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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,865

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.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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21

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
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Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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That was easy. I solved it in Paint for~1min

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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{3,4}?

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
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Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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(9,6)?

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
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Correct, mathsyperson!

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,865

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.

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**Patrick****Real Member**- Registered: 2006-02-24
- Posts: 1,005

CG# 3

a = -6?

*Last edited by Patrick (2006-03-01 06:17:20)*

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**ganesh****Moderator**- Registered: 2005-06-28
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You're correct, Patrick! Well done!

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,865

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.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

El que pega primero pega dos veces.

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**ganesh****Moderator**- Registered: 2005-06-28
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They are the only equations for which the conditions work

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**Dionysus****Member**- Registered: 2006-03-06
- Posts: 9

*Last edited by Dionysus (2006-03-06 16:22:49)*

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**ganesh****Moderator**- Registered: 2005-06-28
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*Excellent, Dionysus! *

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,865

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.

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**Dionysus****Member**- Registered: 2006-03-06
- Posts: 9

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**ganesh****Moderator**- Registered: 2005-06-28
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Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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.

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**ganesh****Moderator**- Registered: 2005-06-28
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**Very good, mathsyperson!**

Character is who you are when no one is looking.

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