You are not logged in.
#1 2006-02-28 15:44:14
- ganesh
- Moderator
Offline
Coordinate Geometry
CG # 1
Find the area of the triangle whose vertices are
(2, -3), (-1, 3) and (4, 7).
Character is who you are when no one is looking.
#2 2006-02-28 17:32:13
- krassi_holmz
- Real Member
Offline
Re: Coordinate Geometry
21
IPBLE: Increasing Performance By Lowering Expectations.
#4 2006-02-28 19:17:05
- krassi_holmz
- Real Member
Offline
Re: Coordinate Geometry
That was easy. I solved it in Paint for~1min
IPBLE: Increasing Performance By Lowering Expectations.
#5 2006-03-01 15:17:19
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 2
Find (x,y) if (3,2), (6,3), (x,y) and (6,5) are the vertices of a parellologram taken in order.
Character is who you are when no one is looking.
#6 2006-03-01 16:59:51
- krassi_holmz
- Real Member
Offline
Re: Coordinate Geometry
{3,4}?
IPBLE: Increasing Performance By Lowering Expectations.
#8 2006-03-02 02:37:59
- mathsyperson
- Moderator
Offline
Re: Coordinate Geometry
(9,6)?
Why did the vector cross the road?
It wanted to be normal.
#9 2006-03-02 02:46:38
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
Correct, mathsyperson! 
Character is who you are when no one is looking.
#10 2006-03-02 03:00:29
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 3
If 3x-2y+4=0 and ax+4y+3=0 are parallel lines, find the value of a.
Character is who you are when no one is looking.
#11 2006-03-02 05:17:02
- Patrick
- Real Member
-
Offline
Re: Coordinate Geometry
CG# 3
a = -6?
Last edited by Patrick (2006-03-02 05:17:20)
#12 2006-03-02 14:59:53
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
You're correct, Patrick! Well done! 
Character is who you are when no one is looking.
#13 2006-03-02 15:13:04
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 4
Find the equations of the lines which pass through the point (3,4) and make intercepts on the axes such that their sum is 14.
Character is who you are when no one is looking.
#15 2006-03-02 17:25:24
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
They are the only equations for which the conditions work
Character is who you are when no one is looking.
#16 2006-03-03 16:39:04
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 5
Find the equation of the perpendicular bisector of the line joining the points A(-1,2) and B(3,-4).
Character is who you are when no one is looking.
#18 2006-03-07 15:50:37
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
Excellent, Dionysus! 
Character is who you are when no one is looking.
#19 2006-03-07 17:31:40
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 6
Prove that the lines 3x-4y+5=0, 7x-8y+5=0 and 4x+5y=45 are concurrent.
Character is who you are when no one is looking.
#20 2006-03-09 19:02:03
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 7
Show that the points (4, -4), (-4, 4), and (4√3, 4√3) are the vertices of an equilateral triangle.
Character is who you are when no one is looking.
#23 2006-03-11 03:27:36
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
CG # 8
Find the equation of the straight line making an angle 135° with the positive direction of x-axis and cutting an intercept 5 on the y-axis.
Character is who you are when no one is looking.
#24 2006-03-11 03:31:37
- mathsyperson
- Moderator
Offline
Re: Coordinate Geometry
tan 135° = -1, so that is the gradient of the line. We are told that the y-intercept is 5, so the equation is y = 5-x.
Why did the vector cross the road?
It wanted to be normal.
#25 2006-03-11 14:50:29
- ganesh
- Moderator
Offline
Re: Coordinate Geometry
Very good, mathsyperson!
Character is who you are when no one is looking.