Chapter 1.10, Problem 63E

(a)

To determine

To prove: that if x and y intercept is nonzero the equation will be of two intercept form.

Answer to Problem 63E

The equation of two intercept form or $\frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{=1}$ .

Explanation of Solution

Given:

  $\frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{=1}$ .

Concept used:

During intercept of $\text{x-axis}$ :

  $\text{y=0}$ .

During intercept of $\text{y-axis}$ :

  $\text{x=0}$ .

Calculation:

Start from the general equation of a line:

  $\text{Ax+By+C=0}$ .

Find the x-intercept of the graph of this equation, set $\text{y=0}$ and solve for x.

  $\text{Ax+By+C=0}$ .

  $\Rightarrow \text{Ax+B}\left(0\right)\text{+C=0}$

  $\Rightarrow \text{Ax=-C}$ .

The intercept of $\text{x}$ is $\text{x=}\frac{\text{-C}}{\text{A}}$ .

Find the $\text{y}$ - intercept of the graph of this equation, set $\text{x=0}$ and solve for $\text{y}$ :

  $\text{Ax+By+C=0}$ .

  $\Rightarrow \text{By=-C}$ .

  $\Rightarrow \text{y=}\frac{\text{-C}}{\text{B}}$ .

The intercept of $\text{y}$ is $\text{y=}\frac{\text{-C}}{\text{B}}$ .

Let $\text{a}$ the intercept so $\text{a=}\frac{\text{-C}}{\text{A}}$ .

Let $\text{b}$ the intercept so $\text{b=}\frac{\text{-C}}{\text{B}}$ .

Back to the general equation of a line:

  $\text{Ax+By+C=0}$ .

  $\Rightarrow \text{Ax+By=-C}$ .

  $\Rightarrow \frac{\text{A}}{\text{-C}}\text{x+}\frac{\text{B}}{\text{-C}}\text{y=1}$ .

  $\Rightarrow \frac{\text{x}}{\left(\frac{\text{-C}}{\text{A}}\right)}\text{+}\frac{\text{y}}{\left(\frac{\text{-C}}{\text{B}}\right)}\text{=1}$ .

  $\Rightarrow \frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{=1}$ .

(b)

To determine

To find:the equation using two intercept form given below.

Answer to Problem 63E

Theequation is $\text{4x-3y-24=0}$ .

Explanation of Solution

Given: $\frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{=1}$ and Set $\text{a=6 and b=-8}$ .

Concept used:

Two intercept form:

During intercept of $\text{x-axis}$ :

  $\text{y=0}$ .

During intercept of $\text{y-axis}$ :

  $\text{x=0}$ .

Calculation:

For the equation of the indicated line use the two- intercept form of linear equation:

  $\frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{=1}$ .

Set $\text{a=6 and b=-8}$ .

  $\Rightarrow \frac{\text{x}}{6}\text{+}\frac{\text{y}}{8}\text{=1}$ .

  $\Rightarrow \frac{\text{24x}}{6}\text{+}\frac{\text{24y}}{\left(-8\right)}\text{-24}\left(1\right)\text{=0}$ .

  $\Rightarrow \text{4x-3y-24=0}$ .

Hence the equation is $\text{4x-3y-24=0}$ .