Chapter 6.3, Problem 66E

Solution

a)

To find: graph of the parametric equation for the values of $a$ , $b$ , $c$ , and $d$ .

The graph of the curve $x=at+b$ , $y=ct+d$ for the values of $a=2$ , $b=3$ , $c=-1$ , and $d=2$ is given in the figure (1).

Given information:

The expression is $x=at+b$ , $y=ct+d$ , where $a$ and $c$ are not both zero.

Also given that $a=2$ , $b=3$ , $c=-1$ , and $d=2$ .

Calculation:

By substituting the values $a=2$ , $b=3$ , $c=-1$ , and $d=2$ , in the expression:

  $x=at+b$ , $y=ct+d$

Then the parametric equations are $x=2t+3$ , $y=-t+2$ .

The following table gives the points $\left(x,y\right)$ by substituting different values of $t$ in the parametric equation $x=2t+3$ , $y=-t+2$ .

$t$	$-3$	0	1	3
$x=2t+3$	$-3$	3	5	9
$y=-t+2$	5	2	1	$-1$

By joining the points $\left(-3,5\right)$ , $\left(3,2\right)$ , $\left(5,1\right)$ , and $\left(9,-1\right)$ , obtain a line. The graph can be seen in the figure (1).

  

Figure (1)

b)

To find: graph of the parametric equation for the values of $a$ , $b$ , $c$ , and $d$ .

The graph of the curve $x=at+b$ , $y=ct+d$ for the values of $a=3$ , $b=4$ , $c=1$ , and $d=3$ is given in the figure (2).

Given information:

The expression is $x=at+b$ , $y=ct+d$ , where $a$ and $c$ are not both zero.

Also given that $a=3$ , $b=4$ , $c=1$ , and $d=3$ .

Calculation:

By substituting the values $a=3$ , $b=4$ , $c=1$ , and $d=3$ , in the expression:

  $x=at+b$ , $y=ct+d$

Then the parametric equations are $x=3t+4$ , $y=t+3$ .

The following table gives the points $\left(x,y\right)$ by substituting different values of $t$ in the parametric equation $x=3t+4$ , $y=t+3$ .

$t$	$-4$	0	1	4
$x=3t+4$	$-8$	4	7	16
$y=t+3$	$-1$	3	4	$7$

By joining the points $\left(-8,-1\right)$ , $\left(4,3\right)$ , $\left(7,4\right)$ , and $\left(16,7\right)$ , obtain a straight line. The graph can be seen in the figure (2).

  

Figure (2)

c)

To find: the equation in $x$ and $y$ after eliminating $t$ .

The equation in $x$ and $y$ after eliminating $t$ is $y=\frac{c}{a}x+\frac{ad-bc}{a}$ .

Given information:

The expression is $x=at+b$ , $y=ct+d$ , where $a$ and $c$ are not both zero.

Calculation:

Elimination of $t$:

Solve for $t$ using the equation: $x=at+b$

  $\begin{array}{c}x=at+b\\ at=x-b\\ t=\frac{x-b}{a}\end{array}$

Substitute this value in the equation $y=ct+d$ .

  $\begin{array}{c}y=ct+d\\ =c\left(\frac{x-b}{a}\right)+d\\ =\frac{c}{a}x-\frac{bc}{a}+d\\ =\frac{c}{a}x+\frac{ad-bc}{a}\end{array}$

Since $a$ and $c$ are not both zero, the expression $y=\frac{c}{a}x+\frac{ad-bc}{a}$ , $a\ne 0$ represents a straight line having slope $\frac{c}{a}$ .

d)

To find: the slope, $y-$ intercept, and $x-$ intercept of the line.

Slope $\frac{c}{a}$ if $a\ne 0$ ; $x-$ intercept $\left(\frac{bc-ad}{c},0\right)$ if $c\ne 0$ , and $y-$ intercept $\left(0,\frac{ad-bc}{a}\right)$ if $a\ne 0$ .

Given information:

From part (c), the expression is $y=\frac{c}{a}x+\frac{ad-bc}{a}$ , $a\ne 0$ .

Formula:

A line equation of the form $y=mx+c$ has slope $m$ , $x-$ intercept $\left(-\frac{c}{m},0\right)$ and $y-$ intercept $\left(0,c\right)$ .

Calculation:

Compare given expression $y=\frac{c}{a}x+\frac{ad-bc}{a}$ with $y=mx+c$ , conclude that the slope of the line is slope $\frac{c}{a}$ if $a\ne 0$ ; $x-$ intercept $\left(\frac{bc-ad}{c},0\right)$ if $c\ne 0$ , and $y-$ intercept $\left(0,\frac{ad-bc}{a}\right)$ if $a\ne 0$ .

e)

To find: conditions for which given expression is the line is horizontal? Vertical?

If $c=0$ then the line is horizontal and if $a=0$ then the line is vertical.

Given information:

From part (c), the expression is $y=\frac{c}{a}x+\frac{ad-bc}{a}$ , $a\ne 0$ .

Formula:

A equation of the form $y=a$ represents a line parallel to $x-$ axis (horizontal line) and the equation of the form $x=b$ represents a line parallel to $y-$ axis (vertical line).

Calculation:

In given expression $y=\frac{c}{a}x+\frac{ad-bc}{a}$ , if $c=0$ then $y=d$ .

Thus, if $c=0$ then the line is horizontal.

Rewrite the expression $y=\frac{c}{a}x+\frac{ad-bc}{a}$ as $ay=cx+ad-bc$ .

Substitute 0 for $a$ in the equation $ay=cx+ad-bc$ :

  $\begin{array}{c}ay=cx+ad-bc\\ cx=ay-ad+bc\\ cx=-0+0+bc\\ cx=bc\\ x=b\end{array}$

Thus, if $a=0$ then the line is vertical.