Chapter 0.1, Problem 40E

(a)

To determine

To explain why $c$ and $d$ are the $x$ -intercept and $y$ -intercept, respectively, of the line $\frac{x}{c}+\frac{y}{d}=1$ .

Answer to Problem 40E

The $x$ -intercept and the $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=1$ are found by substituting $y=0$ and $x=0$ in the equation $\frac{x}{c}+\frac{y}{d}=1$ and are found to be $x=c$ and $y=d$ , respectively.

Explanation of Solution

Given the equation of a line:

  $\frac{x}{c}+\frac{y}{d}=1$ .

Concept Used:

Given a function $y=f\left(x\right)$ .

  $x$-intercept:

The $x$ -intercepts of the function $y=f\left(x\right)$ are the values of $x$ through which the graph of the function passes. It is obtained by substituting $y=0$ in the equation of the function $y=f\left(x\right)$ .

  $y$-intercept:

The $y$ -intercept of the function $y=f\left(x\right)$ is the values of $y$ through which the graph of the function passes. It is obtained by substituting $x=0$ in the equation of the function $y=f\left(x\right)$ .

Calculation:

Find the $x$-intercept of the line $\frac{x}{c}+\frac{y}{d}=1$:

Substitute $y=0$ in the equation of the line $\frac{x}{c}+\frac{y}{d}=1$ :

  $\frac{x}{c}+\frac{0}{d}=1$ .

Solve the equation $\frac{x}{c}+\frac{0}{d}=1$ for $x$ :

  $\begin{array}{c}\frac{x}{c}+\frac{0}{d}=1\\ \frac{x}{c}+0=1\\ \frac{x}{c}=1\\ x=c\end{array}$ .

Thus, the $x$ -intercept of the line is $x=c$ .

Find the $y$-intercept of the line $\frac{x}{c}+\frac{y}{d}=1$:

Substitute $x=0$ in the equation of the line $\frac{x}{c}+\frac{y}{d}=1$ :

  $\frac{0}{c}+\frac{y}{d}=1$ .

Solve the equation $\frac{0}{c}+\frac{y}{d}=1$ for $y$ :

  $\begin{array}{c}\frac{0}{c}+\frac{y}{d}=1\\ 0+\frac{y}{d}=1\\ \frac{y}{d}=1\\ y=d\end{array}$ .

Thus, the $y$ -intercept of the line is $y=d$ .

Conclusion:

The $x$ -intercept and the $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=1$ are found by substituting $y=0$ and $x=0$ in the equation $\frac{x}{c}+\frac{y}{d}=1$ and are found to be $x=c$ and $y=d$ , respectively.

(b)

To determine

To determine how the $x$ -intercept and the $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ are related with $c$ and $d$ .

Answer to Problem 40E

The $x$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is found to be:

  $x=2c$ .

Thus, the $x$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is twice $c$ .

The $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is found to be:

  $y=2d$ .

Thus, the $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is twice $d$ .

Explanation of Solution

Given the equation of a line:

  $\frac{x}{c}+\frac{y}{d}=2$

Calculation:

Find the $x$-intercept of the line $\frac{x}{c}+\frac{y}{d}=2$:

Substitute $y=0$ in the equation of the line $\frac{x}{c}+\frac{y}{d}=2$ :

  $\frac{x}{c}+\frac{0}{d}=2$ .

Solve the equation $\frac{x}{c}+\frac{0}{d}=2$ for $x$ :

  $\begin{array}{c}\frac{x}{c}+\frac{0}{d}=2\\ \frac{x}{c}+0=2\\ \frac{x}{c}=2\\ x=2c\end{array}$ .

Thus, the $x$ -intercept of the line is $x=2c$ .

Find the $y$-intercept of the line $\frac{x}{c}+\frac{y}{d}=2$:

Substitute $x=0$ in the equation of the line $\frac{x}{c}+\frac{y}{d}=2$ :

  $\frac{0}{c}+\frac{y}{d}=2$ .

Solve the equation $\frac{0}{c}+\frac{y}{d}=2$ for $y$ :

  $\begin{array}{c}\frac{0}{c}+\frac{y}{d}=2\\ 0+\frac{y}{d}=2\\ \frac{y}{d}=2\\ y=2d\end{array}$ .

Thus, the $y$ -intercept of the line is $y=2d$ .

Conclusion:

The $x$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is found to be:

  $x=2c$ .

Thus, the $x$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is twice $c$ .

The $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is found to be:

  $y=2d$ .

Thus, the $y$ -intercept of the line $\frac{x}{c}+\frac{y}{d}=2$ is twice $d$ .