# Math Help - Geometry Question Sheet (Need someone to confirm I did it right)

1. ## Geometry Question Sheet (Need someone to confirm I did it right)

Alright, So since I tried posting a picture in my last thread and that didn't go so well, I'm just going to write out the questions on this sheet I have. There's one question I'm stuck on, and the rest I just want confirmation that it's been done properly. Thanks in advance!

24) The endpoints of a line segment are A(-7, -1) and B(5, 5).

a) Determine the midpoint of AB.
b) Determine the equation of the right bisector of AB.
c) Show that the origin, (0, 0), lies on the right bisector of AB.
d) Determine the equation of the circle with center (0,0) and passing through both A and B.

25)

a) Consider the following statement: "If a circle with its center at the origin has a chord with slope m, the equation of the right bisector of the chord is y = mx." Is this statement true or false? Give detailed reasons for your answer.

b) C(3, 2) is the center of a circle with a radius of 5.

I) Which of the following points lie on the circle?
P(-2, 2), Q(0, 6), R(7, 4), S(-1, -1), and T(6, 6.5)

II) Which point is outside the circle?

III) State the equation of the circle with the same radius, if its center is at the origin.

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Now these are my answers, and where I'm unsure / stuck. Please confirm where I am right, and explain (as simply as possible) how to do the things I'm unsure of. Thanks a million!

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24) a)

(-7 + 5 -1 + 5)
M = (------ , ------)
( 2 2 )

(-2 4 )
M = (--- , ---)
( 2 2 )

M = (-1,2)

Therefore, the midpoint of AB is (-1, 2)
------------------------------------------

b)

Mab = 5 - (-1)
--------
5 - (-7)

Mab = 6/12 Or if simplified: 1/2

Therefore, the slope of the right bisector of line segment AB is -2 since it is the negative reciprocal of Mab
y = mx + b
2 = -2(-1) + b
2 - 2 = b
0 = b

Therefore, the equation of the right bisector of AB is y = -2x
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c)

y = 2x
0 = -2(0)
0 = 0

This proves that the origin (0,0) satisfies this equation, and therefore proves that the right bisector lies on the origin.
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d)

x2 + y2 = r2

For point A(-7, -1),

-72 + -12 = r2
49 + 1 = r2
50 = r2

For point B(5, 5),

52 + 52 = r2
25 + 25 = r2
50 = r2
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25) a)

False, because m in the equation y = mx would have to be the negative reciprocal of the chord, whereas in this equation, it's the slope of the the chord itself, not the negative reciprocal of the chord.

***FROM HERE DOWN IS WHERE I REALLY NEED THE HELP***

Note to readers - when they changed the coordinates of the origin from (0,0) to (3,2), they never did any examples or anything on what exactly changes, and I'm really just confused as to how this works. All I did was draw a graph of a circle with its origin at 3,2 to find the following answers, if I had to try and find the answers and explain it through equations or formula's, that's where I'm stuck.

25) b) I) Points P(-2, 2), Q(0, 6), 5(-1, -1), lie on the circle.
II) T(6, 6.5) is outside the circle.
III) I'm stuck on this last question completely.

2. ## Re: Geometry Question Sheet (Need someone to confirm I did it right)

[QUOTE=Asthuran00;790791
25 b) C(3, 2) is the center of a circle with a radius of 5.

I) Which of the following points lie on the circle?
P(-2, 2), Q(0, 6), R(7, 4), S(-1, -1), and T(6, 6.5)

II) Which point is outside the circle?

III) State the equation of the circle with the same radius, if its center is at the origin.

***FROM HERE DOWN IS WHERE I REALLY NEED THE HELP***

Note to readers - when they changed the coordinates of the origin from (0,0) to (3,2), they never did any examples or anything on what exactly changes, and I'm really just confused as to how this works. All I did was draw a graph of a circle with its origin at 3,2 to find the following answers, if I had to try and find the answers and explain it through equations or formula's, that's where I'm stuck.

25) b) I) Points P(-2, 2), Q(0, 6), 5(-1, -1), lie on the circle.
II) T(6, 6.5) is outside the circle.
III) I'm stuck on this last question completely.[/QUOTE]

The equation of the circle is $(x-3)^2+(y-2)^2=25$.
I. All to do is substitute the points in to see which are on the circle.

II. Substitute the points into $(x-3)^2+(y-2)^2$ it the result is $>25$ the point is outside the circle.

III. Any circle centered at the origin looks like $x^2+y^2=r^2$.

3. ## Re: Geometry Question Sheet (Need someone to confirm I did it right)

Thanks a bunch Plato.

I just re-read my original post, sorry that all of the equations I wrote got all messy, but you really cleared up question 25 for me, they never told me to substitute, although now when you've said that it makes perfect sense!

Appreciated.