Linear Equations in Two Variables Pattern - Parallel Perpendicular

Finding the slope of a line parallel or perpendicular to a given line

This pattern tests two rules about slopes in the $xy$-plane: parallel lines have equal slopes, and perpendicular lines have slopes that are negative reciprocals of each other. If one line has slope $m$, a parallel line also has slope $m$, and a perpendicular line has slope $-\dfrac{1}{m}$. On easier questions the slope is handed to you in $y = mx + b$ form; on harder ones you must first extract it from standard form ($Ax + By = C$) or from unusual arrangements, and on the hardest questions you use the slope to build a full equation and find a missing coordinate.

 

Parallel Lines — Slope from $y = mx + b$

When the given line is already in slope-intercept form, finding the slope of a parallel line is immediate — just read off the coefficient of $x$.

The equation $y = 1.5x - 2$ defines line $g$. Line $h$ is parallel to line $g$. What is the slope of line $h$?

The slope of $g$ is $1.5$ (the coefficient of $x$). Parallel lines share the same slope, so the slope of $h$ is $1.5$ (or equivalently $\dfrac{3}{2}$).

Sometimes the question asks you to write the full equation of the parallel line, given a point it passes through — usually the $y$-intercept:

What is the equation of the line parallel to $y = -3x + 2$ that passes through $(0, -4)$?

The slope of the given line is $-3$. A parallel line has the same slope. Since the new line passes through $(0, -4)$, its $y$-intercept is $-4$. Plug into $y = mx + b$: the equation is $y = -3x - 4$.

 

Perpendicular Lines — Slope from $y = mx + b$

For perpendicular lines, flip the fraction and change the sign. If the original slope is $\dfrac{a}{b}$, the perpendicular slope is $-\dfrac{b}{a}$.

Line $L$ has the equation $y = \dfrac{1}{4}x - 2$. Line $M$ is perpendicular to $L$. What is the slope of line $M$?

The slope of $L$ is $\dfrac{1}{4}$. The negative reciprocal is $-\dfrac{1}{1/4} = -4$. The slope of $M$ is $-4$.

Line $p$ has the equation $y = \dfrac{5}{8}x - 2$. Line $q$ is perpendicular to $p$. What is the slope of line $q$?

The slope of $p$ is $\dfrac{5}{8}$. Flip and negate: $-\dfrac{8}{5}$. The slope of $q$ is $-\dfrac{8}{5}$ (or $-1.6$).

 

Extracting the Slope from Standard Form

Many medium and hard questions give the equation in a form like $Ax + By = C$ or a rearranged variant. You must solve for $y$ first to read off the slope. The quick shortcut: for $Ax + By = C$, the slope is $-\dfrac{A}{B}$.

Line $f$ has the equation $8y - 3x = 1$. Line $g$ is parallel to $f$. What is the slope of line $g$?

Solve for $y$: $8y = 3x + 1$, so $y = \dfrac{3}{8}x + \dfrac{1}{8}$. The slope of $f$ is $\dfrac{3}{8}$. Since $g$ is parallel, the slope of $g$ is $\dfrac{3}{8}$.

Line $k$ has the equation $5x - 4y = 12$. Line $j$ is perpendicular to $k$. What is the slope of line $j$?

Solve for $y$: $-4y = -5x + 12$, so $y = \dfrac{5}{4}x - 3$. The slope of $k$ is $\dfrac{5}{4}$. The perpendicular slope is $-\dfrac{4}{5}$. The slope of $j$ is $-\dfrac{4}{5}$.

Watch for non-standard arrangements where $x$ is isolated instead of $y$:

Line $m$ is defined by $x = 3y + 5$. Line $n$ is perpendicular to $m$. What is the slope of line $n$?

Solve for $y$: $x - 5 = 3y$, so $y = \dfrac{1}{3}x - \dfrac{5}{3}$. The slope of $m$ is $\dfrac{1}{3}$. The perpendicular slope is $-3$. The slope of $n$ is $-3$.

