Area of a Triangle (Coordinates)

\( A=\tfrac{1}{2}\,\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big| \)
Coordinate Geometry GCSE

Find the area of the triangle with vertices (2,3), (7,4), (5,9).

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Substitute carefully

Explanation

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Statement

The area of a triangle can be calculated directly from the coordinates of its three vertices. The formula is:

\[ A = \tfrac{1}{2} \Big| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \Big| \]

Here, the triangle has vertices \((x_1,y_1)\), \((x_2,y_2)\), and \((x_3,y_3)\). The absolute value ensures the area is always positive.

Why it’s true (short reason)

  • The formula comes from coordinate geometry, specifically the determinant method for polygon areas.
  • It can be derived by using the shoelace method or by splitting the polygon into trapezia against the x-axis.
  • The factor of \(\tfrac{1}{2}\) corresponds to the general triangle area rule (half base times height).

Recipe (how to use it)

  1. Write down the coordinates of the three vertices clearly.
  2. Substitute them into the formula: \[ A = \tfrac{1}{2} \Big| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \Big|. \]
  3. Carry out the multiplications and additions inside the brackets carefully.
  4. Take the absolute value so that the area is positive.
  5. Simplify to the final area, with squared units if coordinates represent lengths.

Spotting it

This formula should be used when:

  • You are given three vertices of a triangle on a coordinate grid.
  • No side lengths or perpendicular heights are provided directly.
  • The question explicitly mentions finding area from coordinates.

Common pairings

  • The distance formula, when checking side lengths for validation.
  • The midpoint formula, sometimes used when exploring medians and centroids.
  • The shoelace formula, which generalises this to polygons.

Mini examples

  1. Given: Points \((0,0)\), \((4,0)\), \((0,3)\). Find: Area. Answer: \(A = \tfrac{1}{2}|0(0-3)+4(3-0)+0(0-0)| = \tfrac{1}{2}(12) = 6\).
  2. Given: Points \((1,1)\), \((4,1)\), \((1,5)\). Find: Area. Answer: \(A = \tfrac{1}{2}|1(1-5)+4(5-1)+1(1-1)| = \tfrac{1}{2}(12) = 6\).

Pitfalls

  • Forgetting the absolute value, which could give a negative result.
  • Mixing up coordinates (e.g. confusing \(x_2\) with \(y_2\)).
  • Leaving the answer as a fraction of negative sign, instead of taking half of the absolute value.
  • Not writing coordinates in consistent order when substituting.

Exam strategy

  • Write coordinates in a table before substitution to avoid mistakes.
  • Check if the triangle is right-angled — sometimes a quicker method exists, but the coordinate formula always works.
  • Double-check sign changes when subtracting coordinates.
  • Always take the absolute value at the end.

Summary

The coordinate geometry formula for the area of a triangle is powerful because it works for any set of three vertices, whether or not the triangle has a simple base and height. It comes from the shoelace method and ensures accuracy even when the triangle is slanted. Always substitute carefully, calculate step by step, and take the absolute value to ensure a positive area.

Worked examples

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  1. Find the area of the triangle with vertices (0,0), (4,0), (0,3).
    1. \( Substitute: A = 1/2 |0(0-3)+4(3-0)+0(0-0)| \)
    2. \( Simplify: A = 1/2 |12| \)
    3. \( A = 6 \)
    Answer: 6
  2. Find the area of the triangle with vertices (1,1), (4,1), (1,5).
    1. \( A = 1/2 |1(1-5)+4(5-1)+1(1-1)| \)
    2. \( A = 1/2 |-4+16+0| \)
    3. \( A = 1/2 × 12 = 6 \)
    Answer: 6
  3. Find the area of the triangle with vertices (2,3), (5,7), (8,3).
    1. \( A = 1/2 |2(7-3)+5(3-3)+8(3-7)| \)
    2. \( A = 1/2 |8+0-32| \)
    3. \( A = 1/2 × 24 = 12 \)
    Answer: 12
  4. Vertices: (0,0), (6,0), (3,9). Find area.
    1. \( A = 1/2 |0(0-9)+6(9-0)+3(0-0)| \)
    2. \( A = 1/2 |0+54+0| \)
    3. \( A = 27 \)
    Answer: 27
  5. Find the area of the triangle with vertices (1,2), (4,6), (6,2).
    1. \( A = 1/2 |1(6-2)+4(2-2)+6(2-6)| \)
    2. \( A = 1/2 |4+0-24| \)
    3. \( A = 1/2 × 20 = 10 \)
    Answer: 10
  6. Vertices: (2,1), (7,4), (5,9). Find the area.
    1. \( A = 1/2 |2(4-9)+7(9-1)+5(1-4)| \)
    2. \( A = 1/2 |-10+56-15| \)
    3. \( A = 1/2 × 31 = 15.5 \)
    Answer: 15.5
  7. Find the area of the triangle with vertices (3,2), (8,5), (6,10).
    1. \( A = 1/2 |3(5-10)+8(10-2)+6(2-5)| \)
    2. \( A = 1/2 |-15+64-18| \)
    3. \( A = 1/2 × 31 = 15.5 \)
    Answer: 15.5
  8. Vertices: (-2,0), (4,0), (1,6). Find the area.
    1. \( A = 1/2 |-2(0-6)+4(6-0)+1(0-0)| \)
    2. \( A = 1/2 |12+24+0| \)
    3. \( A = 18 \)
    Answer: 18
  9. Vertices: (0,0), (5,2), (3,7). Find the area.
    1. \( A = 1/2 |0(2-7)+5(7-0)+3(0-2)| \)
    2. \( A = 1/2 |0+35-6| \)
    3. \( A = 1/2 × 29 = 14.5 \)
    Answer: 14.5
  10. Find the area of the triangle with vertices (2,2), (6,5), (4,9).
    1. \( A = 1/2 |2(5-9)+6(9-2)+4(2-5)| \)
    2. \( A = 1/2 |-8+42-12| \)
    3. \( A = 1/2 × 22 = 11 \)
    Answer: 11