Slope describes direction, so it can tell you how two lines sit relative to each other. You will use slopes and point-slope form together here.
Why it works
Parallel lines never meet, so neither can be steeper than the other: they have the same slope (and different intercepts). For perpendicular lines, picture a slope triangle with run $3$ and rise $2$, so slope $\frac{2}{3}$. Turn it a quarter turn: the run becomes $2$ and the rise becomes $-3$, so the new slope is $-\frac{3}{2}$. Turning a quarter turn always swaps rise and run and changes one sign. So perpendicular slopes are negative reciprocals, and their product is $-1$.
Key idea
Parallel: $m_1 = m_2$. Perpendicular: $m_1 \cdot m_2 = -1$, so $m_2 = -\frac{1}{m_1}$ (flip and change the sign).
Worked examples
Worked example: A parallel line
Write the line through $(3, 1)$ parallel to $y = -2x + 5$.
- Parallel means the same slope: $m = -2$.
- $y - 1 = -2(x - 3)$, so $y = -2x + 7$.
Answer: $y = -2x + 7$
Worked example: A perpendicular line
Write the line through $(4, -1)$ perpendicular to $y = \frac{2}{3}x - 1$.
- Negative reciprocal of $\frac{2}{3}$: flip it to $\frac{3}{2}$ and change the sign: $-\frac{3}{2}$.
- $y + 1 = -\frac{3}{2}(x - 4)$, so $y = -\frac{3}{2}x + 6 - 1 = -\frac{3}{2}x + 5$.
- Check: $\frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -1$.
Answer: $y = -\frac{3}{2}x + 5$
Worked example: Classify
Are $y = 4x - 1$ and $x + 4y = 8$ parallel, perpendicular, or neither?
- Second line: $4y = -x + 8$, so $y = -\frac{1}{4}x + 2$.
- Slopes $4$ and $-\frac{1}{4}$: product $-1$.
Answer: Perpendicular.
Common mistake
Doing only half of “negative reciprocal”: changing the sign but not flipping, or flipping but not changing the sign. The perpendicular slope to $\frac{2}{3}$ is neither $-\frac{2}{3}$ nor $\frac{3}{2}$; it is $-\frac{3}{2}$. Check by multiplying the two slopes: you must get $-1$.