 

Fractional Coefficients in Standard Form (Harder)

Hard questions use fractional coefficients that create messy algebra. The process is exactly the same — isolate $y$ and simplify.

The equation $\dfrac{2}{5}x + \dfrac{3}{4}y - 10 = 0$ defines line $a$. Line $b$ is perpendicular to $a$. What is the slope of line $b$?

Isolate the $y$-term: $\dfrac{3}{4}y = -\dfrac{2}{5}x + 10$. Multiply both sides by $\dfrac{4}{3}$:

$y = -\dfrac{4}{3} \cdot \dfrac{2}{5}\,x + \dfrac{4}{3} \cdot 10 = -\dfrac{8}{15}x + \dfrac{40}{3}$

The slope of $a$ is $-\dfrac{8}{15}$. The perpendicular slope is the negative reciprocal: $-\dfrac{1}{-8/15} = \dfrac{15}{8}$. The slope of $b$ is $\dfrac{15}{8}$.

 

Finding a Missing Coordinate

Some questions go a step further: after finding the parallel or perpendicular slope, you must use a given point to write the full equation, then plug in a second point to find an unknown coordinate.

Line $h$ passes through the origin and is parallel to $y = \dfrac{2}{3}x - 1$. If $h$ also passes through $(a, 10)$, what is the value of $a$?

The given line has slope $\dfrac{2}{3}$. Parallel means same slope. Since $h$ passes through $(0, 0)$, its equation is $y = \dfrac{2}{3}x$. Substitute the point $(a, 10)$:

$10 = \dfrac{2}{3}a \implies a = 10 \cdot \dfrac{3}{2} = 15$

The value of $a$ is $15$.

Here's one using perpendicular lines and a point that isn't the origin:

Line $m$ is perpendicular to $y = 2x + 9$ and passes through $(4, 1)$. If $m$ also passes through $(7, k)$, what is the value of $k$?

The given slope is $2$. The perpendicular slope is $-\dfrac{1}{2}$. Use point-slope form with $(4, 1)$:

$y - 1 = -\dfrac{1}{2}(x - 4) \implies y = -\dfrac{1}{2}x + 2 + 1 = -\dfrac{1}{2}x + 3$

Substitute $x = 7$: $k = -\dfrac{7}{2} + 3 = -\dfrac{7}{2} + \dfrac{6}{2} = -\dfrac{1}{2}$

The value of $k$ is $-\dfrac{1}{2}$ (or $-0.5$).

 

The SLOPE RELATIONSHIP Method

1. Get the slope of the given line. If the equation is in $y = mx + b$ form, just read $m$. If it's in $Ax + By = C$ form, the slope is $-\dfrac{A}{B}$. For any other arrangement, solve for $y$ first.
2. Apply the rule. Parallel → same slope. Perpendicular → flip and negate (negative reciprocal).
3. If the problem asks for a coordinate, use the slope from step 2 and a given point to write the equation of the new line ($y = mx + b$ or point-slope form), then substitute the second point to solve for the unknown.

 

Watch Out For

• Confusing parallel and perpendicular. Parallel means same slope; perpendicular means negative reciprocal. Common wrong answers use the reciprocal without negating, or use the negative without reciprocating.
• Reading the slope straight from standard form. In $5x + 2y = 11$, the slope is NOT $5$ — it's $-\dfrac{5}{2}$. You must isolate $y$ (or use $-\dfrac{A}{B}$).
• Sign errors on negative reciprocals. If the slope is $-\dfrac{1}{3}$, the negative reciprocal is $+3$, not $-3$. The two negatives cancel. Double-check: the product of perpendicular slopes must equal $-1$.
• Horizontal and vertical lines. The line $y = 8$ has slope $0$. A parallel line also has slope $0$. A perpendicular line would be vertical (undefined slope), though this is rarely tested.