Question

Find the area of a triangle with vertices (–8, 10), (6, 17), and (2, –4).

Answer 1

@zepp

Answer 2

@abstracted

Answer 3

with A(-2,4) B(-8,10) C(6,7)
from trig we know that,\[h=AB \times \sin(atB)\]\[AB=\sqrt{(-8--2)^{2}+(10-4)^{2}}=6\sqrt{2}\]Now for the angle we will need vectors BA and BC\[BA=A-B=(6,-6)\]\[BC=C-B=(14,-3)\]

Answer 4

So the height it 6 root 2?

And so we use trig functions to solve for this?

Answer 5

\[angleBA=\arctan(-6/6)=-\pi/4\]\[angleBC=\arctan(-3/14)\approx-0.211 rad\]So angle at B=\[|-\pi/4-0.211|\approx0.5743rad\]

Answer 6

thus\[h \approx 6\sqrt{2} \times sen(0.5743)\approx4.60\]Now Area=h x BC/2

Answer 7

Can this question also be solved through a matrix? Because i think thats what i'm supposed to use... xD I'm on the matrix lesson, which is where this came up.

Answer 8

hum... a matrix... let me think about it, anyway you just needed to know how much from b to c and you'd have your area

Answer 9

Alright xD

Answer 10

hey here put your neurons to work too xd,
For any given triangle Area = BC x AB times sen (angle at b) over 2
now... with that we need to make a matrix....

Answer 11

I'm trying haha i'm going over the lesson thing to see if they had this anywhere

Answer 12

bummer... just put it into a matrix calculate det and divide by 2

Answer 13

@Syderitic You don't even need matrices to solve this problem xD There's an easier way :)

Answer 14

Thats what it said.