example, if one line has a slope of \frac{7}{2}, then a line perpendicular to it must have a slope of -\frac{2}{7}.

In this case, none of the lines are given to you in slope-intercept form, so your first step should be to put at least the first line you’re given into that form.

    \begin{align*}-2x+3y&=6\\3y&=2x+6\\y&=\frac{2}{3}x+2\end{align*}

That tells you that the slope of the given line is \frac{2}{3}, so the slope you need to find in the answer choices is -\frac{3}{2}.

They do you a real favor in the answer choices by only using 3 and 2 as coefficients together in one choice. So just from the get-go you should be leaning choice A. Let’s just put it in slope-intercept form to be sure.

    \begin{align*}3x+2y&=6\\2y&=-3x+6\\y&=-\frac{3}{2}x+3\end{align*}

Yep, that’s what we wanted. We’re all done!