P(7,6), Q(1,6), R(4,2)
We have PQ parallel to the x axis. We'll call that the base,
b = 7 - 1 = 6
The altitude is then the y difference h = 6 - 2 = 4
The area is
![\frac 1 2 bh = \frac 1 2 (6)(4) = 12](https://img.qammunity.org/2019/formulas/mathematics/high-school/ukt44rydbfsbc4tmcnwvg4tbrygyldchgu.png)
Answer: 12 square units
In general we can use the shoelace formula for the area of any polygon given coordinates. We write the points like this:
(7,6), (1,6), (4,2)
(1,6), (4,2), (7,6)
The area is then half the absolute value of the sum of the cross products:
![A = \frac 1 2 | 7(6)-6(1) + 1(2)-6(4) + 4(6)-2(7) | = \frac 1 2 |24| = 12](https://img.qammunity.org/2019/formulas/mathematics/high-school/go0flwfoa373va21atfg3bps575qfuq13y.png)