# Coordinate Geometry Problems - Page 3

Coordinate Geometry Problems
• Page 3
21.
ABCD is a trapezoid with $\stackrel{‾}{\mathrm{AB}}$ parallel to $\stackrel{‾}{\mathrm{DC}}$. P and Q are the midpoints of $\stackrel{‾}{\mathrm{AC}}$ and $\stackrel{‾}{\mathrm{BD}}$ respectively. The length of PQ is equal to __________.

 a. $\frac{1}{2}$ (AB+DC) b. (AB+DC) c. $\frac{1}{2}$|AB - DC| d. |AB-DC|

#### Solution:

Use the midpoint formula to find the coordinates of P and Q.

Coordinates of P: (0 + 2b2 , 0 + 2c2) = (b, c)

Coordinates of Q: (2a + 2d2 , 0 + 2c2) = (a + d, c)

Use the distance formula, d = (x2 -x1)² + (y2 -y1)² to find the lengths of AB, DC and PQ.

AB = (2a - 0)2 +(0 - 0)2 = (2a)2 +02 = 2a

DC = (2b - 2d)2 +(2c - 2c)2 = (2b - 2d)2 +02 = (2b - 2d)

PQ = (a + d  - b)2 +(c - c)2 = (a + d - b)2 +02 = (a + d - b)

|AB - DC| = |2a - (2b - 2d)| = |2a - 2b + 2d |

= 2|a + d - b|= 2PQ

So, PQ = 12|AB - DC|