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Coordinate Geometry - Equation of a Straight Line (y = mx + c)

Grade 12IGCSE

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

🔑Concepts

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Gradient (m): Represents the steepness of the line, calculated as the ratio of 'rise over run'.

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

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Parallel Lines: Two lines are parallel if they have the same gradient (m1=m2m_1 = m_2).

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Perpendicular Lines: Two lines are perpendicular if the product of their gradients is -1 (m1×m2=−1m_1 \times m_2 = -1).

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Collinear Points: Points that lie on the same straight line, sharing the same gradient between any two points.

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Horizontal and Vertical Lines: Horizontal lines have the equation y=ky = k (gradient 0); vertical lines have the equation x=kx = k (gradient undefined).

📐Formulae

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

Equation of a straight line (Slope-intercept form): y=mx+cy = mx + c

Point-gradient form: y−y1=m(x−x1)y - y_1 = m(x - x_1)

Midpoint of a line segment: 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}

Perpendicular gradient: m⊥=−1mm_{\perp} = -\frac{1}{m}

💡Examples

Problem 1:

Find the equation of the line passing through the points A(2,5)A(2, 5) and B(4,13)B(4, 13).

Solution:

m=13−54−2=82=4m = \frac{13 - 5}{4 - 2} = \frac{8}{2} = 4. Using y=mx+cy = mx + c: 5=4(2)+c⇒5=8+c⇒c=−35 = 4(2) + c \Rightarrow 5 = 8 + c \Rightarrow c = -3. Equation: y=4x−3y = 4x - 3.

Explanation:

First, calculate the gradient using the two-point formula. Then, substitute one point and the gradient into the slope-intercept form to solve for the y-intercept (c).

Problem 2:

Find the equation of the line perpendicular to y=2x+5y = 2x + 5 that passes through the point (6,2)(6, 2).

Solution:

Gradient of given line m1=2m_1 = 2. Perpendicular gradient m2=−12m_2 = -\frac{1}{2}. Using y−y1=m(x−x1)y - y_1 = m(x - x_1): y−2=−12(x−6)⇒y−2=−0.5x+3⇒y=−0.5x+5y - 2 = -\frac{1}{2}(x - 6) \Rightarrow y - 2 = -0.5x + 3 \Rightarrow y = -0.5x + 5.

Explanation:

Perpendicular lines have negative reciprocal gradients. After finding the new gradient, use the point-slope formula with the given coordinate.

Problem 3:

A line segment PQPQ has endpoints P(−2,3)P(-2, 3) and Q(4,7)Q(4, 7). Find the equation of the perpendicular bisector of PQPQ.

Solution:

Midpoint M=(−2+42,3+72)=(1,5)M = (\frac{-2+4}{2}, \frac{3+7}{2}) = (1, 5). Gradient PQ=7−34−(−2)=46=23PQ = \frac{7-3}{4-(-2)} = \frac{4}{6} = \frac{2}{3}. Perpendicular gradient m=−32m = -\frac{3}{2}. Equation: y−5=−32(x−1)⇒2y−10=−3x+3⇒3x+2y=13y - 5 = -\frac{3}{2}(x - 1) \Rightarrow 2y - 10 = -3x + 3 \Rightarrow 3x + 2y = 13.

Explanation:

The perpendicular bisector passes through the midpoint of the segment at a right angle. Find the midpoint, the gradient of the segment, the perpendicular gradient, and then the equation.