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Geometry and Trigonometry - Gradients of parallel and perpendicular lines

Grade 9IB

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

🔑Concepts

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Parallel lines have identical gradients (m1=m2m_1 = m_2). This means they maintain a constant distance from each other and never intersect. For example, if two lines both have a gradient of 22, they will rise and run at the same rate.

Graph showing two parallel lines with the same gradient.
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Perpendicular lines intersect at a right angle (90∘90^\circ). Their gradients are negative reciprocals of each other, satisfying the condition m1×m2=−1m_1 \times m_2 = -1. Visually, if one line goes up, the other must go down at a corresponding steepness.

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To find the equation of a line parallel to a given line passing through a specific point (x1,y1)(x_1, y_1), identify the gradient mm from the original equation and use the point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1).

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To find the equation of a line perpendicular to a given line, first determine the gradient m1m_1 of the original line. Calculate the new gradient m2=−1m1m_2 = -\frac{1}{m_1}, and then use the given coordinates to find the yy-intercept.

📐Formulae

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

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

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

m2=−1m1 (negative reciprocal)m_2 = -\frac{1}{m_1} \text{ (negative reciprocal)}

💡Examples

Problem 1:

Find the equation of a line that is parallel to y=3x−5y = 3x - 5 and passes through the point (2,10)(2, 10).

Solution:

  1. Identify the gradient of the given line: m=3m = 3.
  2. Since the lines are parallel, the new line also has m=3m = 3.
  3. Use the point-slope form or y=mx+cy = mx + c with the point (2,10)(2, 10): 10=3(2)+c10 = 3(2) + c 10=6+c10 = 6 + c c=4c = 4
  4. The equation is y=3x+4y = 3x + 4.

Explanation:

Parallel lines share the same gradient. We used the given gradient 33 and the point (2,10)(2, 10) to solve for the yy-intercept cc.

Problem 2:

Line L1L_1 passes through the points (0,4)(0, 4) and (2,8)(2, 8). Find the gradient of a line L2L_2 that is perpendicular to L1L_1.

Solution:

  1. Find the gradient of L1L_1 using m1=y2−y1x2−x1m_1 = \frac{y_2 - y_1}{x_2 - x_1}: m1=8−42−0=42=2m_1 = \frac{8 - 4}{2 - 0} = \frac{4}{2} = 2
  2. Use the perpendicular condition m1×m2=−1m_1 \times m_2 = -1: 2×m2=−12 \times m_2 = -1 m2=−12m_2 = -\frac{1}{2}
  3. The gradient of L2L_2 is −12-\frac{1}{2}.

Explanation:

First, calculate the slope of the first line. Then, find its negative reciprocal to determine the slope of any line perpendicular to it.

Problem 3:

Determine if the lines y=23x+1y = \frac{2}{3}x + 1 and 3x+2y=63x + 2y = 6 are perpendicular.

Solution:

  1. The gradient of the first line is m1=23m_1 = \frac{2}{3}.
  2. Rewrite the second equation in y=mx+cy = mx + c form: 2y=−3x+62y = -3x + 6 y=−32x+3y = -\frac{3}{2}x + 3 So, m2=−32m_2 = -\frac{3}{2}.
  3. Multiply the gradients: m1×m2=23×(−32)=−66=−1m_1 \times m_2 = \frac{2}{3} \times \left(-\frac{3}{2}\right) = -\frac{6}{6} = -1
  4. Since the product is −1-1, the lines are perpendicular.

Explanation:

By converting both equations to gradient-intercept form, we can compare their gradients. Since their product is −1-1, they are confirmed to be perpendicular.

Problem 4:

Line AA passes through (−2,−1)(-2, -1) and (2,1)(2, 1). Line BB is parallel to Line AA and passes through the point (0,2)(0, 2). Determine the equation of Line BB.

Two parallel lines with gradient 0.5, one passing through the origin and the other through (0,2).

Solution:

  1. Find gradient of Line AA: m=1−(−1)2−(−2)=24=0.5m = \frac{1 - (-1)}{2 - (-2)} = \frac{2}{4} = 0.5.
  2. Since Line BB is parallel, mB=0.5m_B = 0.5.
  3. Line BB passes through (0,2)(0, 2), which is the yy-intercept (c=2c = 2).
  4. Equation: y=0.5x+2y = 0.5x + 2.

Explanation:

First calculate the slope of the first line. Parallel lines share this slope. Because the second line passes through (0,2)(0,2), we can immediately identify 22 as the yy-intercept in the y=mx+cy=mx+c form.

Problem 5:

Find the equation of the line L1L_1 that passes through the point P(2,4)P(2, 4) and is perpendicular to the line L2L_2 which passes through the points A(0,0)A(0, 0) and B(4,2)B(4, 2).

A coordinate plane showing two perpendicular lines. Line L2 passes through the origin and (4,2). Line L1 passes through (2,4) and is perpendicular to L2.

Solution:

  1. First, calculate the gradient of line L2L_2 (m2m_2) using points A(0,0)A(0, 0) and B(4,2)B(4, 2): m2=y2−y1x2−x1=2−04−0=24=12m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{4 - 0} = \frac{2}{4} = \frac{1}{2}

  2. Determine the gradient of line L1L_1 (m1m_1). Since L1⊥L2L_1 \perp L_2: m1=−1m2=−11/2=−2m_1 = -\frac{1}{m_2} = -\frac{1}{1/2} = -2

  3. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point P(2,4)P(2, 4) and m=−2m = -2: y−4=−2(x−2)y - 4 = -2(x - 2) y−4=−2x+4y - 4 = -2x + 4 y=−2x+8y = -2x + 8

Explanation:

To find the equation of a perpendicular line, we first find the gradient of the original line. The perpendicular gradient is the negative reciprocal of the original gradient. Once we have the new gradient and a point on the line, we use the linear equation formula to find the final equation.