[Ashrak]

i have 2 equasions x+y+2=o and 2x+5y-3=0 (these are lines) and i need to find the corner in between


how?

[Petrarch of the VBB]

x+y=-2
2x+5y=3

As far as I can go.

[Ashrak]

Originally posted by Petrarch of the VBB
x+y=-2
2x+5y=3

As far as I can go.

erm not what i meant i need the corner between those lines

[Petrarch of the VBB]

I know. but the chance are that the fact they are lines does not matter. You probably have to treat them as quadratic (or simultaneous) equations, to work out x and y, then draw the lines.

If A) x+y=-2
   B) 2x+5y=3

Then multiply A*2

A) 2x+2y=-4

A) - B)
2x+2y=-4
2x+5y=3
--------------
-3y=-7
-y=-7/3 = -2.333333333333 etc.

Substitute in A) x+ (-2.333333333etc) = -2
then x = -0.333333333333

Check in B) 2(0.333333333333)+5(-2.33333333333) =
0.666666666+ -11.666666666= -11

Oh dear. Theory=yes. Practice = no.

[Stealth]

set them both equal to "y"

now put them both equal to each other

now solve for "x" (put "x" on one side and the numbers on the other)

now you've got an "X" position, plug it in to the original equation(s)

[Petrarch of the VBB]

Originally posted by Stealth
set them both equal to "y"

now put them both equal to each other

now solve for "x" (put "x" on one side and the numbers on the other)

now you've got an "X" position, plug it in to the original equation(s)

I think that's what I was tring to do, but it went hoorribly wrong.

[Tiara]

Originally posted by Stealth
set them both equal to "y"
 

If possible, always do that first

[Mr. Vega]

Petrarch, that's a +2.33333333333, not -.

x+7/3=-2
x=-13/3 (-4.3333333333333333)

2(-13/3)+5(7/3)=3

So it does work.

[Top Gun]

Do you just want to find where they intesect or the ancle of intersection?

[Petrarch of the VBB]

Ahh, silly mistakes.

And I've got Maths GCSEs in about 2 weeks!

[Ashrak]

ok now you told me how to get a cordinate of a point now can you tell me how to get the Angle corner or whatever you call it that is created when those 2 straight lines cross eachother i need the size of the angle between the crossed lines .....

[Petrarch of the VBB]

The best way would be to draw the lines on a grid, and measure the angle.

[Ashrak]

cant i have to calculate it degrees minutes and seconds

[Petrarch of the VBB]

Then you're on your own. I leave you alone with my pity, as I cannot help you. Even my power is as nothing next to DMS.

[Darkage]

Where is CP when you need him?

[Ashrak]

got it



S1 = (-B1;A1) S2 = (-B2;A2)

Vektor S1 times vektor S2 divided by absolut Vektor S1 times Absolut vektor S2 = Cos a

[beatspete]

Originally posted by Petrarch of the VBB
The best way would be to draw the lines on a grid, and measure the angle.

y=mx+c is your friend.

Rearrange, find the gradients (m)...  Then i think you multiply them, and take tan-1 of that. I got 21.8 degrees.

[DragonClaw]

Originally posted by Ashrak
ok now you told me how to get a cordinate of a point now can you tell me how to get the Angle corner or whatever you call it that is created when those 2 straight lines cross eachother i need the size of the angle between the crossed lines .....

use the distance formula between on the 3 points you now have. It'll give you the distance between each of the points, then you can use the law of cosines to figure out the measure of the angle created by the intersection of the two lines.

distance formula: Square Root of: (x2-x1)^2+(y2-y1)^2
law of cosines: c^2=a^2+b^2-2ac(cos C)


there's probably an easier way, a lot easier way, but whatever, I like doing things the hard way.

[Petrarch of the VBB]

Originally posted by beatspete




y=mx+c is your friend.
 

I'd forgotten about that. Shows how much notice I take of my maths teacher.

[Stealth]

admit it, we're all a bunch of morons