Chapter 3.4, Problem 45PPS

To determine

Find the line that is intersecting but not perpendicular to $y=\frac{1}{2}x-5$

Answer to Problem 45PPS

Lines p, n, r

Explanation of Solution

Given:

  

Formula Used:

• The slope of the line $ax+by+c=0$ is $m=\frac{rise}{run}=-\frac{a}{b}$
• The product of slopes of two perpendicular lines is -1.
• Slopes of parallel lines are equal.

Calculation :

  $\begin{array}{l}y=\frac{1}{2}x-5\\ \Rightarrow \frac{1}{2}x-y-5=0\end{array}$

Comparing the line with standard form of linear equation $ax+by+c=0$ , we have,

a = $\frac{1}{2}$ , b = -1 , c = -5

So, slope of the equation is $m=-\frac{a}{b}=-\frac{\left(\frac{1}{2}\right)}{(-1)}=\frac{1}{2}$

As we know that product of slopes of two perpendicular lines is -1.

So, the line that is perpendicular to $y=\frac{1}{2}x-5$ will have slope $m=\frac{rise}{run}=-\frac{2}{1}=-2$ , which is negative.

And the line parallel to $y=\frac{1}{2}x-5$ will not intersect it and will have slope $\frac{1}{2}$ .

So, we need a line that intersects $y=\frac{1}{2}x-5$ and is not perpendicular to it.

Hence , any line that does not have slope $\frac{1}{2}$ or -2 will satisfy the condition.

First, find the line with slope -2:

Find the line with negative slope , then in the graph , go down 2 units and then run 1 unit to the right , to find the slope .Line l has the slope -2 , which is perpendicular to $y=\frac{1}{2}x+7$ .

Now, find the line with slope $\frac{1}{2}$ :

Find the line with positive slope, then , in the graph , rise 1 unit and run 2 units to the right , to find the slope.Line q has slope $\frac{1}{2}$ , which is parallel to $y=\frac{1}{2}x+7$ .

Hence all the lines other than q and l will intersect $y=\frac{1}{2}x+7$ and are not perpendicular to it .

So, the lines are p, n, r with slopes 2 , $-\frac{1}{2}$ , $\frac{1}{3}$ respectively.