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5.4 Parallel & Perpendicular Lines and Direct Variation

Key Takeaways

  • Parallel lines lie in the same coordinate plane and never intersect, possessing identical slopes (m1=m2m_1 = m_2) and distinct yy-intercepts (b1≠b2b_1 \neq b_2).

  • Perpendicular lines intersect at a right angle (90∘90^\circ) and have negative reciprocal slopes (m1⋅m2=−1  ⟺  m2=−1m1m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}); any horizontal line (m=0m = 0) is perpendicular to any vertical line (m=undefinedm = \text{undefined}).

  • Direct variation represents a strictly proportional linear relationship modeled by y=kxy = kx (or yx=k\frac{y}{x} = k), where k≠0k \neq 0 is the constant of variation.

  • The graph of a direct variation relationship is always a straight line passing through the origin (0,0)(0, 0) with a yy-intercept of b=0b = 0.

  • Proportional linear relationships (y=kxy = kx) maintain a constant ratio yx=k\frac{y}{x} = k across all pairs, whereas non-proportional linear equations (y=mx+by = mx + b with b≠0b \neq 0) do not.

Last updated: September 2026

Parallel & Perpendicular Lines and Direct Variation

Quick Summary: Parallel lines have identical slopes (m1=m2m_1 = m_2) and different yy-intercepts, meaning they never intersect. Perpendicular lines intersect at right angles (90∘90^\circ) and have negative reciprocal slopes (m1⋅m2=−1m_1 \cdot m_2 = -1). Direct variation (y=kxy = kx) describes proportional relationships where the line passes directly through the origin (0,0)(0,0) with a constant ratio yx=k\frac{y}{x} = k.

Geometric relationships between lines are governed entirely by their slopes. On the HiSET, you will frequently need to write equations for lines that are parallel or perpendicular to a given line, or solve word problems based on proportional direct variation.


1. Parallel Lines in the Coordinate Plane

Two non-vertical lines in a plane are parallel (denoted L1∥L2L_1 \parallel L_2) if and only if they have the exact same slope and different yy-intercepts:

m1=m2andb1≠b2m_1 = m_2 \quad \text{and} \quad b_1 \neq b_2
   Parallel Lines (m₁ = m₂ = 2/3):           Perpendicular Lines (m₁ = 2/3, m₂ = -3/2):
          y                                         y
          │      / Line 1: y = (2/3)x + 3           │      / Line 1: y = (2/3)x + 1
          │     /                                   │     / 
          │    /   / Line 2: y = (2/3)x - 1         │    / ┌─┐
     ─────┼───/───/──────► x                   ─────┼───/──┴─┼──► x
          │  /   /                                  │  / \   │
          │ /   /                                   │ /   \  ▼ Line 2: y = -(3/2)x - 2
          │/   /                                    │/     \ (90° Right Angle)

Coincident Lines vs. Parallel Lines

  • If m1=m2m_1 = m_2 and b1≠b2b_1 \neq b_2, the lines are parallel and have zero points of intersection.
  • If m1=m2m_1 = m_2 and b1=b2b_1 = b_2, the equations represent the exact same line (coincident lines), sharing infinitely many solutions.

Horizontal and Vertical Parallel Lines

  • Any two horizontal lines (y=c1y = c_1 and y=c2y = c_2) are parallel because both have slope m=0m = 0.
  • Any two vertical lines (x=k1x = k_1 and x=k2x = k_2) are parallel because both have an undefined slope.

2. Perpendicular Lines in the Coordinate Plane

Two non-vertical lines are perpendicular (denoted L1⊥L2L_1 \perp L_2) if they intersect at a right angle (90∘90^\circ). Mathematically, their slopes are negative reciprocals (also called opposite reciprocals) of each other:

m1⋅m2=−1⟺m2=−1m1m_1 \cdot m_2 = -1 \quad \Longleftrightarrow \quad m_2 = -\frac{1}{m_1}

The Negative Reciprocal Rule

To find the perpendicular slope of any given non-zero slope mm:

  1. Invert the fraction (take the reciprocal): Flip the numerator and denominator (ab→ba\frac{a}{b} \to \frac{b}{a}).
  2. Change the sign: Make positive numbers negative, and negative numbers positive.
Original Slope (m1m_1)Reciprocal (1m1\frac{1}{m_1})Perpendicular Slope (m2=−1m1m_2 = -\frac{1}{m_1})Verification (m1⋅m2m_1 \cdot m_2)
23\frac{2}{3}32\frac{3}{2}−32-\frac{3}{2}(23)(−32)=−1\left(\frac{2}{3}\right)\left(-\frac{3}{2}\right) = -1
−45-\frac{4}{5}−54-\frac{5}{4}+54+\frac{5}{4}(−45)(54)=−1\left(-\frac{4}{5}\right)\left(\frac{5}{4}\right) = -1
3=313 = \frac{3}{1}13\frac{1}{3}−13-\frac{1}{3}(3)(−13)=−1(3)\left(-\frac{1}{3}\right) = -1
−1=−11-1 = -\frac{1}{1}−1-1+1+1(−1)(1)=−1(-1)(1) = -1
00 (Horizontal: y=cy = c)UndefinedUndefined\text{Undefined} (Vertical: x=kx = k)Horizontal and vertical lines are perpendicular
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Geometric Relationships Between Two Lines

Step-by-Step Writing of Parallel and Perpendicular Equations

Worked Example 1: Line Parallel to a Given Equation

Write the equation of the line in slope-intercept form that is parallel to 6x−2y=146x - 2y = 14 and passes through the point (2,−5)(2, -5).

