Find area of triangle RTS when $R=(-3,-4),$ $S=(3,-1),$ $T=(2,4)$

Question

I can't from find this answer I tried whole day please help

Answer 1

You can use the Pick's theorem:

$$A=i+{b\over2}-1$$

where $i$ are the interior points with integer coordinates and $b$ the points on the boundary (with integer coordinates).

So we get:

$$A=15+{5\over2}-1={33\over2}$$

Answer 2

The area of a triangle is given by $\frac{1}{2}|\vec{a} \times \vec{b}|$.

We define $\vec{a}$ to be the vector from $R$ to $S$, and $\vec{b}$ to be the vector from $S$ to $T$.

We obtain $\vec{a}= \begin{pmatrix} 6 \\ 3 \end{pmatrix}$ and $\vec{b}= \begin{pmatrix} -1 \\ 5 \end{pmatrix}$.

Evaluate the cross product of these two vectors. $\vec{a} \times \vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 3 & 0 \\ -1 & 5 & 0 \end{vmatrix}=33\hat{k}$.

Now, evaluate the modulus to be $|\vec{a} \times \vec{b}|=33$, and divide by $2$.

Hence the area of triangle $\Delta RTS$ is $16.5$ units.

Answer 3

You can find area of a triangle using shoelace formula:

\begin{align*} \text{Area} &= \frac12 \big| (x_1 - x_3) (y_2 - y_1) - (x_1 - x_2) (y_3 - y_1) \big|\\ &= \frac12 \big| x_1 y_2 + x_2 y_3 + x_3 y_1 - x_1 y_3 - x_3 y_2 - x_2 y_1 \big|. \end{align*}

In your case $x_1=-3, x_2=3, x_3=2, y_1=-4, y_2=-1, y_3=4$

$= \frac12 \big| (-3) * (-1) + (3) * (4) + (2) * (-4) - (-3) * (4) - (2) * (-1) - (3) * (-4) \big|$

$= \frac12 \big| 3 + 12 - 8 + 12 + 2 + 12\big|$

$= \frac12 \big| 33\big|$

= 16.5 square units.

Or

You can solve it using 1/2 of the following determinant.

$$ \begin{vmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{vmatrix} $$