A. ABCD has vertices at A(3,4), B(-1,2), C(-1,-1) and D(a,b), given that ABCD is a parallelogram, use vector method to find the coordinates of D.

The position vectors are:

$$\vec{A} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}, \quad \vec{C} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}, \quad \vec{D} = \begin{pmatrix} a \\ b \end{pmatrix}$$

For a parallelogram ABCD, the vector $\vec{AB}$ must be equal to the vector $\vec{DC}$.

$$\vec{AB} = \vec{B} - \vec{A} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 - 3 \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} -4 \\ -2 \end{pmatrix}$$ $$\vec{DC} = \vec{C} - \vec{D} = \begin{pmatrix} -1 \\ -1 \end{pmatrix} - \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -1 - a \\ -1 - b \end{pmatrix}$$

$$\begin{pmatrix} -4 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 - a \\ -1 - b \end{pmatrix}$$

Equating the x-components:

$$-4 = -1 - a$$ $$a = -1 + 4$$ $$a = 3$$

Equating the y-components:

$$-2 = -1 - b$$ $$b = -1 + 2$$ $$b = 1$$

The coordinates of D are $(a, b)$.

$$D = (3, 1)$$

The coordinates of D are $*(3, 1)*$.