Question

two vertices of rectangle abcd are located at a (-4, -2) and b (5, -2). if the midpoint of (overline{ad}) is located 5 units down from a, what are the coordinates of the vertices c and d? enter the correct coordinates in the boxes.

Explanation:

Step1: Find midpoint of \( \overline{AD} \)

A is at \((-4, -2)\). Moving 5 units down (y - coordinate decreases by 5), so midpoint \( M \) has y - coordinate \(-2 - 5=-7\), x - coordinate same as A: \(-4\). So midpoint \( M(-4, -7)\).

Step2: Find coordinates of D

Let \( D=(x,y) \). Midpoint formula: \( M = (\frac{-4 + x}{2},\frac{-2 + y}{2}) \). We know \( M(-4, -7) \).
For x - coordinate: \(\frac{-4 + x}{2}=-4\) → \(-4 + x=-8\) → \( x=-4 \).
For y - coordinate: \(\frac{-2 + y}{2}=-7\) → \(-2 + y=-14\) → \( y=-12 \). So \( D(-4, -12) \).

Step3: Find coordinates of C

In rectangle \( ABCD \), \( \overrightarrow{AB}=\overrightarrow{DC} \). \( \overrightarrow{AB}=(5 - (-4),-2 - (-2))=(9,0) \). Let \( C=(p,q) \), \( D(-4, -12) \). So \( (p - (-4),q - (-12))=(9,0) \).
For x - coordinate: \( p + 4 = 9 \) → \( p = 5 \).
For y - coordinate: \( q + 12 = 0 \) → \( q=-12 \). So \( C(5, -12) \).

Answer:

\( C(5, -12) \), \( D(-4, -12) \)