Math help

[classified]

Originally Posted by T.H.C

no, its 5!

going back in time i realised i carried the 13 wrong. you are correct

[Dr-Vn]

[alex.w *//]

lol i remember ppl asking this on vancouver xchange, under the school subforum.

they actually got help cuz probably 90% of the members there were highschoolers

[sas]

I won't do it for you, but..

determine your centre for your circle. H, K

Use the standard or general formula, maybe midpoint forumula since your given 4 points, etc.. (its been a long time) (x-h) + (y-k) = r

[chouchou]

Originally Posted by CruisingDream

this is like gr 12 math? pretty easy stuff man

grade 10 :P
hahaha

[Adrenaline Rush]

[viper11885]

Originally Posted by HB20

I've seen the others in this thread. But this one is new to me

[wuuhoo]

Originally Posted by Dr-Vn

lmao

[!Aznboi128]

[Gary Oak]

ahaha dont do it...
your never gonna use this in life

[twitchyzero]

yeah math 12 stuff...it's one of the easiest chapter.

Math 11 was so much harder than math 12 i thought.

for the OP..i suggest getting the "Byng book" of old provincial exams..it sorts out a lot of old exam Q by its section. If it wasnt for that book, I wouldn't be in university right now hahah

[tamiya_s2000]

zomg rocket science!!!!
lol lazy shit.. answers are:
1. 6
2. (F4,1L)
3. 3.2km

Originally Posted by HB20

lol nice

[cctw]

i'll teach..$30/hr lol =)

[slammer111]

Since nobody's helping, ah what the hell, I'm bored. No guarantees on the accuracy, but here goes. Damn I miss doing this stuff.

Just wondering, what grade Math is this?

Not gonna show the full work (how do you type equations lol), but here are the solutions.

1. Make a centre point C(h,4). By eyeballing it or drawing it on grid paper (and yes you can prove it using Pythogoras as well), you'll find that h = 3, so C(3,4) is the centre. Use Pythagoras again, and you'll find R^2 = 13. Stick that into General Form. h = 3, k = 4, R^2 = 13.
2. Since you know y = 2x - 2, substitute that into "y" in your circle equation to get 5x^2 - 18x + 13 = 0. Solve the quadratic and you'll get x1 = 2.6, x2 = 1. Plug into line (or circle if you want but line's way easier) equation to get intersection points P1(2.6,3.2) and P2(2,0).

3. Change the line to General Form y = mx + b, or y = (-3/4)x + 6. Find the line (with slope m2) that's perpendicular between the existing line (which has slope m1 = -0.75). Remember that perpendicular lines have slopes m1 * m2 = -1, therefore m2 = 4/3. Plug coordinates of P into the equation, so 5 = (4/3)(6) + b, so b = -3. Now find the intersection points of these 2 lines y = (4/3)x - 3, y = (-3/4)x + 6. Solve to find intersection point A(102/25,69/25) or A(4.32,2.76). Then use Pythagoras to get the distance between A and P, d = 2.8.

[noots]

Originally Posted by slammer111

Since nobody's helping, ah what the hell, I'm bored. No guarantees on the accuracy, but here goes. Damn I miss doing this stuff.

Just wondering, what grade Math is this?

Not gonna show the full work (how do you type equations lol), but here are the solutions.

1. Make a centre point C(h,4). By eyeballing it or drawing it on grid paper (and yes you can prove it using Pythogoras as well), you'll find that h = 3, so C(3,4) is the centre. Use Pythagoras again, and you'll find R^2 = 13. Stick that into General Form. h = 3, k = 4, R^2 = 13.
2. Since you know y = 2x - 2, substitute that into "y" in your circle equation to get 5x^2 - 18x + 13 = 0. Solve the quadratic and you'll get x1 = 2.6, x2 = 1. Plug into line (or circle if you want but line's way easier) equation to get intersection points P1(2.6,3.2) and P2(2,0).

