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Algebra - Equations of Parallel and Perpendicular Lines

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The slope-intercept form of a linear equation is y=mx+cy = mx + c, where mm represents the gradient (slope) and cc represents the yy-intercept. Lines that never intersect and are always the same distance apart are called parallel lines.

Graph showing two parallel lines with identical gradients.
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Parallel lines have the exact same gradient. If line L1L_1 has gradient m1m_1 and line L2L_2 has gradient m2m_2, then L1∥L2L_1 \parallel L_2 implies m1=m2m_1 = m_2.

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Perpendicular lines intersect at a right angle (90∘90^\circ). The product of the gradients of two perpendicular lines is always −1-1. This is expressed as m1×m2=−1m_1 \times m_2 = -1 or m2=−1m1m_2 = -\frac{1}{m_1}.

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To find the equation of a new line, first identify the gradient of the reference line, apply the parallel/perpendicular rule to find the new gradient mm, then use the point-slope formula y−y1=m(x−x1)y - y_1 = m(x - x_1) with the given coordinates.

📐Formulae

y=mx+cy = mx + c

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

mparallel=moriginalm_{parallel} = m_{original}

mperpendicular=−1moriginalm_{perpendicular} = -\frac{1}{m_{original}}

y−y1=m(x−x1)y - y_1 = m(x - x_1)

💡Examples

Problem 1:

Find the equation of the line that is parallel to y=4x−5y = 4x - 5 and passes through the point (2,10)(2, 10).

Solution:

  1. Identify the gradient of the given line: m=4m = 4.
  2. Since the lines are parallel, the new line also has m=4m = 4.
  3. Use the point-slope form: y−10=4(x−2)y - 10 = 4(x - 2).
  4. Expand and simplify: y−10=4x−8y - 10 = 4x - 8 y=4x−8+10y = 4x - 8 + 10 y=4x+2y = 4x + 2

Explanation:

Parallel lines share the same slope. By identifying the slope from the given equation and substituting the coordinates of the point into the line equation, we find the specific yy-intercept for the new line.

Problem 2:

Find the equation of the line perpendicular to y=−13x+7y = -\frac{1}{3}x + 7 that passes through the point (1,5)(1, 5).

Solution:

  1. Identify the gradient of the given line: m1=−13m_1 = -\frac{1}{3}.
  2. Find the perpendicular gradient: m2=−1m1=−1−1/3=3m_2 = -\frac{1}{m_1} = -\frac{1}{-1/3} = 3.
  3. Use the slope-intercept form y=mx+cy = mx + c with the point (1,5)(1, 5): 5=3(1)+c5 = 3(1) + c 5=3+c5 = 3 + c c=5−3=2c = 5 - 3 = 2
  4. The equation is y=3x+2y = 3x + 2.

Explanation:

Perpendicular gradients are negative reciprocals. We flip the fraction and change the sign of the original slope to get 33, then solve for the yy-intercept using the provided point.

Problem 3:

Determine if the line passing through (0,2)(0, 2) and (2,6)(2, 6) is perpendicular to the line y=−12x+10y = -\frac{1}{2}x + 10.

Solution:

  1. Calculate the gradient of the first line (m1m_1): m1=6−22−0=42=2m_1 = \frac{6 - 2}{2 - 0} = \frac{4}{2} = 2
  2. Identify the gradient of the second line: m2=−12m_2 = -\frac{1}{2}.
  3. Multiply the gradients: m1×m2=2×(−12)=−1m_1 \times m_2 = 2 \times \left(-\frac{1}{2}\right) = -1
  4. Since the product is −1-1, the lines are perpendicular.

Explanation:

By calculating the slope between two points and comparing it to the slope of the second equation, we use the property m1×m2=−1m_1 \times m_2 = -1 to confirm perpendicularity.

Problem 4:

Find the equation of the line that is parallel to the line passing through A(0,−2)A(0, -2) and B(4,0)B(4, 0) and passes through the point (2,4)(2, 4).

Graph showing the reference line through A and B and the new parallel line through P.

Solution:

  1. Find the gradient of line ABAB: m=0−(−2)4−0=24=12m = \frac{0 - (-2)}{4 - 0} = \frac{2}{4} = \frac{1}{2}
  2. Since the new line is parallel, its gradient is also m=12m = \frac{1}{2}.
  3. Use the point (2,4)(2, 4) in the equation y−y1=m(x−x1)y - y_1 = m(x - x_1): y−4=12(x−2)y - 4 = \frac{1}{2}(x - 2) y−4=12x−1y - 4 = \frac{1}{2}x - 1 y=12x+3y = \frac{1}{2}x + 3

Explanation:

Parallel lines share the same slope. By calculating the slope between the two given points, we define the slope for our target line and then substitute the specific point into the linear equation formula.

Problem 5:

Find the equation of the line that passes through the point (3,−2)(3, -2) and is perpendicular to the line y=32x+1y = \frac{3}{2}x + 1. Express your final answer in the form y=mx+cy = mx + c.

Graph showing two perpendicular lines intersecting. Line L1 has a positive gradient and L2 has a negative gradient passing through point (3,-2).

Solution:

  1. Identify the gradient of the given line: m1=32m_1 = \frac{3}{2}

  2. Determine the gradient of the perpendicular line (m2m_2): Since the lines are perpendicular, m2=−1m1m_2 = -\frac{1}{m_1} m2=−13/2=−23m_2 = -\frac{1}{3/2} = -\frac{2}{3}

  3. Use the point-gradient formula y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (3,−2)(3, -2): y−(−2)=−23(x−3)y - (-2) = -\frac{2}{3}(x - 3) y+2=−23x+2y + 2 = -\frac{2}{3}x + 2

  4. Solve for yy: y=−23x+2−2y = -\frac{2}{3}x + 2 - 2 y=−23xy = -\frac{2}{3}x

Explanation:

To find a perpendicular line, we use the property that the product of their gradients is −1-1. Once the new gradient is found, we substitute the given coordinates into the linear equation formula to find the specific cc value (y-intercept).