# Thread: Area of a triangle. Formed by two axes and a line. Small question.

1. ## Area of a triangle. Formed by two axes and a line. Small question.

Hello everyone,

Heres the problem; A right triangle is formed in the first quadrant by the x and y axis and a line through point (2,1). Write the area A of the triangle as a function of x.

Using the information above I can conjure these three points;
(0,y)
(2,1)
(x,0)

Furthermore, I know area of a triangle is equal to .5(bxh) or in this case (y*x)/2.

The problem is that I need to express this area as a function of x. Which sounds simple. Unfortunately, that means I need to get y in terms of x. This is where I am stuck.

How would I put y in terms of x?

Thanks,
Taylor S. Amarel
Living is learning.

2. Originally Posted by tsa256
Hello everyone,

Heres the problem; A right triangle is formed in the first quadrant by the x and y axis and a line through point (2,1). Write the area A of the triangle as a function of x.

Using the information above I can conjure these three points;
(0,y)
(2,1)
(x,0)

Furthermore, I know area of a triangle is equal to .5(bxh) or in this case (y*x)/2.

The problem is that I need to express this area as a function of x. Which sounds simple. Unfortunately, that means I need to get y in terms of x. This is where I am stuck.

How would I put y in terms of x?

Thanks,
Taylor S. Amarel
Living is learning.
You can split the triangle up into two "internal" right-triangles and a rectangle.

The rectangle area is 2(1)=2 square units.

The area of the inner triangle to the right of the rectangle is

$\displaystyle\frac{(x-2)(1)}{2}$

The area of the inner triangle above the rectangle is

$\displaystyle\frac{(y-1)(2)}{2}$

Then, writing y as a function of x....

$\displaystyle\frac{y-1}{0-2}=\frac{1-0}{2-x}$

from the slope of the line.

If you use that relationship to write y in terms of x,
then you can write the triangle area in terms of x only, without y.

Of course, then you can ignore the partitioning of the triangle!

3. area = xy / 2

equating the slopes: y = x / (x-2)

area = x[x / (x-2)] / 2 = x^2 / [2(x-2)]