Chapter 1.5, Problem 32E

To determine

To show: the given equations have no point in common.

Answer to Problem 32E

Since parallel lines do not have a common point. Therefore, the given condition is true for the given line equations.

Explanation of Solution

Given information:

The given equations are as follows:

  $\begin{array}{l}y=m_{1}x+b_1\mathrm{...}\left(1\right)\\ y=m_{2}x+b_2\mathrm{...}\left(2\right)\end{array}$

When $m_1=m_2\text{ and }{b}_1\ne b_2$ , then both the equations representing the line are parallel.

There is a point (let’s say P ) common between the two lines.

Calculation:

Let the coordinates of point P be $\left(P_x,P_y\right)$ .

Put the point coordinates in equation (1) and evaluate the value of $b_1$ as shown below.

  $\begin{array}{c}y=m_{1}x+b_1\\ P_y=m_1{P}_x+b_1\left[\text{Substituting }\left(P_x,P_y\right)\right]\\ b_1=P_y-m_1{P}_x\end{array}$

Put the point coordinates and $m_2=m_1$ in equation (2) and evaluate the value of $b_2$ as shown below.

  $\begin{array}{c}y=m_{2}x+b_2\\ P_y=m_1{P}_x+b_2\left[\text{Substituting }\left(P_x,P_y\right)\text{ and }{m}_2=m_1\right]\\ b_2=P_y-m_1{P}_x\end{array}$

When $b_1=b_2$ , the given two lines cannot be parallel because there is no common point in the parallel lines.

Therefore, at $m_1=m_2\text{ and }{b}_1\ne b_2$ , the two line will be parallel to each other.