Question

The diameter of a circle has endpoints P(3, 7) and A(-3, -1). What is the length of the diameter?

Answer 1

Use the formula for the distance between two arbitrary points \((x_1, y_1), (x_2,y_2)\):
\[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

If this looks suspiciously like the Pythagorean theorem to you, give yourself a pat on the back.

Answer 2

Thanks :)

Answer 3

What do you get for an answer? It should be a nice round number.

Answer 4

Im doing it right now

Answer 5

Also, can you see why it doesn't matter which point you call \((x_1, y_1)\) and which point you call \((x_2, y_2)\)?

Answer 6

no, so it doesn't matter where i put it

Answer 7

Is it 8

Answer 8

Mmm....no. Can you show your work?

Answer 9

(3 - 7)^2 + (-3 - -1)^2
-4^2 + -2^2
16 + 4
square root 20 = 4

Answer 10

but 8 is the closest to my multiple choice answers so i choose that one

Answer 11

Ah, but what if you are doing a real problem, where you don't get to choose from 4 answers (some of which are probably designed to trick you)?

Okay, your points are (3,-7) and (-3, -1). We'll take the first one as x1, y1, and the second as x2, y2. Watch my hands carefully :-)

\[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{((-3)-(3))^2 + {((-1)-(7))^2}}\]\[\sqrt{((-3)-(3))^2 + {((-1)-(7))^2}} = \sqrt{(-6)^2 + (-8)^2}\]
Can you do the rest?

Answer 12

Your mistake was that you did \((x_1-y_1)\) and \((x_2-y_2)\) instead of \((x_1-x_2)\) and \((x_2-y_2)\)...

Answer 13

As I just reversed the order of \(x_1\) and \(x_2\) I might as well demonstrate why it doesn't matter:

\[d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{((-x_1 -(-x_2))^2+(-y_1-(-y_2))^2} =\]\[\sqrt{((-1)(x_1 -x_2))^2+((-1)(y_1-y_2))^2} =\sqrt{(-1)^2(x_1-x_2)^2 + (-1)^2(y_1-y_2)^2}\]But \((-1)^2 =1\) so we can just ignore that and write \[d=\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\] and that's what we set out to demonstrate.

Answer 14

oh okay well thanks I wasn't able to turn it in on time but thanks for the future... really appreciate your time :)

Answer 15

The answer for your question was \[d=\sqrt{(-6)^2+(-8)^2} = \sqrt{36+64} = \sqrt{100} = 10\]