Chapter 1.4, Problem 67E

To determine

Find the area of triangle as a function of $x$ and domain of the function.

Answer to Problem 67E

  $\boxed{A=\frac{x^2}{2\left(x-2\right)}}$

The domain of the function is all real values of $x$ where $\boxed{x>2}$

Explanation of Solution

Given information:

A right triangle is formed in the first quadrant by the $x-axis$ and $y-axis$ and a line through the point $\left(2,1\right)$ (see figure). Write the area of the triangle as a function of $x$ and determine the domain of the function.

  

Calculation:

Consider the given triangle,

  

A line is passing through the points $\left(2,1\right)$ .

Consider base of given right triangle $x$ and height $y$ ,

Area of triangle A is $\frac{1}{2}base\times height$

  $A=\frac{1}{2}x\times y$

Now find the function of area in terms of $x$ .

Now to get $y$ in terms of $x$

The slope of line is $\frac{y_2-y_1}{x_2-x_1}$ .

Slope of line with coordinates $\left(0,y\right)\&\left(2,1\right)is\frac{y-1}{0-2}$

Slope of line with coordinates $\left(x,0\right)\&\left(2,1\right)is\frac{1-0}{2-x}$

  $\begin{array}{l}\frac{y-1}{0-2}=\frac{1-0}{2-x}\\ y-1=\frac{-2}{2-x}\\ y=\frac{2}{x-2}+1\\ y=\frac{2+x-2}{x-2}\\ y=\frac{x}{x-2}\end{array}$

Thus, $A=\frac{1}{2}x\times y$

  $\begin{array}{l}A=\frac{1}{2}x\left(\frac{x}{x-2}\right)\\ \boxed{A=\frac{x^2}{2\left(x-2\right)}}\end{array}$

Hence area is $\boxed{A=\frac{x^2}{2\left(x-2\right)}}$

Now determine the domain of function,

As the triangle is in first quadrant so then $x$ takes positive values $x>0$

We will include those values in the domain where denominator is non zero.

  $\begin{array}{l}x-2=0\\ x=2\end{array}$

For $x=2$ , the denominator becomes zero. So exclude $x=2$

Hence the domain of the function is all real values of $x$ where $\boxed{x>2}$