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Coordinate Geometry - Equations of Straight Lines (y = mx + c)

Grade 10IGCSE

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

🔑Concepts

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The gradient (m) represents the steepness of the line, calculated as 'rise over run'.

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The y-intercept (c) is the point where the line crosses the y-axis (where x = 0).

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Parallel lines have the same gradient (m1=m2m_1 = m_2).

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Perpendicular lines have gradients that are negative reciprocals (m1imesm2=−1m_1 imes m_2 = -1).

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To find the equation of a line, you need at least one point on the line and the gradient, or two points on the line.

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Horizontal lines have a gradient of 0 and equations in the form y=ky = k.

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Vertical lines have an undefined gradient and equations in the form x=kx = k.

📐Formulae

General Equation: y=mx+cy = mx + c

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

Point-Slope Form: y−y1=m(x−x1)y - y_1 = m(x - x_1)

Midpoint: M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Distance between two points: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Condition for Perpendicular Lines: m2=−1m1m_2 = -\frac{1}{m_1}

💡Examples

Problem 1:

Find the equation of the line passing through the points A(2, 5) and B(4, 9).

Solution:

y=2x+1y = 2x + 1

Explanation:

  1. Find the gradient: m=9−54−2=42=2m = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2. 2. Use the point (2, 5) in y=mx+cy = mx + c: 5=2(2)+c5 = 2(2) + c. 3. Solve for c: 5=4+c⇒c=15 = 4 + c \Rightarrow c = 1. 4. Substitute m and c into the general form.

Problem 2:

Line L1 has the equation y=3x−4y = 3x - 4. Find the equation of line L2 which is perpendicular to L1 and passes through the point (6, 1).

Solution:

y=−13x+3y = -\frac{1}{3}x + 3

Explanation:

  1. Identify the gradient of L1 (m1=3m_1 = 3). 2. Find the perpendicular gradient: m2=−13m_2 = -\frac{1}{3}. 3. Use the point (6, 1) in y=mx+cy = mx + c: 1=−13(6)+c1 = -\frac{1}{3}(6) + c. 4. Simplify: 1=−2+c⇒c=31 = -2 + c \Rightarrow c = 3. 5. Write the final equation.

Problem 3:

Rearrange the equation 2y−4x=102y - 4x = 10 into the form y=mx+cy = mx + c and state the gradient and y-intercept.

Solution:

y=2x+5y = 2x + 5; Gradient = 2, Y-intercept = 5

Explanation:

  1. Add 4x to both sides: 2y=4x+102y = 4x + 10. 2. Divide every term by 2 to isolate y: y=42x+102y = \frac{4}{2}x + \frac{10}{2}. 3. Simplify to get y=2x+5y = 2x + 5.