Since I know we have a bunch of math wizzes here, I am stuck on a question in Coordinate Geometry.

Question:
Determine the coordinates of the points of intersection of each line and cricle.

i) y = 2x
x²+y²=5

These are the instructions I am given.

[quote]
Notice that the method of "substitution" is the best method for solving this system. The example in the text substitutes for "y" because it was easier to change the linear equation to y =.... However, sometimes it may be easier to change the linear equation to x = ..... In this case you simply substitute for x then, instead of y.
Part of the solution to this problem involves working with a quadratic. Remember to create a quadratic equation equal to zero, and then solve the equation by factoring (or other suitable method). Once you have the (generally two) values for one variable, determine the value(s) of the other variable. Generally these problems have two solutions (the two points of intersection).
[/quote]

As near as I can tell, the answer would be:
y = 2x
x²+y²=5

x² + (2x)² = 5
x² + 4x² = 5
5x² = 5
x² = 1
x = +/-1

y = 2x, x = 1
y = 2
(1, 2)

y = 2x, x = -1
y = -2
(-1,-2)
Someone else should probably verify.
Edit: Formatting

Thanks Alot, could you "explain" the steps you took into solving this problem? Because I have to do another similar question and I still do not fully understand the solution.

My second question is:

y = 2x - 5
x² + y² = 5

I guess you would take the same sort of steps, but this one is a little different :(

[quote author=ChR0NiC link=board=36;threadid=6891;start=0#msg60998 date=1085078985]
Thanks Alot, could you "explain" the steps you took into solving this problem? Because I have to do another similar question and I still do not fully understand the solution.

My second question is:

y = 2x - 5
x² + y² = 5

I guess you would take the same sort of steps, but this one is a little different :(
[/quote]Okay, sure
So you have two equations, two unknowns.
One equation is y = 2x - 5, so you can substitute in 2x - 5 into the other equation wherever you see y. Substituting for y gives us
x² + (2x-5)² = 5
Now, just solve for x:
x² + 4x² + 25 = 5
5x² + 25 = 5
x² + 5 = 1
x² = -4

You can't take the square root of a negative number, therefore this problem has no real solutions. Lets say x² = 4, just so I can show you the rest of the steps.
x² = 4
x = +/-2
The line intersects the circle at x = 2 and x = -2
Substitute these values into the original line equation to get y.
One coordinate:
y = 2x - 5, x = 2
y = 2(2) - 5
y = -1
(2, -1)
The other:
y = 2x - 5, x = -2
y = 2(-2) - 5
y = -9
(-2,-9)

Nice, thanks alot. You helped me very much. I wish I could give you +1

Edit: Hit another problem, anyone feel free to help out....

Graph the circle: (x-3)²+(y-1)²=5

How much sense does that make?? NONE !!!

well, the standard form for a circle is
(x - h)[sup]2[/sup] + (y - k)[sup]2[/sup] = r[sup]2[/sup]

where (h, k) is the center and r is the radius. that's all you need to know! It's a circle with center 3,1 and radius sqrt(5)

While you math wizzes are on a roll, I've got one last one for you.

[quote]
From a lighthouse, the range of visiblity on a clear day is 40km. On a coordinate system, where O(0, 0) represents the lighthouse, a ship is travelling on a course represented by y=2x+80. Between which two points on the course can the ship be seen from the lighthouse?
[/quote]

the distance between two points is sqrt( (x[sub]2[/sub] - x[sub]1[/sub])[sup]2[/sup] + (y[sub]2[/sub] - y[sub]1[/sub])[sup]2[/sup] ). You want the distance to be less than 40. x1, y1 is the light house at (0,0), and the location we're solving for is at (x, 2x + 80) so, we'll set it equal to 40 to find the endpoints.

40 = sqrt( x[sup]2[/sup] + (2x + 80)[sup]2[/sup] ).
40 = sqrt( x[sup]2[/sup] + 4x[sup]2[/sup] + 320x + 6400).
40 = sqrt( 5x[sup]2[/sup] + 320x + 6400).
1600 = 5x[sup]2[/sup] + 320x + 6400
0 = 5x[sup]2[/sup] + 320x + 4800
run that through the quadratic equation:
x = -24, x = -40
plug that into the equation to find y:
between (-40, 0) and (-24, 32)

you should probably check my algebra.