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Coordinate Geometry - Gradient and Midpoint of a Line

Grade 12IGCSE

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

🔑Concepts

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The gradient (m) measures the steepness of a line, defined as the 'rise over run' between two points.

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A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards.

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The midpoint is the exact center of a line segment, found by taking the arithmetic mean of the x and y coordinates of the endpoints.

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Parallel lines have identical gradients (m1=m2m_1 = m_2).

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

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Points are collinear if the gradient between any two pairs of points is the same.

📐Formulae

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

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

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

💡Examples

Problem 1:

Find the gradient and the midpoint of the line segment connecting the points A(−4,7)A(-4, 7) and B(2,1)B(2, 1).

Solution:

Gradient m=1−72−(−4)=−66=−1m = \frac{1 - 7}{2 - (-4)} = \frac{-6}{6} = -1. Midpoint M=(−4+22,7+12)=(−22,82)=(−1,4)M = \left( \frac{-4 + 2}{2}, \frac{7 + 1}{2} \right) = \left( \frac{-2}{2}, \frac{8}{2} \right) = (-1, 4).

Explanation:

Apply the gradient formula by subtracting the y-coordinates and x-coordinates. For the midpoint, calculate the average of the x-values and the average of the y-values.

Problem 2:

The midpoint of a line XYXY is M(3,−2)M(3, -2). If the coordinates of XX are (7,4)(7, 4), find the coordinates of point YY.

Solution:

3=7+xY2  ⟹  6=7+xY  ⟹  xY=−13 = \frac{7 + x_Y}{2} \implies 6 = 7 + x_Y \implies x_Y = -1. −2=4+yY2  ⟹  −4=4+yY  ⟹  yY=−8-2 = \frac{4 + y_Y}{2} \implies -4 = 4 + y_Y \implies y_Y = -8. Point Y=(−1,−8)Y = (-1, -8).

Explanation:

Use the midpoint formula as an equation where the midpoint is known. Solve for the unknown coordinates xYx_Y and yYy_Y individually.

Problem 3:

Given line L1L_1 passes through (1,2)(1, 2) and (4,8)(4, 8). Find the gradient of a line L2L_2 that is perpendicular to L1L_1.

Solution:

Gradient of L1(m1)=8−24−1=63=2L_1 (m_1) = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2. Since L2⊥L1L_2 \perp L_1, m2=−1m1=−12m_2 = -\frac{1}{m_1} = -\frac{1}{2}.

Explanation:

First, calculate the gradient of the first line using the two given points. Then, apply the perpendicular gradient rule: take the negative reciprocal of m1m_1.