Pythagorean theorem?

Question

Use the Pythagorean theorem to determine the length of each line segment. Round to the nearest hundredth where appropriate.
I do not understand this in the slightest.
I kow that the two points are (3,2) and (1,-1).
Please help?!

Answer 1

You need a third point. Pythagorean theorem only works with right triangles. Triangles need 3 points.

Answer 2

first, start by picturing (or actually drawing) the points on a graph

one point (1,-1) is just to the right and just below the origin
the other point (3,2) is over 3 to the right of the origin and up two

if you connect the two points you have a line that slants across the graph

if you start at (1,-1) and draw straight sideways until you get to (3,-1) then you will be directly under (3,2) and you can draw straight up to (3,2)

now you have a right triangle

the Pythagorean theorem (I hope you know) states that the square of the length of the hypoteneuse of a right triangle (the first slanting line) is equal to the sum of the squares of the other two sides (c^2=a^2+b^2)

with just a little study, we can easily get the length of the verticle and horizontal lines

the horizontal line starts at (1,-1) and goes right to (3, -1) which is clearly a distance of 2

the vertical line starts at (3,-1) and goes up to (3,2) which is a distance of 3

now we have the two non-hypoteneus sides and we can write the Pythagorian theorem with the lengths inserted

c^2=2^2+3^2
c^2=4+9=13
c=3.61 (rounded as directed)

drawing a picture is almost always a good way to start any geometry problem

finding the right triangle is the way to proceed with any pythagorean application

keep at it

Answer 3

First it may be helpful if you graphed these two points on a graph.

Then realize that you can draw the hypotenuse of a triangle by connecting the two points. Then if you draw straigh up and straight left, you can draw a right triangle.

Then just plug into the equation of a^2+b^2=c^2
We will call a the distance straight up from 1, -1 to 3,2. That would be 3.

Then b would be the distance going left from 3,2 to 1, -1. That would be 2.

So we get 3^2 + 2^2 = c^2
9+4=c^2
13=c^2
c=square root of 13

Answer 4

You can draw the two points on a graph, then connect them with a segment, and then make a right triangle with that segment as its hypotenuse. Then just count how long each leg is, and use these for the Pythagorean Theorem.

Answer 5

I think the two line segments are from (0,0) to (3,2) and from (0,0) to (1,-1).

Using the pythagorean theorem,

the first has a length of (3^2 + 2^2)^0.5 = 13^0.5 or the square root of 13;

the second has a length of (1^2 + 1^2)^0.5 or 2^.05 or the square root of 2.

Answer 6

The pythagorean theorem states that A^2xB^2=C^2
To do the problem graph the points and use the distance formula to discoer the length of the segment. It should create a right triangle and use A=length of leg A, B=Length of leg B, and C=length of hypotenuse, if you need further help email me

Answer 7

When in doubt, draw a picture!
It helps so much. You will see that the third point is (1,2).
or (3,-1), either one will work.
Then you can figure out the lengths of the two legs
a and b and use the the formula a^2 + b^2 = c^2
to figure out the hypoteneuse c, which is the length of the
line segment you are supposed to find.

Answer 8

c = 3.61
a = 3
b = 2

Don't listen to these bozos.

1) Make a grid.
2) Plot the points and connect them.
3) Draw orthogonal lines from both points toward a common intersection.
4) Count the grid number of each non-hypotenuse side. One will be 2, the other will be 3. The hypotenuse side will be the square root of the sum of the squares of the other two sides.

Answer 9

The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry.

Answer 10

Pythagorean Theorem

c² = a² + b²

Directly Related to Triangles

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - --

Ordered Pairs

(3, 2) (1, - 1)

Can be used in the distance formula

d = √(x₂- x₁) + (y₂- y₁)

Slope of a line

m = y₂- y ₁/ x ₂- x₁

Point Slope Form

y - y₁= m(x - x₁)