Triangles Help?

Question

I need help getting started with this one... please!
Verify that c(2,-1), a (8,7) and t(4,10) are vertices of a right triangle.
do i plot these out? thanks very much!

Answer 1

The two sides that form the right angle are the sides CA and AT. Look at the slopes of each of these sides.

Slope of side CA = (8 / 6)
Slope of side AT = - (3 / 4)

8/6 and -3/4 are negative reciprocals. Negative reciprocal slopes imply that two lines are perpendicular, hence a right angle.

If you really want to get technical you also have to verify that this is indeed a triangle. Use the triangle inequaltiy. Find distances of each segment, then show that
(1) AC + CT >= AT
(2) AT + CT >= AC
(3) AT + AC >= CT

Answer 2

Each two points describes a line. Each line is in a certain orientation the others. Describe the lines and find two that have a negative inverse of the other, ie perpendicular; if they are perp, then you have a right triangle.
Line ca:
yca(x) = mx + b
yca(2) = m(2) + b = -1
yca(8) = m(8) + b = 7
Subtract the first from the second to get rid of b, then find m:
6m = 8 and m = 8/6 = 4/3

Line ct:
yct(2) = 2m + b = -1
yct(4) = 4m + b = 10
2m = 11, m = 11/2 which is not the negative invers of line ca so onward and upward

Line at:
yat(8) = 8m + b = 7
yat(4) = 4m + b = 10
4m = -3
m = -3/4

So lines ca and at have the negative inverse of each others slope; hence, 90 degrees/perpendicular or a right triangle.

Answer 3

You should ALWAYS plot things out. You can show this since the slope from a to t is -3/4, while the slope from from c to a is 4/3. That is sufficient to establish the right angle.

Answer 4

you can either plot these coordinates on a graph paper and prove that it is a right angled triangle, OR you can generate line equations out of the coordinates given nad then prove that one pair of lines out of the 3 are mutually perpendicular.

hope this helps

Answer 5

get the distance of each point then try pythagorean theorem;

a^2 + b^2 = c^2

if it satisfies the given equation, then that is a right triangle.