What is the value of x in a triangle with the verticies (6,5) (8,2) (x,11) and is 15 square units
Im getting ready to upload a picture, this is not the easiest way, but the fastest way i know of to tackle this problem. it uses vectors and a matrix. If you want i can make another solution using the distance formula and law of sines/cosines/whatever, although it will be a lot longer than my current solution
impress me joe
the idea for this comes from the fact that the det of 2x2 matrix gives you the area of the parallelogram formed by the vectors. the triangle is half that area.
did you try the draw the triangle with vertices (-8,11), (6,5) and (8,2)
hmm...trying to verify the answer and its not working...maybe my idea is a fluke >.< one sec <.<
i think x has to be positive
how is x = -8 in that picture?
satellite!
http://en.wikipedia.org/wiki/Determinant#2-by-2_matrices check this out, jump to 2x2 matrices part
hello my myininaya.
i got x=11 but i dont know if it is right
it is not 4 or eight
11 is not a option
is it 12? x = 12
do you know the way joe was talking about with matrices?
-8 and 12 works
oh really? sweet :P
so its multiplie choice of those choices the answer is most likely 12 since i cannot draw a traingle with x=-8
method? or guess and check?
it comes as -8 or 12 as one answer
\[\left| \det(A) \right| = area \] of the parallelogram, thats what generates 2 answers, i forgot that in my sketch
i forgot the absolute value
nice work
im jealous
gj joe
One of my professors has an extreme passion for all that is Linear Algebra, and it rubbed off on me, so ive been reading books on the subject. its a beatiful subject :)
Thanks so much joe and the other 2 of you :)
satellite and i were just here for looks
i give you medals because you both are awesome :)
how sweet :)
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