Question: Parallelogram \( DEFG \) is shown below. Give the coordinates of \( D \). \[…

Parallelogram \( DEFG \) is shown below. Give the coordinates of \( D \).

\[ \begin{array}{c} \begin{tikzpicture} \draw[->] (-2, 0) — (2, 0) node[right] {$x$}; \draw[->] (0, -1) — (0, 3) node[above] {$y$}; \draw[thick, blue] (-1, 0) — (1, 0) — (2, 2) — (0, 2) — cycle; \draw[fill=blue] (-1, 0) circle (1pt) node[below] {$F (-p, 0)$}; \draw[fill=blue] (1, 0) circle (1pt) node[below] {$G (q, 0)$}; \draw[fill=blue] (0, 2) circle (1pt) node[above] {$E (0, n)$}; \draw[fill=blue] (2, 2) circle (1pt) node[above] {$D (?, ?)$}; \end{tikzpicture} \end{array} \]

Solution

In a parallelogram, opposite sides are parallel and have the same length. To find the coordinates of point \( D \), we can use the property that the vectors for opposite sides are equal. The coordinates of \( F \) are \( (-p, 0) \). The coordinates of \( G \) are \( (q, 0) \). The coordinates of \( E \) are \( (0, n) \). Since \( DEFG \) is a parallelogram, vector \( \overrightarrow{EF} = \overrightarrow{DG} \). First, find vector \( \overrightarrow{EF} \): \[ \overrightarrow{EF} = (-p - 0, 0 - n) = (-p, -n) \] Since \( \overrightarrow{DG} = \overrightarrow{EF} \), set them equal: \[ \overrightarrow{DG} = (q - x, 0 - y) = (-p, -n) \] Set the components equal: \[ \begin{cases} q - x = -p & \\ 0 - y = -n & \end{cases} \] Solve for \( x \) and \( y \): For the first equation: \[ q - x = -p \] \[ x = q + p \] For the second equation: \[ -y = -n \] \[ y = n \] Thus, the coordinates of \( D \) are \( (q + p, n) \).