1. ## Triangle vertices problem

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?

2. 15 sq. units

3. Originally Posted by sri340
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]

4. Originally Posted by ukorov
15 sq. units
just giving them the answer doesnt help them one bit.

5. 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..

6. Originally Posted by orange gold
just giving them the answer doesnt help them one bit.
Simply put the 3 points on a coordinate plane and he should see a right-angled triangle with BC = 6 and AB = 5