Chapter P.4, Problem 60E

Solution

a.

To show that $ax+by=c\text{ and }ax+by=d$ are parallel line for some $c\ne d$ and a , b are not both zero.

If $b=0$ , both lines are vertical else both are parallel as they have same slope as $m=-\frac{a}{b}$ .

If c and d are equal, the lines will be coincident and have infinitely many solutions.

Given:

Two lines $ax+by=c\text{ and }ax+by=d$ .

Concept Used:

• The slope-intercept form of a line is $y=mx+b$, where m is the slope of the line and b is the y -intercept of the line.
• Slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
• The slopes of two perpendicular lines are negative reciprocal of each other.
• The slopes of two parallel lines are equal to each other.

Calculation:

In order to show that the equations are parallel, find their slopes.

Write the equation $ax+by=c$ in slope intercept form to get

  $\begin{array}{c}ax+by=c\\ by=-ax+c\\ y=-\frac{a}{b}x+\frac{c}{b}\end{array}$

Comparing with the slope intercept form of line $y=mx+b$ , the slope of above line is $m=-\frac{a}{b}$ .

By following the same procedure, the slope of the line $ax+by=d$ is $m=-\frac{a}{b}$ .

Since the slopes of both lines are same, the lines are parallel.

If $b=0$ , both lines are vertical else both are parallel as they have same slope as $m=-\frac{a}{b}$ .

If c and d are equal, the lines will be coincident and have infinitely many solutions.

b.

To show that $ax+by=c\text{ and }bx-ay=d$ are perpendicular line for a , and b are not both zero.

If either a or b equals 0, one line is horizontal and other is vertical line, otherwise the lines are perpendicular as their slopes are negative reciprocal of each other.

Given:

Two lines with negative slope.

Concept Used:

• The slope-intercept form of a line is $y=mx+b$, where m is the slope of the line and b is the y -intercept of the line.
• Slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
• The slopes of two perpendicular lines are negative reciprocal of each other.

Calculation:

In order to show that the equations are perpendicular, find their slopes.

Write the equation $ax+by=c$ in slope intercept form to get

  $\begin{array}{c}ax+by=c\\ by=-ax+c\\ y=-\frac{a}{b}x+\frac{c}{b}\end{array}$

Comparing with the slope intercept form of line $y=mx+b$ , the slope of above line is $m=-\frac{a}{b}$ .

Similarly, the slope of the line $bx-ay=d$ is given by

  $\begin{array}{c}bx-ay=d\\ ay=bx-d\\ y=\frac{b}{a}x-\frac{d}{a}\end{array}$

Comparing with the slope intercept form of line $y=mx+b$ , the slope of above line is $m=\frac{b}{a}$ .

Since the slopes of both lines are negative reciprocal of each other, the lines are perpendicular.

If either a or b equals 0, one line is horizontal and other is vertical line, otherwise the lines are perpendicular as their slopes are negative reciprocal of each other.