# Geometry triangle help?

• Dec 28th 2008, 08:46 AM
Ifailatmaths
Geometry triangle help?
So, the question is:

The straight line l1 passes through points A(2,2) and B(6,0).

a)Find an equation for this

I found this to be: y=-1/2x+3

The straight line l2 with equation x + y = 8 cuts the y-axis at point C. The lines l1 and l2 intersect at point D.

b)Calculate the cordinates of D

I found this to be (4,4)

c) Calculate the area of traingle ACD.

I have no idea how to do this, I know point A=(2,2) C=(0,8) and D=(4,4) but that's all I know. Any help would be great.
• Dec 28th 2008, 09:41 AM
masters
Quote:

Originally Posted by Ifailatmaths
So, the question is:

The straight line l1 passes through points A(2,2) and B(6,0).

a)Find an equation for this

I found this to be: y=-1/2x+3

The straight line l2 with equation x + y = 8 cuts the y-axis at point C. The lines l1 and l2 intersect at point D.

b)Calculate the cordinates of D

I found this to be (4,4)

c) Calculate the area of traingle ACD.

I have no idea how to do this, I know point A=(2,2) C=(0,8) and D=(4,4) but that's all I know. Any help would be great.

I don't see how you found the coordinates of D, the intersection of $\displaystyle x+y=8$ and $\displaystyle y=-\frac{1}{2}x+3$.

This is what I came up with when I solved for the intersection of those two lines.

[1] $\displaystyle x+y=8$

[2] $\displaystyle y=-\frac{1}{2}x+3$

Substituting [2] into [1], we get

$\displaystyle x+\left(-\frac{1}{2}x+3\right)=8$

$\displaystyle 2x-x+6=16$

$\displaystyle x=10$

Substituting this into [2], we get

$\displaystyle y=-\frac{1}{2}(10)+3$

$\displaystyle y=-5+3$

$\displaystyle y=-2$

So, D(10, -2) is the intersection.

Using A(2, 2), C(0, 8), D(10, -2), you can find the area.

Here are a couple of ways you might want to consider to find the area of triangle ACD.

(1) Using determinants

$\displaystyle A=\frac{1}{2}\left|\begin{array}{ccc}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|$

Remember to take the absolute value of this determinant to ensure a positive area.

(2) Heron's formula

$\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}$ where $\displaystyle s=\frac{1}{2}(a+b+c)$ and a, b, c are the lengths of the sides of the triangle.

Here, you'll have to find the lengths of each side using the distance formula. Then use Heron's formula to determine the area:

• Dec 29th 2008, 04:45 AM
earboth
Quote:

Originally Posted by Ifailatmaths
So, the question is:

...

c) Calculate the area of traingle ACD.

I have no idea how to do this, I know point A=(2,2) C=(0,8) and D=(4,4) but that's all I know. Any help would be great.

Here is another method to calculate the area of a polygon in a coordinate-system: Chopoff method:

(I use masters result to draw the triangle)

1. Construct a rectangle such that the area in question is completely inside this rectangle.

2. Divide the area which doesn't belong to the triangle into right triangles or trapeziums.

3. Subtract all areas of #2 from the area of the overall container:

$\displaystyle A_{container} = 10 \cdot 10 = 100$
$\displaystyle A_1 = \dfrac{10+4}2 \cdot 2 = 14$
$\displaystyle A_2 = \dfrac12 \cdot 4 \cdot 8 = 16$
$\displaystyle A_3 = \dfrac12 \cdot 10 \cdot 10 = 50$
$\displaystyle A_{triangle} = 100-14-16-50 = 20$