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Algebra - Parallel and perpendicular lines

Grade 9IB

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

🔑Concepts

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Lines are parallel if they have the same gradient. This means for two lines with equations y=m1x+c1y = m_1x + c_1 and y=m2x+c2y = m_2x + c_2, they are parallel if m1=m2m_1 = m_2. Parallel lines never intersect.

Graph showing two parallel lines with the same gradient of 1.
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Lines are perpendicular if they meet at a right angle (90∘90^\circ). The product of their gradients is −1-1, expressed as m1×m2=−1m_1 \times m_2 = -1. Alternatively, one gradient is the negative reciprocal of the other: m2=−1m1m_2 = -\frac{1}{m_1}.

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The yy-intercept (cc) does not affect whether lines are parallel or perpendicular; it only shifts the line vertically. Parallel lines must have different yy-intercepts to be distinct.

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To find the equation of a new line, first determine its gradient based on the relationship (parallel or perpendicular) to a known line, then substitute a given point into the point-gradient formula y−y1=m(x−x1)y - y_1 = m(x - x_1).

📐Formulae

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

y=mx+cy = mx + c

m1=m2 (Parallel lines)m_1 = m_2 \text{ (Parallel lines)}

m1×m2=−1 (Perpendicular lines)m_1 \times m_2 = -1 \text{ (Perpendicular lines)}

y−y1=m(x−x1) (Point-gradient form)y - y_1 = m(x - x_1) \text{ (Point-gradient form)}

💡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:

The given line has a gradient m=4m = 4. Since the new line is parallel, its gradient is also m=4m = 4. Using the point-gradient formula with (x1,y1)=(2,10)(x_1, y_1) = (2, 10): y−10=4(x−2)y - 10 = 4(x - 2) y−10=4x−8y - 10 = 4x - 8 y=4x+2y = 4x + 2

Explanation:

Parallel lines share the same gradient. We identify the gradient from the reference equation and use the given point to solve for the new yy-intercept.

Problem 2:

Determine the equation of the line perpendicular to y=−13x+7y = -\frac{1}{3}x + 7 that passes through the point (3,−2)(3, -2).

Solution:

The gradient of the given line is m1=−13m_1 = -\frac{1}{3}. The perpendicular gradient m2m_2 is the negative reciprocal: m2=−1m1=−1−1/3=3m_2 = -\frac{1}{m_1} = -\frac{1}{-1/3} = 3 Now use the point (3,−2)(3, -2) in y=mx+cy = mx + c: −2=3(3)+c-2 = 3(3) + c −2=9+c-2 = 9 + c c=−11c = -11 The equation is y=3x−11y = 3x - 11.

Explanation:

To find a perpendicular gradient, flip the fraction and change the sign. Then, substitute the given coordinates to find the constant cc.

Problem 3:

Line AA passes through (0,4)(0, 4) and (2,8)(2, 8). Line BB is defined by the equation 2y+x=102y + x = 10. Determine if these lines are parallel, perpendicular, or neither.

Solution:

Find gradient of Line AA: mA=8−42−0=42=2m_A = \frac{8 - 4}{2 - 0} = \frac{4}{2} = 2 Find gradient of Line BB by rearranging to y=mx+cy = mx + c: 2y=−x+102y = -x + 10 y=−12x+5y = -\frac{1}{2}x + 5 So, mB=−12m_B = -\frac{1}{2}. Since 2×(−12)=−12 \times (-\frac{1}{2}) = -1, the lines are perpendicular.

Explanation:

By calculating the gradient of the first line and rearranging the second equation into y=mx+cy = mx + c form, we can compare their gradients. Their product is −1-1, confirming they are perpendicular.

Problem 4:

Find the equation of the line passing through point P(0,−2)P(0, -2) that is parallel to the line L1L_1 shown in the graph, which passes through (0,2)(0, 2) and (2,3)(2, 3).

Graph showing line L1 and a point P below it.

Solution:

  1. Find the gradient of L1L_1: m1=3−22−0=12m_1 = \frac{3 - 2}{2 - 0} = \frac{1}{2}
  2. Since the new line L2L_2 is parallel, m2=m1=12m_2 = m_1 = \frac{1}{2}.
  3. Use the yy-intercept c=−2c = -2 from point (0,−2)(0, -2).
  4. The equation is y=12x−2y = \frac{1}{2}x - 2.

Explanation:

Parallel lines share the same gradient. We calculate the gradient from the first line and apply it to the second line using the provided intercept.

Problem 5:

Find the equation of the line passing through the point Q(4,3)Q(4, 3) that is perpendicular to the line L2L_2, which passes through the points (0,0)(0, 0) and (2,4)(2, 4). Show your answer in the form y=mx+cy = mx + c.

Coordinate plane showing line L2 passing through the origin and a perpendicular line passing through point Q(4,3).

Solution:

  1. Find the gradient (m2m_2) of line L2L_2: m2=y2−y1x2−x1=4−02−0=2m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{2 - 0} = 2

  2. Determine the gradient (m⊥m_{\perp}) of the perpendicular line. Since the lines are perpendicular, m1×m2=−1m_1 \times m_2 = -1: m⊥=−1m2=−12m_{\perp} = -\frac{1}{m_2} = -\frac{1}{2}

  3. Use the point-gradient form with point Q(4,3)Q(4, 3) and m⊥=−12m_{\perp} = -\frac{1}{2}: y−3=−12(x−4)y - 3 = -\frac{1}{2}(x - 4) y−3=−12x+2y - 3 = -\frac{1}{2}x + 2 y=−12x+5y = -\frac{1}{2}x + 5

Explanation:

To find a perpendicular line, we first identify the gradient of the reference line. The product of gradients for perpendicular lines is always −1-1. Once the new gradient is found, the specific equation is determined by substituting the coordinates of the given point into the linear equation.