Find the area of the right triangle with vertices A, B, C.?

Question

Point A = (-6,-3)
Point B = (3,5)
POint c = (-1,-5)

Answer 1

It really helps to draw a picture.
Find the slope between AC and BC. This will tell you that it is a right triangle.
Use the distance formula to find the length of AC and the length of BC
Multiply the two lengths together, then divide by 2. That will be your answer.

Answer 2

the answer is 29
to work it out
first draw a grid the plot the triangle. surround the triangle with a box touching all the corners of the triangle.then for each part of the box which isn't overlapped by the triangle draw a smaller square do this for all three sides then subtract half of all the squares area from the area of the main box
alternatively gain the area of the triangles surrounding the original by calculation (1/2bxh) the subtract that from the total area of the box.

Answer 3

Well, I would apply the correct formula, which is A = 1/2(L*H)

So determine the Legnth of Segment AC, and the Legnth of Segment BC and plug the numbers into the formula.

By Plotting the points on a graph you should be able to determine where the right angle will be.

Hope this helps!

Answer 4

(-6,-3) and (3,5)
D = sqrt((5 - (-3))^2 + (3 - (-6))^2)
D = sqrt((5 + 3)^2 + (3 + 6)^2)
D = sqrt(8^2 + 9^2)
D = sqrt(64 + 81)
D = sqrt(145)

(3,5) and (-1,-5)
D = sqrt((-5 - 5)^2 + (-1 - 3)^2)
D = sqrt((-5 + (-5))^2 + (-1 + (-3))^2)
D = sqrt((-10)^2 + (-4)^2)
D = sqrt(100 + 16)
D = sqrt(116)

(-6,-3) and (-1,-5)
D = sqrt((-5 - (-3))^2 + (-1 - (-6))^2)
D = sqrt((-2)^2 + 5^2)
D = sqrt(4 + 25)
D = sqrt(29)

a^2 + b^2 = c^2

The legs are AC and BC
Hypothenuse is AB

Area = (AC * BC)/2
Area = (sqrt(29) * sqrt(116))/2
Area = (sqrt(3364))/2
Area = (58/2)
Area = 29

ANS : 29 units^2

Answer 5

first you have to prove it's a right triangle. then you need the length of the two shortest sides. then you are on your way to getting the answer