  1. Find the slope of the given line by solving for yy: −2y=−6x+14  ⟹  y=3x−7-2y = -6x + 14 \implies y = 3x - 7 The given slope is m=3m = 3.
  2. Identify the parallel slope: m∥=3m_{\parallel} = 3
  3. Use point-slope form with m=3m = 3 and (x1,y1)=(2,−5)(x_1, y_1) = (2, -5): y−(−5)=3(x−2)  ⟹  y+5=3x−6y - (-5) = 3(x - 2) \implies y + 5 = 3x - 6
  4. Isolate yy: y=3x−11y = 3x - 11

Worked Example 2: Line Perpendicular to a Given Equation

Write the equation of the line that is perpendicular to y=−23x+4y = -\frac{2}{3}x + 4 and passes through the point (−4,1)(-4, 1).

  1. Identify the original slope: m1=−23m_1 = -\frac{2}{3}.
  2. Determine the perpendicular slope (negative reciprocal): m⊥=−1−23=+32m_{\perp} = -\frac{1}{-\frac{2}{3}} = +\frac{3}{2}
  3. Apply point-slope form with m=32m = \frac{3}{2} and point (−4,1)(-4, 1): y−1=32(x−(−4))  ⟹  y−1=32(x+4)y - 1 = \frac{3}{2}(x - (-4)) \implies y - 1 = \frac{3}{2}(x + 4)
  4. Distribute and simplify: y−1=32x+6  ⟹  y=32x+7y - 1 = \frac{3}{2}x + 6 \implies y = \frac{3}{2}x + 7

Direct Variation (Proportional Relationships)

Direct variation describes a special linear relationship in which the ratio of the dependent variable to the independent variable is strictly constant.

y=kx⟺yx=ky = kx \quad \Longleftrightarrow \quad \frac{y}{x} = k
  • kk (Constant of Variation / Constant of Proportionality): A non-zero constant representing the slope of the line (m=km = k).
  • Graph Behavior: The graph of every direct variation equation is a straight line that always passes through the origin (0,0)(0, 0).
  • yy-Intercept: For direct variation, the yy-intercept is always b=0b = 0.

Proportional vs. Non-Proportional Linear Comparison

PropertyDirect Variation (Proportional)Non-Proportional Linear Relation
General Equationy=kxy = kxy=mx+by = mx + b (b≠0b \neq 0)
Passes Through Origin?YES, always passes through (0,0)(0, 0)NO, passes through (0,b)(0, b) where b≠0b \neq 0
Constant Ratio yx\frac{y}{x}?YES, y1x1=y2x2=k\frac{y_1}{x_1} = \frac{y_2}{x_2} = kNO, ratio yx\frac{y}{x} changes at every point
Real-World ExampleHourly pay without base salary (y=20xy = 20x)Taxi ride with base flag-drop fee (y=2.50x+5.00y = 2.50x + 5.00)

Solving Direct Variation Problems: The Two-Step Method

  1. Step 1: Find the constant kk: Substitute the given initial pair (x1,y1)(x_1, y_1) into k=y1x1k = \frac{y_1}{x_1}.
  2. Step 2: Solve for the unknown quantity: Use y=kxy = kx with the newly determined kk to compute the unknown variable.

Worked Example: Direct Variation

The variable yy varies directly as xx. When x=12x = 12, y=42y = 42. What is the value of yy when x=20x = 20?

  1. Calculate the constant of variation kk: k=yx=4212=72=3.5k = \frac{y}{x} = \frac{42}{12} = \frac{7}{2} = 3.5
  2. Write the direct variation equation: y=3.5xy = 3.5x
  3. Substitute x=20x = 20: y=3.5(20)=70y = 3.5(20) = 70
  4. Proportion Alternative: 4212=y20  ⟹  12y=42(20)  ⟹  12y=840  ⟹  y=84012=70\frac{42}{12} = \frac{y}{20} \implies 12y = 42(20) \implies 12y = 840 \implies y = \frac{840}{12} = 70
Test Your Knowledge

Which of the following represents the equation of a line that is parallel to 6x - 2y = 14 and passes through the point (2, -5)?

A

y = -3x + 1

B

y = -1/3 x - 13/3

C

y = 3x + 11

D

y = 3x - 11

Test Your Knowledge

What is the equation of the line that is perpendicular to the line y = -2/3 x + 4 and passes through the point (-4, 1)?

A

y = -2/3 x - 5/3

B

y = 3/2 x + 7

C

y = -3/2 x - 5

D

y = 3/2 x - 7

Test Your Knowledge

The variable y varies directly with x. If y = 42 when x = 12, what is the value of y when x = 20?

A

y = 50

B

y = 60

C

y = 70

D

y = 84

Test Your Knowledge

Consider the two linear equations L1: 2x - 5y = 15 and L2: 10x + 4y = -8. What is the geometric relationship between the two lines on the coordinate plane?

A

The lines are perpendicular because their slopes are negative reciprocals.

B

The lines are parallel because their slopes are equal.

C

The lines are coincident and represent the exact same line.

D

The lines intersect at an acute angle but are not perpendicular.

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