Triangle ABC has its vertices at A(0,4), B(0,9) and C(6,9). What is the area of the triangle, in square units?
Look at the points this way (repeat the first coord pair at the end):
$\displaystyle \frac{0}{4} ; \dfrac{0}{9} ; \dfrac{6}{9} ; \;\; \dfrac{0}{4}$
Multiply the numerators by the next following denominator:
0*9 + 0*9 + 6*4 = 24
Then multiply the denominators by the next following numerator:
4*0 + 9*6 + 9*0 = 54
Subtract those two sums & take half:
(54 - 24 )/2 = 15
15 sq. units [that agrees with ukorov]
anyways since you have 2 points alligned on the x - coordinates
(0,4) and (0,9) then the line on that side of the triangle is 5... 9-4 = 5
it happens again with (0,9) and (6,9) ... the line on that side is 6... 6-0 = 6
a triangles area is length times width divided by 2.. or half its base times its height...
6*5 = 30.... 30/2 = 15..
-or-
(0.5)(5)(6) = 15..
15 sq units..