krit.club logo

Geometry and Measurement - Coordinate Geometry: Distance, Midpoint, and Gradient

Grade 8IB

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

🔑Concepts

•

The Distance formula calculates the length of the straight line segment between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) using the Pythagorean theorem where d2=(x2−x1)2+(y2−y1)2d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.

A right-angled triangle formed between two points in the coordinate plane to illustrate the distance formula.
•

The Midpoint is the exact center point between two coordinates, found by averaging the xx-values and the yy-values of the endpoints.

A line segment AB with a center point M labeled as the midpoint.
•

The Gradient (slope) measures the steepness of a line as the ratio of 'rise' over 'run'. A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards.

Diagram showing the vertical rise and horizontal run of a line segment.
•

Horizontal lines have a gradient of 00 because the change in yy is 00. Vertical lines have an undefined gradient because the change in xx is 00 (division by zero).

📐Formulae

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

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

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

💡Examples

Problem 1:

Find the distance, midpoint, and gradient for the line segment joining points P(2,3)P(2, 3) and Q(8,11)Q(8, 11).

Solution:

  1. Distance: d=(8−2)2+(11−3)2=62+82=36+64=100=10d = \sqrt{(8 - 2)^2 + (11 - 3)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 units.
  2. Midpoint: M=(2+82,3+112)=(102,142)=(5,7)M = \left( \frac{2 + 8}{2}, \frac{3 + 11}{2} \right) = \left( \frac{10}{2}, \frac{14}{2} \right) = (5, 7).
  3. Gradient: m=11−38−2=86=43≈1.33m = \frac{11 - 3}{8 - 2} = \frac{8}{6} = \frac{4}{3} \approx 1.33.

Explanation:

Substitute x1=2,y1=3,x2=8,y2=11x_1=2, y_1=3, x_2=8, y_2=11 into the coordinate geometry formulae. For distance, we use the square root of the sum of squares of the differences. For midpoint, we average the coordinates. For gradient, we divide the change in yy by the change in xx.

Problem 2:

A line passes through A(−3,5)A(-3, 5) and B(1,−3)B(1, -3). Calculate the gradient of the line.

Solution:

m=−3−51−(−3)=−81+3=−84=−2m = \frac{-3 - 5}{1 - (-3)} = \frac{-8}{1 + 3} = \frac{-8}{4} = -2

Explanation:

The gradient is negative, which means the line slopes downwards as it moves from left to right. Note how subtracting a negative number in the denominator becomes addition: 1−(−3)=1+31 - (-3) = 1 + 3.

Problem 3:

The midpoint of a line segment JKJK is M(4,2)M(4, 2). If point JJ is (1,−1)(1, -1), find the coordinates of point K(x,y)K(x, y).

Solution:

Using the midpoint formula for each coordinate: For xx: 4=1+x2  ⟹  8=1+x  ⟹  x=74 = \frac{1 + x}{2} \implies 8 = 1 + x \implies x = 7 For yy: 2=−1+y2  ⟹  4=−1+y  ⟹  y=52 = \frac{-1 + y}{2} \implies 4 = -1 + y \implies y = 5 Therefore, K=(7,5)K = (7, 5).

Explanation:

Since the midpoint is known, set up two separate algebraic equations (one for xx and one for yy) to solve for the missing endpoint coordinates.

Problem 4:

Determine the distance between point R(−2,−1)R(-2, -1) and point S(4,7)S(4, 7).

Coordinate plot showing points R and S connected by a line segment of length 10.

Solution:

d=(4−(−2))2+(7−(−1))2d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} d=(6)2+(8)2d = \sqrt{(6)^2 + (8)^2} d=36+64d = \sqrt{36 + 64} d=100d = \sqrt{100} d=10d = 10 units

Explanation:

Substitute the coordinates into the distance formula. The difference in xx is 66 and the difference in yy is 88. Squaring these gives 3636 and 6464. The square root of their sum (100100) results in a distance of 1010.

Problem 5:

A line segment connects C(2,8)C(2, 8) and D(8,2)D(8, 2). Calculate the gradient of the line and the coordinates of the midpoint MM.

Graph showing a line segment with a negative slope from (2,8) to (8,2) and its midpoint at (5,5).

Solution:

Gradient: m=2−88−2=−66=−1m = \frac{2 - 8}{8 - 2} = \frac{-6}{6} = -1

Midpoint: M=(2+82,8+22)=(102,102)=(5,5)M = \left( \frac{2 + 8}{2}, \frac{8 + 2}{2} \right) = \left( \frac{10}{2}, \frac{10}{2} \right) = (5, 5)

Explanation:

To find the gradient, divide the change in yy (−6-6) by the change in xx (66). For the midpoint, calculate the average of the xx-coordinates and the yy-coordinates separately.