Example: Finding the Missing Point in a Parallelogram

Given a parallelogram $ABCD$ with sides defined by position vectors:

• $A = 2i + j$
• $B = 6i$
• $C = 4i - 2j$ Find the position vector of point $D$.

Steps to Solve:

1. Convert Position Vectors to Column Form:

• $A = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
• $B = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$
• $C = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$

2. Finding Vector $\overrightarrow{BC}$:

• $\overrightarrow{BC} = C - B$
• $\overrightarrow{BC} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} - \begin{pmatrix} 6 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \end{pmatrix}$

3. Finding Position Vector of $D$:

• Since $ABCD$ is a parallelogram, $\overrightarrow{BC} = \overrightarrow{AD}$.
• Therefore, $D = A + \overrightarrow{BC}$
• $D = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} -2 \\ -2 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$

• Rectangle: A 4-sided shape with two pairs of parallel sides and right angles.
• Parallelogram: A 4-sided shape with two pairs of parallel sides.
• Rhombus: A shape with four sides of the same length.
• Square: A shape with all four sides of the same length and all angles as right angles.

Example Problem: Prove ABCD is a Parallelogram

Given points:

• $A = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
• $B = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$
• $C = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$
• $D = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$

4. Finding Vectors:

• $\overrightarrow{AB} = B - A = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$
• $\overrightarrow{CD} = D - C =\begin{pmatrix} 4 \\ -1 \end{pmatrix}$

5. Finding Vectors:

• $\overrightarrow{BC} = C - B =\begin{pmatrix} -2 \\ -2 \end{pmatrix}$
• $\overrightarrow{AD} = D - A =\begin{pmatrix} -2 \\ -2 \end{pmatrix}$

6. Conclusion:

• Since $\overrightarrow{AB} = \overrightarrow{CD}$ and $\overrightarrow{BC} = \overrightarrow{AD}$, $AB \parallel CD$ and $BC \parallel AD$.
• Therefore, $ABCD$ is a parallelogram.

Prove $ABCD$ is not a Rectangle (Hint: Use Gradients)

1. Calculate Gradient of $BC$:

2. Calculate Gradient of $CD$:

3. Conclusion: Since $1 \times -\frac{1}{4} \neq -1$, the lines are not perpendicular. Therefore, $ABCD$ is not a rectangle.