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Coordinate Geometry - Gradient, Midpoint, and Length of a Line

Grade 10IGCSE

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

🔑Concepts

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Coordinates represent positions on a 2D plane as (x, y).

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The Gradient (m) measures the steepness of a line, defined as the 'rise over run'.

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The Midpoint is the central point of a line segment, found by averaging the x and y coordinates of the endpoints.

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The Length (Distance) of a line segment is calculated using Pythagoras' Theorem.

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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).

📐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)

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

Equation of a straight line: y=mx+cy = mx + c

💡Examples

Problem 1:

Find the gradient, midpoint, and length of the line segment joining points A(2,−3)A(2, -3) and B(6,5)B(6, 5).

Solution:

Gradient: m=5−(−3)6−2=84=2m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2. Midpoint: M=(2+62,−3+52)=(4,1)M = (\frac{2+6}{2}, \frac{-3+5}{2}) = (4, 1). Length: d=(6−2)2+(5−(−3))2=42+82=16+64=80≈8.94d = \sqrt{(6-2)^2 + (5 - (-3))^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} \approx 8.94 units.

Explanation:

To find the gradient, subtract the y-coordinates and divide by the difference in x-coordinates. For the midpoint, find the average of the x and y values. For the length, apply the distance formula derived from Pythagoras' theorem.

Problem 2:

The line joining P(3,k)P(3, k) and Q(7,10)Q(7, 10) has a gradient of 0.50.5. Find the value of kk.

Solution:

0.5=10−k7−3⇒0.5=10−k4⇒2=10−k⇒k=80.5 = \frac{10 - k}{7 - 3} \Rightarrow 0.5 = \frac{10 - k}{4} \Rightarrow 2 = 10 - k \Rightarrow k = 8.

Explanation:

Substitute the known coordinates and the given gradient into the gradient formula. Solve the resulting linear equation for the unknown variable kk.

Problem 3:

Given point A(1,4)A(1, 4) and midpoint M(4,6)M(4, 6), find the coordinates of point B(x,y)B(x, y).

Solution:

4=1+x2⇒8=1+x⇒x=74 = \frac{1 + x}{2} \Rightarrow 8 = 1 + x \Rightarrow x = 7; 6=4+y2⇒12=4+y⇒y=86 = \frac{4 + y}{2} \Rightarrow 12 = 4 + y \Rightarrow y = 8. Point BB is (7,8)(7, 8).

Explanation:

Since the midpoint is the average of the endpoints, set up two separate equations (one for x and one for y) and solve for the missing endpoint coordinates.