3. Change the line to General Form y = mx + b, or y = (-3/4)x + 6. Find the line (with slope m2) that's perpendicular between the existing line (which has slope m1 = -0.75). Remember that perpendicular lines have slopes m1 * m2 = -1, therefore m2 = 4/3. Plug coordinates of P into the equation, so 5 = (4/3)(6) + b, so b = -3. Now find the intersection points of these 2 lines y = (4/3)x - 3, y = (-3/4)x + 6. Solve to find intersection point A(102/25,69/25) or A(4.32,2.76). Then use Pythagoras to get the distance between A and P, d = 2.8.

what he said

[chouchou]

^damn ninja'd.. i knew that..:P

[cctw]

Originally Posted by slammer111

Since nobody's helping, ah what the hell, I'm bored. No guarantees on the accuracy, but here goes. Damn I miss doing this stuff.

Just wondering, what grade Math is this?

Not gonna show the full work (how do you type equations lol), but here are the solutions.

1. Make a centre point C(h,4). By eyeballing it or drawing it on grid paper (and yes you can prove it using Pythogoras as well), you'll find that h = 3, so C(3,4) is the centre. Use Pythagoras again, and you'll find R^2 = 13. Stick that into General Form. h = 3, k = 4, R^2 = 13.
2. Since you know y = 2x - 2, substitute that into "y" in your circle equation to get 5x^2 - 18x + 13 = 0. Solve the quadratic and you'll get x1 = 2.6, x2 = 1. Plug into line (or circle if you want but line's way easier) equation to get intersection points P1(2.6,3.2) and P2(2,0).

3. Change the line to General Form y = mx + b, or y = (-3/4)x + 6. Find the line (with slope m2) that's perpendicular between the existing line (which has slope m1 = -0.75). Remember that perpendicular lines have slopes m1 * m2 = -1, therefore m2 = 4/3. Plug coordinates of P into the equation, so 5 = (4/3)(6) + b, so b = -3. Now find the intersection points of these 2 lines y = (4/3)x - 3, y = (-3/4)x + 6. Solve to find intersection point A(102/25,69/25) or A(4.32,2.76). Then use Pythagoras to get the distance between A and P, d = 2.8.

i'm bored so i'll make corrections...
#1: R^2 = 20 not 13..just plug one of the coordinates back into your equation (x-3)^2 + (y-4)^2 = R^2 and you'll see

#2: its P2(1,0)..

#3: its A(108/25, 69/25)..decimal answers are correct though

[StaxBundlez]

hahhahahahahah revscene... no help at all

[slammer111]

Originally Posted by cctw

i'm bored so i'll make corrections...
#1: R^2 = 20 not 13..just plug one of the coordinates back into your equation (x-3)^2 + (y-4)^2 = R^2 and you'll see

#2: its P2(1,0)..

#3: its A(108/25, 69/25)..decimal answers are correct though

Good call. Yeah I checked, and you're right on all 3 corrections Oh well, hopefully the OP gets the gist of it at least.

[J-Wangsta]

hey! its pretty foggy out there eh?

[no_mercy]

so in addition to the answers above

#1. since you know (h,4)
the equation of a circle must be (x+h)^2 + (y-4)^2 = r^2
now plug in your first coordinates (x=1, y=0) into the equation of the circle you'll get

(1+h)^2 + (0-4)^2 = r^2
h^2+2h+1+16 = r^2
h^2+2h+17=r^2 <--- equation A

now plug in your second coordinates (x=5, y=0) into the equation of the circle and you'll get
(5+h)^2+(0-4)^2 = r^2
h^2+10h+25+16 = r^2
h^2+10h+41 = r^2 <---- equation B

Since you know that those two coordinates lie on the circle, their radius must be the same so you can equate the two equations together and eliminate variables OR set them up in a linear system and reduce it (linear algebra)

h^2+2h+17=r^2 <--- equation A
h^2+10h+41 = r^2 <---- equation B

so setting the radius equal to each other of equation A and equation B (A=B) we get

h^2+2h+17 = h^2+10h+41
2h+17=10h+41
-24=8h
h=-3

Now you can plug the h back into the equation of a circle

(x-3)^2 + (y-4)^2 = r^2

Your centre is (3,4) describe in the equation and your radius is unknown but that is easy since you have an X and Y coordinate of your choice (x=1,y=0) or (x=5,y=0)
plugging those numbers in you'll find that r^2 = 20

[sIN]

thank you very much. \I would not be able to do this with out the help