Area of a Triangle

Area of a Triangle#

Coordinate Geometry Basics Revision#

Theory#

Before calculating areas, let’s review key coordinate geometry concepts:

  • Distance between two points: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

  • Slope of a line: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

  • These concepts help verify triangle properties

Application#

For triangle with vertices \(A(1, 2)\), \(B(4, 6)\), \(C(7, 3)\):

  • Side \(AB = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5\)

  • This confirms the triangle exists (non-collinear points)

Area of a Triangle in Coordinate Geometry#

Theory#

The area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using:

Determinant Formula#

\[\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\]

Alternative Matrix Form#

\[\begin{split}\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|\end{split}\]

Shoelace Formula#

\[\text{Area} = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3|\]

Special Cases#

  • Collinear Points: If area = 0, the three points lie on the same line

  • Right Triangle: Can verify using perpendicular slopes

  • Isosceles/Equilateral: Can verify using equal side lengths

Interactive Visualization: Triangle Area Explorer#

Application#

Examples#

Example 1: Basic Triangle Area Find the area of the triangle with vertices \(A(1, 2)\), \(B(4, 6)\), and \(C(7, 3)\).

Solution: Using the determinant formula: $\(\text{Area} = \frac{1}{2}|1(6 - 3) + 4(3 - 2) + 7(2 - 6)|\)\( \)\(= \frac{1}{2}|1(3) + 4(1) + 7(-4)|\)\( \)\(= \frac{1}{2}|3 + 4 - 28|\)\( \)\(= \frac{1}{2}|-21|\)\( \)\(= \frac{21}{2} = 10.5 \text{ square units}\)$

Example 2: Checking for Collinearity Show that the points \(P(2, 3)\), \(Q(4, 7)\), and \(R(5, 9)\) are collinear.

Solution: Calculate the area of triangle \(PQR\): $\(\text{Area} = \frac{1}{2}|2(7 - 9) + 4(9 - 3) + 5(3 - 7)|\)\( \)\(= \frac{1}{2}|2(-2) + 4(6) + 5(-4)|\)\( \)\(= \frac{1}{2}|-4 + 24 - 20|\)\( \)\(= \frac{1}{2}|0| = 0\)$

Since the area is 0, the points are collinear.

Example 3: Finding a Missing Vertex A triangle has vertices at \(A(2, 1)\) and \(B(5, 4)\). If the area is 6 square units and the third vertex \(C\) lies on the line \(x = 8\), find the coordinates of \(C\).

Solution: Let \(C = (8, y)\). Using the area formula: $\(6 = \frac{1}{2}|2(4 - y) + 5(y - 1) + 8(1 - 4)|\)\( \)\(12 = |8 - 2y + 5y - 5 + 8(-3)|\)\( \)\(12 = |8 - 2y + 5y - 5 - 24|\)\( \)\(12 = |3y - 21|\)$

This gives us: \(3y - 21 = 12\) or \(3y - 21 = -12\)

Case 1: \(3y = 33\), so \(y = 11\) Case 2: \(3y = 9\), so \(y = 3\)

Therefore, \(C = (8, 11)\) or \(C = (8, 3)\).

Multiple Choice Questions#

Sector Specific Questions: Triangle Area Applications#

Key Takeaways#

Important

Area of a Triangle - Essential Formulas

  1. Determinant Formula: Area = \(\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)

  2. Matrix Form: Area = \(\frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|\)

  3. Collinearity Test: If area = 0, the three points are collinear

  4. Applications: Used in surveying, computer graphics, engineering design, and geographic mapping

  5. Remember: The absolute value ensures area is always positive, regardless of vertex order

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