Chapter 4, Problem 12CT

To determine

To determine whether the lines $\text{Line 1: }y-c=ax,\text{ Line 2: }ay=-x-b,\text{ Line 3: }ax+y=d$ are parallel or perpendicular.

Answer to Problem 12CT

the Line 1 and Line 2 are perpendicular, Line 1 and Line 3 are parallel and Line 2 and Line 3 are neither perpendicular nor parallel.

Explanation of Solution

Given information : Given three equation of lines

  $\text{Line 1: }y-c=ax,\text{ Line 2: }ay=-x-b,\text{ Line 3: }ax+y=d$ .

To check whether the lines are parallel or perpendicular, we need to find their slope. If the product of their slope is $-1$ , then lines are parallel and if their slopes are integral multiple then, they are parallel.

Line 1 can be written as $y=ax+c$ .

Therefore, its slope is $a$ .

Line 2 can be written as $y=-\frac{1}{a}x-\frac{b}{a}$ .

Therefore, its slope is $-\frac{1}{a}$ .

Line 3 can be written as $y=-ax+d$ .

Therefore, its slope is $-a$ .

Product of slope of Line 1 and Line 2

  $\begin{array}{l}=a\times \left(-\frac{1}{a}\right)\\ =-a\times \frac{1}{a}\\ =-1\end{array}$

Therefore, Line 1 and Line 2 are perpendicular.

Slope of Line 1 $=-1\times$ Slope of Line 3 .

i.e. $a=-1\left(-a\right)$ .

Therefore, Line 1 and Line 3 are parallel.

In case of Line 2 and Line 3, products of their slopes are not equal to $-1$ . Also, it is not integral multiple of each other.

Therefore, they are neither perpendicular nor parallel.

Thus, the Line 1 and Line 2 are perpendicular, Line 1 and Line 3 are parallel and Line 2 and Line 3 are neither perpendicular nor parallel.