Write the equation of a line through a given point parallel or perpendicular to a given line, and write and solve equations involving direct variation (TEKS A.2D, A.2E, A.2F).

What this topic is asking

TEKS A.2D, A.2E, and A.2F extend equation-writing to two special relationships. A.2E and A.2F ask for lines parallel or perpendicular to a given line through a stated point, which turns on the slope relationships. A.2D asks you to write and solve direct variation, the proportional relationship $y = kx$. All three are reliable points in Reporting Category 2, and the direct-variation formula must be memorized because it is not on the reference sheet.

Parallel lines: equal slopes

Two non-vertical lines are parallel exactly when they have the same slope and different $y$-intercepts. To write a line parallel to $y = 3x - 2$ through $(1, 4)$: keep $m = 3$, then point-slope gives $y - 4 = 3(x - 1)$, so $y = 3x + 1$.

Perpendicular lines: negative-reciprocal slopes

Two lines are perpendicular when their slopes multiply to $-1$, that is, each slope is the negative reciprocal of the other. If one slope is $m$, the perpendicular slope is $-\frac{1}{m}$: flip the fraction and change the sign.

A common slip is doing only half the transformation: negating without flipping, or flipping without negating. Both steps are required.

Direct variation: y equals kx

Direct variation means $y$ is a constant multiple of $x$: $y = kx$, where $k$ is the constant of proportionality. Its graph is a line through the origin with slope $k$. Find $k$ from any known pair using $k = \frac{y}{x}$, then use the relationship to predict.

For $y$ varying directly with $x$, with $y = 21$ when $x = 7$: $k = \frac{21}{7} = 3$, so $y = 3x$. When $x = 12$, $y = 36$.

How STAAR examines this topic

• Multiple choice. Identify a parallel or perpendicular line, with "flip only" and "negate only" distractors, or recognize a direct-variation table (constant ratio $\frac{y}{x}$).
• Number entry and equation editor. Solve a direct-variation problem, or build the equation of a parallel or perpendicular line.
• Inline choice. Classify a pair of lines as parallel, perpendicular, or neither.

A clarifying idea is that direct variation is just a line with $b = 0$, so it shares everything with slope-intercept form except that it is pinned to the origin. That is why a relationship that does not pass through $(0, 0)$ is not a direct variation, even if it is linear.

Handling equations not already solved for y

When a line is given in standard form, find its slope before comparing. Rewrite $4x + 2y = 6$ as $y = -2x + 3$, so the slope is $-2$; a parallel line has slope $-2$ and a perpendicular line has slope $\frac{1}{2}$. Skipping this step and reading a coefficient straight from standard form is a common error, because the slope is not simply the number in front of $x$ in $Ax + By = C$ (it is $-\frac{A}{B}$). Solving for $y$ first removes all ambiguity.

For direct variation, the same care applies to recognizing it from data: compute $\frac{y}{x}$ for every pair, not just one. If the ratio is constant across the whole table, it is direct variation and that constant is $k$; if the ratio changes, the relationship may still be linear but with a nonzero intercept, which is not direct variation. This distinction is exactly what an inline-choice item testing "direct variation or not" is checking.

Try this

Q1. Write the line parallel to $y = -2x + 3$ through $(0, 5)$. [1 point]

• Cue. Same slope $-2$, intercept 5: $y = -2x + 5$.

Q2. $y$ varies directly with $x$; $y = 10$ when $x = 4$. Find $y$ when $x = 6$. [2 points]

• Cue. $k = \frac{10}{4} = 2.5$, so $y = 2.5(6) = 15$.