I need help figuring out the area of the triangle with the following vertices.?

Question

(3, 0), (1, 3), (6, 2)

So then my answer is 6.5 or 13/2 and it would be positive because area can only be positive?

Answer 1

The area of a triangle is half base times height.

Could you properly define the length of the base, height and hypotenuse?

Answer 2

If you can use slope, the slope (rise over run) of the line from (1,3) to (3,0) is m=3/2 and from (3,0) to (6,2) is m=2/3. These are reciprocals, so they are perpendicular lines (IE. you have a right triangle with rt angle at the point (3,0) ). Then use the distance formula for these two lines and the area formula for a rt triangle.

d= √[(x2-x1)^2+(y2-y1)^2]
with (1,3) and (3,0)
d=√[(3-1)^2+(0-3)^2], d=√[2^2+(-3)^2], d=√(4+9) or √13
now with (3,0) and (6,2)
it is very similar and you get √13 also.
So with the rt triangle area formula where base = √13 and h =√13 you get:
Area=1/2*b*h or 1/2*√13*√13 = 1/2*13 = 6.5 square units

Answer 3

The answer is six (6).

It's simple: You have to enclose the whole triangle into a big rectangle, which has an area of 15 (=5 x 3). And then get the areas of the triangles outside your main triangle which are 3 each. And then subtract its sum from 15. The answer is 15 - 9 = 6.


I hope I got it right.!

Answer 4

First, you need to plot the 3 points on a graph. You should get that the height is 3 and the base is 6 after you plot the points. Then you use the formula A=1/2(b)(h), and plug in the height and base and you've figured it out.

Answer 5

There is a formula for the calculation of the area of a triangle from the coordinates of its vertices. It is 0.5 times the determinant formed from the following row vectors:

Top row [x1,,,x2,,,x3]; second row [y1...y2...y3]; bottom row [1...1...1]

For your problem, the top row of the determinant is [3...1...6], and the second row is [0...3...2]

EDIT:

In answer to your additional comment, yes the answer is 6.5. If you look at the reference the formula calls for the absolute value of the determinant. (I should have said that.)

Answer 6

First, you graph the standards. on the grounds that your factors all have 0s in it, that makes the calculations undemanding. The formula for area of a triangle is 0.5bh (base x top). Your top is 8 and your base is 7. The equation is 0.5 (7) (8) = 28. And bear in ideas the instruments for area are squared, ex. 28 sq. feet. or 28 sq. miles.

Answer 7

Let A(3,0), B(1,3) and C(6,2) be the 3 points and D be the point where a perpendicular from point C meets AB.
Area = 1/2 length AB X length CD.
(This is generally not required but it would be a good step if we can ascertain all these points are not linear find slopes of AB,AC,BC and you will find they are not linear hence area is not '0') Now I tell you the method - I wont solve since you mentioned help.
Find Slope AB and using coordinates of either point A or B find equation of line AB. (using point-slope form)
As CD is perpendicular to AB find slope CD (using slope AB X slope CD = -1)
Find equation of CD using above slope of CD and coordinate of point C.
Solve equation of line CD and equation of line AB to find point D.
Find length AB and length CD using distance formula.
Area of triangle is 1/2 [length AB X length CD]
Method 2 - using determinants
we can find area using
A = (1/2) determinant (x1,y1, 1) ( x2,y2,1) and (x3,y3,1) in 3 rows
here (x1,y1) (x2,y2),(x3 y3) represent different points A,B,C


Subhash

Answer 8

I don't have an answer, but maybe you'll appreciate this pick-up line I heard at Hi-Y camp.
"I wish I was a derivative, so I could lie to your curves."
Maybe you can use it on some girl in your math class.
Good luck with that math, sorry I don't have any math courses until next term, and my brain is kind of fried.