Question 37·200 Super-Hard SAT Math Questions·Algebra

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Lines $p$ and $q$ are perpendicular and intersect at the point $(-1,4)$ . Line $p$ has equation $y = kx + 4 + k$ , where $k$ is a nonzero constant.

Which point lies on line $q$ ?

The cue for a perpendicular-lines question is a line equation and a shared point. Read the given line’s slope, take its negative reciprocal, and use that slope as vertical change divided by horizontal change. Scale those changes to reach a coordinate in the choices. A sign error in either change sends you along the wrong line.

Hints

Hint 1
In an equation of the form $y=mx+b$ , the slope is the number multiplying $x$ . Line $p$ has slope $k$ , so what is the negative reciprocal that line $q$ needs?
Hint 2
Three listed points have $y=1$ . From the shared point $(-1,4)$ , the vertical change to that height is down $3$ . What horizontal change would give line $q$ its required slope?
Hint 3
The slope compares vertical change with horizontal change. Check the signs before matching their sizes: if $k$ is positive, which direction must you move horizontally when you move down? What changes if $k$ is negative?

Step-by-step

Use the perpendicular slope as a coordinate change
Step 1
Find line $q$ ’s slope
No Desmos needed. The perpendicular-slope rule gives an exact coordinate change. In $y=kx+4+k$ , the slope of line $p$ is $k$ , the number multiplying $x$ . Perpendicular slopes are negative reciprocals, so line $q$ has slope $-\frac{1}{k}$ . The given nonzero $k$ makes that fraction possible.
Step 2
Find the needed vertical change
Three choices have $y=1$ , so test that height. From the shared point $(-1,4)$ , the rise, or change in $y$ , is $1-4=-3$ . The move is down $3$ , not up $3$ .
Step 3
Match the horizontal change to the slope
Slope means vertical change divided by horizontal change. The slope $-\frac{1}{k}$ pairs a change of $-1$ in $y$ with a change of $k$ in $x$ . Scale both changes by $3$ to keep the same slope. So moving down $3$ requires a horizontal change of $3k$ :
$-\frac{1}{k}=\frac{-3}{3k}.$
Step 4
Locate the point on line $q$
Add that horizontal change to the shared point’s $x$ -coordinate: $-1+3k$ . Its $y$ -coordinate is the target height, $1$ , so the point on line $q$ is $(-1+3k,1)$ . Choice C.

Question 37·200 Super-Hard SAT Math Questions·Algebra

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