Chapter 1.2, Problem 77E

To determine

To obtain: the area of the triangle as a function of the given variable.

Also to find its domain.

Answer to Problem 77E

The area of the triangle is given by $A=\frac{1}{2}\frac{a^2}{a-2}$

The domain of the function is $x=a,a\in \left(2,\infty \right)$.

Explanation of Solution

Given Information:

A right triangle is formed in the first quadrant by x - and y -axes and a line through the point (2,1).

Formula Used (1)The area of a right triangle with base b and height h is given by,

  $A=\frac{1}{2}bh$

(2) The equation of a line with a point (x₁ , y₁ ) and slope m is given by,

  $y-y_1=m\left(x-x_1\right)$

Where the slope of the line is given by,

  $m=\frac{y_2-y_1}{x_2-x_1}$

(2) The domain of any function f (x ) is all real values of x for which the function can be defined.

Calculation Here, the triangle has base given by a and height given as b . Therefore, the area of the triangle is given by,

  $A=\frac{1}{2}ab$

Now, the line with point (2, 1) has the end points as (0, b ) and (a , 0).

The slope of the line can be expressed in terms of (2, 1) and (a , 0) as,

  $\begin{array}{c}m=\frac{0-1}{a-2}\\ =\frac{1}{2-a}\end{array}$

The slope can also be written in terms of the point (2, 1) and (0, b ) as,

  $\begin{array}{c}m=\frac{b-1}{-2}\\ =\frac{1-b}{2}\end{array}$

Comparing the two expressions for slope,

  $\begin{array}{c}\frac{1}{2-a}=\frac{1-b}{2}\\ 1-b=\frac{2}{2-a}\\ b=1-\frac{2}{2-a}\\ =\frac{-a}{2-a}\end{array}$

  $b=\frac{a}{a-2}$

  

Now, the area of the triangle can be written by using the above substitution for b in terms of a as,

  $\begin{array}{c}A=\frac{1}{2}ab\\ =\frac{1}{2}\frac{a^2}{a-2}\end{array}$

This gives the expression for area of the triangle in terms of a.

The domain of the function representing the area of the triangle $A=\frac{1}{2}\frac{a^2}{a-2}$ is given by all those real values of a which makes the function definable. The function is area here which places a restriction on the values of a to be real and positive. Here, the value of a in the denominator should be greater than 2. When it is equal or less than two the function does not satisfy the given condition. Thus, the domain of the function representing the area A is given by $x=a,a\in \left(2,\infty \right)$ .