# Determine the area of a triangle whose vertices are P(8,8), Q(5,-7) and R(-5,-5).

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Let A be the point at (8,8), B (-5,-5) and C (5,-7)

(1)The easiest way to find the area of a triangle with vertices on lattice points in the coordinate plane is to find the determinant of the 3x3 matrix whose row elements are the coordinates and a 1. The area will be `+-1/2` this determinant, making sure that the area is positive.

So we find the determinant of :

| 8 8 1|

| 5 -7 1|

|-5 -5 1|

**The determinant is -156. Multiplying by `-1/2` gives the area as 78.**

(2) You might realize you have a right triangle. Then the area is 1/2 the product of the legs. ** The line through BC has slope `-1/5` while the slope through AC has slope 5, thus they are perpendicular.**

The length of the legs can be found using the pythagorean theorem:

`BC=sqrt(2^2+10^2)=sqrt(104)` and `AC=sqrt(3^2+15^2)=sqrt(234)`

So the area is `1/2BC*AC=1/2sqrt(104)sqrt(234)=1/2(156)=78`

(3) If you plot the points on a coordinate system you can use Pick's theorem (see reference): The area is the number of lattice points in the interior of the triangle plus 1/2 of the lattice points that lie on the boundary of the triangle minus 1.

Thus `A=I+1/2B-1=70+1/2(18)-1=78`