Question Video: Finding the Area of a Triangle given the Coordinates of Its Vertices Mathematics

Video Transcript

Find the area of a triangle 𝐴𝐵𝐶, where 𝐴 has coordinates negative eight, negative nine; 𝐵 has coordinates negative seven, negative eight; and 𝐶 has coordinates nine, negative two.

We could begin by trying to draw the points on a coordinate grid. However, as points 𝐴 and 𝐵 are so close together, it will be difficult to draw this accurately. As a result, we will just sketch the triangle roughly as shown.

We recall that the area of any triangle is equal to half the area of a parallelogram, where the lengths of the sides of the parallelogram are equal to two of the lengths of the triangle. We recall that the area of any parallelogram is equal to the magnitude of the cross product of vector 𝐚 and vector 𝐛.

In this question, the area of parallelogram 𝐴𝐵𝐶𝐷 is equal to the magnitude of the cross product of vectors 𝐁𝐀 and 𝐁𝐂. The area of triangle 𝐴𝐵𝐶 is therefore equal to a half of this. We know that the magnitude of the cross product is equal to the magnitude of vector 𝐚 multiplied by the magnitude of vector 𝐛 multiplied by the magnitude of sin 𝜃, where 𝜃 is the angle between vectors 𝐚 and 𝐛.

The area of triangle 𝐴𝐵𝐶 is therefore equal to a half multiplied by the magnitude of vector 𝐁𝐀 multiplied by the magnitude of vector 𝐁𝐂 multiplied by the magnitude of sin 𝜃. The magnitude of any vector is equal to its length. Therefore, the magnitude of vector 𝐁𝐀 is equal to the square root of negative eight minus negative seven all squared plus negative nine minus negative eight all squared. This is equal to the square root of two.

We can repeat this process to calculate the magnitude of vector 𝐁𝐂. This is equal to nine minus negative seven all squared plus negative two minus negative eight all squared. This is equal to the square root of 292, which is equal to two root 73.

In order to calculate the angle 𝜃 in our triangle, we will firstly need to calculate the length of 𝐴𝐶 and then use the cosine rule. The length of 𝐴𝐶 is equal to the magnitude of vector 𝐀𝐂. And this is equal to the square root of nine minus negative eight all squared plus negative two minus negative nine all squared. This is equal to the square root of 338, which is equal to 13 root two.

The cosine rule states that the cos of angle 𝐵 is equal to 𝑎 squared plus 𝑐 squared minus 𝑏 squared all divided by two 𝑎𝑐. In this question, the sides 𝑎, 𝑏, and 𝑐 are equal to two root 73, 13 root two, and root two, respectively. Angle 𝐵 is the angle 𝜃 we are trying to calculate. Substituting in our values, we see that the cos of 𝜃 is equal to two root 73 squared plus root two squared minus 13 root two squared all divided by two multiplied by two root 73 multiplied by root two.

Typing the right-hand side of the equation into our calculator gives us negative 0.9103 and so on. We can then take the inverse cosine of both sides of this equation such that 𝜃 is equal to 155.556 and so on degrees.

We can now substitute these values back into our equation to calculate the area of the triangle. It is important at this stage that we do not round the angle 155.556 and so on degrees. Typing this into the calculator gives us an answer of five. Therefore, the area of triangle 𝐴𝐵𝐶 is equal to five square units.