Thread: 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?

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)

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?

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

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?

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 ?

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

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

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?

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

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?