Area of triangle with given coordinates of the vertices

Question

The question for my math is:

"Sharon made a scale drawing of a triangular park. The coordinate for the vertices of the park are: $(-10,5)$, $(15,5)$, $(10,12)$. What is the area of the triangular park in square meters?

(I used the formula: $\frac{1}{2}\left( x_1(y_2-y_3)+ x_2(y_3-y_1)+ x_3(y_1-y_2)\right)$ The answer is got is: $92.5$. But the answer is $87.5$ (according to the online practice test).

Answer 1

You don't need to use this formula: the basic 1/2 height × base will do, as you very simply get a base and a height for this triangle.

Answer 2

Using determinant method:

$$\text{Area}=\frac{1}{2} |\det \left[ \begin{array}{} 1&1&1\\-10&15&10\\ 5&5&12\end{array}\right]|$$ $$=\frac{1}{2} |\det \left[ \begin{array}{} 1&0&0\\-10&25&20\\ 5&0&7\end{array}\right]|$$ $$=\frac{1}{2}\times25\times7=87.5$$

Answer 3

When I put the formula in Wolfram Alpha, I get $87.5$, so you probably just put in some of the values incorrectly. However, as Bernard points out, the complicated formula you have is overkill for this problem: the points are actually nice enough the the usual $\frac{1}{2}\times\text{base}\times\text{height}$ formula is simple to apply.

Here's a picture of the situation:

The points $(-10,5)$ and $(15,5)$ both lie on the line $y= 5$ (since they have the same $y$ coordinate). So, if you take the line segment between the two points to be the base, the height will be given by the distance between the line $y=5$ and the point $(10,12)$. Can you find that distance?

Answer 4

Answer:

base = $ \sqrt{(15+10)^2+(5-5)^2} = 25$

This case:

$y_2 - y_1 = \text{height}$

$ 12 - 5 = 7$ is height

area = $\frac{25 * 7}{2} = 87.5$

Answer 5

The actual determinant method is as follows:-

$\dfrac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} &1 \\ x_{3} & y_{3} & 1 \end{vmatrix}=\dfrac{1}{2}\left|x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2-x_3)\right|$