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Coordinate Geometry - Coordinates

Grade 9IGCSE

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

🔑Concepts

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The Cartesian coordinate system uses two perpendicular axes: the horizontal xx-axis and the vertical yy-axis. The point where they intersect is called the origin (0,0)(0, 0). Any point PP is represented as (x,y)(x, y), where xx is the horizontal distance and yy is the vertical distance from the origin.

A coordinate plane showing point P at (3, 2) with dashed lines to the axes.
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The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is derived from Pythagoras' Theorem. By forming a right-angled triangle, the horizontal leg is ∣x2−x1∣|x_2 - x_1| and the vertical leg is ∣y2−y1∣|y_2 - y_1|.

A right-angled triangle formed between two points to demonstrate distance calculation.
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The midpoint is the average of the coordinates. It is the point exactly halfway between two endpoints of a line segment.

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The gradient (or slope) mm measures the steepness of a line. It is the ratio of 'rise' (vertical change) over 'run' (horizontal change). A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards.

📐Formulae

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

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)

💡Examples

Problem 1:

Find the gradient of the line passing through the points A(2,−3)A(2, -3) and B(6,5)B(6, 5).

Solution:

Given (x1,y1)=(2,−3)(x_1, y_1) = (2, -3) and (x2,y2)=(6,5)(x_2, y_2) = (6, 5). Using the gradient formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} m=5−(−3)6−2m = \frac{5 - (-3)}{6 - 2} m=5+34m = \frac{5 + 3}{4} m=84=2m = \frac{8}{4} = 2

Explanation:

Substitute the coordinates of points AA and BB into the gradient formula. Remember that subtracting a negative number results in addition.

Problem 2:

Calculate the distance between the points P(1,2)P(1, 2) and Q(4,6)Q(4, 6).

Solution:

Given (x1,y1)=(1,2)(x_1, y_1) = (1, 2) and (x2,y2)=(4,6)(x_2, y_2) = (4, 6). Using the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d=(4−1)2+(6−2)2d = \sqrt{(4 - 1)^2 + (6 - 2)^2} d=32+42d = \sqrt{3^2 + 4^2} d=9+16d = \sqrt{9 + 16} d=25=5d = \sqrt{25} = 5 units

Explanation:

Apply the distance formula which is based on the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2.

Problem 3:

Find the coordinates of the midpoint of the line segment joining M(−4,8)M(-4, 8) and N(2,4)N(2, 4).

Solution:

Given (x1,y1)=(−4,8)(x_1, y_1) = (-4, 8) and (x2,y2)=(2,4)(x_2, y_2) = (2, 4). Using the midpoint formula: Midpoint=(x1+x22,y1+y22)Midpoint = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) Midpoint=(−4+22,8+42)Midpoint = \left( \frac{-4 + 2}{2}, \frac{8 + 4}{2} \right) Midpoint=(−22,122)Midpoint = \left( \frac{-2}{2}, \frac{12}{2} \right) Midpoint=(−1,6)Midpoint = (-1, 6)

Explanation:

The midpoint is found by taking the average of the xx-coordinates and the average of the yy-coordinates.

Problem 4:

Determine the midpoint MM of the line segment connecting points A(−2,−1)A(-2, -1) and B(4,3)B(4, 3).

Line segment AB with its midpoint M clearly marked at (1, 1).

Solution:

  1. Identify coordinates: (x1,y1)=(−2,−1)(x_1, y_1) = (-2, -1) and (x2,y2)=(4,3)(x_2, y_2) = (4, 3).
  2. Use the midpoint formula: M=(−2+42,−1+32)M = \left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right).
  3. Calculate: M=(22,22)=(1,1)M = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1).

Explanation:

The midpoint is found by averaging the x-coordinates and the y-coordinates separately.

Problem 5:

A line passes through R(0,4)R(0, 4) and S(6,0)S(6, 0). Calculate the gradient mm of the line RSRS.

A line segment sloping downwards from (0, 4) to (6, 0).

Solution:

  1. Identify coordinates: (x1,y1)=(0,4)(x_1, y_1) = (0, 4) and (x2,y2)=(6,0)(x_2, y_2) = (6, 0).
  2. Use the gradient formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.
  3. Substitute values: m=0−46−0m = \frac{0 - 4}{6 - 0}.
  4. Simplify: m=−46=−23m = \frac{-4}{6} = -\frac{2}{3}.

Explanation:

Since the line goes downwards from left to right, the gradient is negative. The vertical drop is 44 units over a horizontal run of 66 units.