# A triangle is formed by the points A(-1,1) B(4,3) and C (0,13)?

Jul 20, 2015

ABC is a triangle, with area $29$

#### Explanation:

Slope $m$ between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by the formula:

$m = \frac{\Delta y}{\Delta x} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

The slope of a line through $A$ and $B$ is:

${m}_{A B} = \frac{3 - 1}{4 - \left(- 1\right)} = \frac{2}{5}$

The slope of a line through $A$ and $C$ is:

${m}_{A C} = \frac{13 - 1}{0 - \left(- 1\right)} = 12$

So these three points are not colinear and do form the vertices of a triangle.

Let ${\underline{v}}_{A B} = B - A = \left(5 , 2\right)$ and ${\underline{v}}_{A C} = C - A = \left(1 , 12\right)$

Then the area of the triangle is:

$\frac{1}{2} \left\mid {\underline{v}}_{A B} \times {\underline{v}}_{A C} \right\mid = \frac{1}{2} \left\mid 5 \cdot 12 - 2 \cdot 1 \right\mid = \frac{1}{2} \left\mid 58 \right\mid = 29$