Question

HEEEELPP ME!!!! The vertices of a right triangle are A(3,3), B(-1,4), C(1,-5) Find the area of the triangle.
Answer: 17

Answer 1

Are you in calculus?

Answer 2

Uh no... ):

Answer 3

locate them on the coordinate plane, see which 2 lines form the right triangle, then area is base x height..you should get it

Answer 4

What do you mean by the two lines to form the right triangle...?

Answer 5

sorry my mistake, 2 lines that form the right angle

Answer 6

I placed it on a grid but it's actually not a right triangle, it's more like an equilateral ):

Answer 7

ohh really, i haven't put them on the coordinate plane yet but i think the right angle is at point A, try that..calculate distance of AB and AC, then multiply them together and see what you get

Answer 8

Alright I'll try it

Answer 9

I got 16... :/

Answer 10

ohh :) let me try on paper, hold on a sec.

Answer 11

Alright, thanks :)

Answer 12

i think answer is right 2x17 = 34 ...since AB = \[\sqrt{1+16 } = \sqrt{17}\]
and AC = \[\sqrt{4 + 64} = \sqrt{68} = \sqrt{4x17} = 2\sqrt{17}\]

so area = 1/2 ABxAC = 1/2 x \[\sqrt{17} x 2\sqrt{17} = 17\]

you can check the work again

Answer 13

forget the part 2x17 = 34...because i haven't divide 2 yet

Answer 14

that x in square root, should be a multiply sign, because i am not familiar with math notation here, when i insert x as multiply, it become letter x

Answer 15

How did you get the 1 + 17? I'm sorry, I'm a slow learner :(

Answer 16

it's \[1^{2} + 4^{2} = 1 + 16 = 17\]
\[2^{2} + 8^{2} = 4 + 64 = 68\]

then take square root of each to get the actual length of AB and AB

Answer 17

What method did you use to get this...? the d=sqrt(x2-x1)^2 + (y2-y1)^2?

Answer 18

yeah, for two points A, B with coordinate of A = (x1,y1) and B = (x2,y2)...

distance of AB =