# Graphing Linear Equations Problem

• Dec 13th 2008, 07:52 AM
lee1001
Graphing Linear Equations Problem
A triangle is formed by the intersection of the lines 2x +3y = 14, 4x -5y = -16 and the x-axis. Find the area of the triangle.

I started by attempting to convert both given equations into slope intercept form, so I could begin to graph the lines

2x -2x +3y = 14 -2x

3y = -2x +14
3 3

y = -2x +4.666666667
3

4x -4x -5y = -16 -4x

-5y = -4x -16
-5 -5

y = -4x + 3.2
5

The y intercept, 4.666666667 is impossible to graph by hand and this is stopping me from going any further.

What am i doing wrong?
• Dec 13th 2008, 08:04 AM
running-gag
Hi

You can start by computing the coordinates of the vertices of the triangle ABC
4x -5y = -16 and y = 0 gives A(-4,0)
2x +3y = 14 and y = 0 gives B(7,0)
2x +3y = 14 and 4x -5y = -16 gives C(1,4)
• Dec 13th 2008, 08:35 AM
lee1001
Quote:

Originally Posted by running-gag
Hi

You can start by computing the coordinates of the vertices of the triangle ABC
4x -5y = -16 and y = 0 gives A(-4,0)
2x +3y = 14 and y = 0 gives B(7,0)
2x +3y = 14 and 4x -5y = -16 gives C(1,4)

I understand how you got Vertices A and B, but how did you compute C?
• Dec 13th 2008, 10:43 AM
running-gag
For C you have to solve the system

2x +3y = 14
4x -5y = -16

2x = 14 - 3y => 4x = 28 - 6y
4x = -16 + 5y

28 - 6y = -16 + 5y => 11y = 44 => y = 4
• Dec 15th 2008, 08:11 AM
lee1001
Thanks for your help so far.

I actually managed to go back and work it out.

I now have a triangle plotted with the area; 7cm x 2.5cm /2 = 8.75cm2

The next part of the question asks me to find the vertices of a triangle with half the area of the triangle just found.

Well the area will be 4.375cm2.

How do I get the vertices from that triangle?
• Dec 15th 2008, 09:03 AM
running-gag
Quote:

Originally Posted by lee1001
Thanks for your help so far.

I actually managed to go back and work it out.

I now have a triangle plotted with the area; 7cm x 2.5cm /2 = 8.75cm2

This is not what I get as area
Are you sure ?
• Dec 15th 2008, 09:14 AM
lee1001
Quote:

Originally Posted by running-gag
This is not what I get as area
Are you sure ?

Well i measured out the base and height of the triangle on 1cm squared paper and got 7cm for the base and 2.5cm on the height. So 7(2.5) /2 = 8.75cm2
• Dec 15th 2008, 09:20 AM
running-gag
Just have a look to the sketch below

The area of the triangle is AB x CH / 2

A(-4,0) and B(7,0) therefore AB = 11
C(1,4) and H(1,0) therefore CH = 4

Area = 11 x 4 / 2 = 22

http://imageshack-france.com/out.php...2_Triangle.JPG
• Dec 15th 2008, 11:40 AM
lee1001
Ok, it took me a little while to see how you did that calculation but the formula is exactly the same as:

Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By) /2

By inputting the vertices co-ordinates I get the same answer.

Thanks!

The next part of the problem asks me to find the co-ordinates of the vertices of a triangle with half the area of the one just found.

Obviously the area of this new triangle will be 11, multiplied by 2 giving 22. How do i begin to deconstruct the formula to find the vertices?
• Dec 15th 2008, 11:45 AM
running-gag
If you have no other constraint, you can use for instance C'(1,2)
In this case the triangle ABC' has an area equal to 11 because
AB = 11 and C'H = 2 therefore area(ABC') = 11 x 2 / 2 = 11
• Dec 18th 2008, 07:11 AM
lee1001
Quote:

Originally Posted by running-gag
If you have no other constraint, you can use for instance C'(1,2)
In this case the triangle ABC' has an area equal to 11 because
AB = 11 and C'H = 2 therefore area(ABC') = 11 x 2 / 2 = 11

I don't understand.

How does that allow me to calculate the vertices?