Question

(3.1) Find the area of the triangle with the given vertices A(1, 3), B(−3, 5), and C with C = 2A.

(3.2) Use (3.1) to find the coordinates of the point D such that the quadrilateral ABCD is a parallelogram. 

Answer 1

(3.1)

$$\overrightarrow{AB}=\langle-3-1, 5-3\rangle=\langle-4, 2\rangle$$

$C=(2(1), 2(3))$

$$\overrightarrow{AC}=\langle2(1)-1, 2(3)-3\rangle=\langle1, 3\rangle$$

$$\overrightarrow{AB}\times \overrightarrow{AC}=\begin{vmatrix} \vec i & \vec j & \vec k \\ -4 & 2 & 0 \\ 1 & 3 & 0 \end{vmatrix}=(-4(3)-2(1))\vec k=-14\vec k$$

$$=(-4(3)-2(1))\vec k=-14\vec k$$

$$Area_{ABC}=\dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=7(units^2)$$

Area of the triangle ABC is 7 square units.

(3.2)

Assume that the fourth vertex of parallelogram is $D(x, y)$

$$\overrightarrow{AB}=\overrightarrow{DC}=\langle-4, 2\rangle$$

$$\langle 2-x, 6-y\rangle=\langle-4, 2\rangle$$

$x=6, y=4$

$D(6, 4)$