Chapter 8, Problem 13CST

To determine

For which values of a are the lines $y=ax+6\text{and }x+2y=4$ parallel?

For which values of a are the lines $y=ax+6\text{and }x+2y=4$ perpendicular? Explain.

Answer to Problem 13CST

Thegiven system of equations is parallel, when the value of a is $-\frac{1}{2}$ .

The given system of equations is perpendicular, when the value of a is $2$ .

Explanation of Solution

Given:

A system of equations:

  $y=ax+6\text{and }x+2y=4$ .

Concept used:

The slopes of two parallel lines are same.

We know that the slope of a perpendicular line to a given line is negative reciprocal of the slope of the given line. The product of slopes of two perpendicular lines is -1.

Calculation:

Let us convert our 2^nd equation in slope-intercept form as shown below:

  $\begin{array}{l}x+2y=4\left(\text{Given equation}\right)\\ x-x+2y=4-x\text{ (Subtracting }x\text{from both sides)}\\ 2y=-x+4\\ \frac{2y}{2}=\frac{-x}{2}+\frac{4}{2}\left(\text{Dividing both sides by 2}\right)\\ y=-\frac{1}{2}x+2\end{array}$

We can see that slope of our given line is $-\frac{1}{2}$ . Since the slopes of two parallel lines are same, therefore, the given system of equations is parallel, when the value of a is $-\frac{1}{2}$ .

Let mbe the slope of perpendicular line to line $y=-\frac{1}{2}x+2$

We will find the slope of perpendicular line to our given line as shown below:

  $\begin{array}{l}-\frac{1}{2}\cdot m=-1\\ -\frac{1}{2}\cdot (-2)\cdot m=-1\cdot (-2)\text{ (Multiplying both sides by }-2\text{)}\\ m=2\end{array}$

Therefore, the given system of equations is perpendicular, when the value of a is $2$ .