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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

CG # 1

Find the area of the triangle whose vertices are

(2, -3), (-1, 3) and (4, 7).

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

21

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

That was easy. I solved it in Paint for~1min

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

CG # 2

Find (x,y) if (3,2), (6,3), (x,y) and (6,5) are the vertices of a parellologram taken in order.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

{3,4}?

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

(9,6)?

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

Correct, mathsyperson!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

CG # 3

If 3x-2y+4=0 and ax+4y+3=0 are parallel lines, find the value of a.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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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****Administrator**- Registered: 2005-06-28
- Posts: 27,896

You're correct, Patrick! Well done!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

El que pega primero pega dos veces.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

They are the only equations for which the conditions work

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- 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).

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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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****Administrator**- Registered: 2005-06-28
- Posts: 27,896

*Excellent, Dionysus! *

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

CG # 6

Prove that the lines 3x-4y+5=0, 7x-8y+5=0 and 4x+5y=45 are concurrent.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

CG # 7

Show that the points (4, -4), (-4, 4), and (4√3, 4√3) are the vertices of an equilateral triangle.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 27,896

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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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****Administrator**- Registered: 2005-06-28
- Posts: 27,896

**Very good, mathsyperson!**

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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