Chapter 1.1, Problem 75E

a.

To determine

Calculate the slope − intercept forms of the equations of the lines through the given points which are parallel to the given line.

Answer to Problem 75E

The slope intercept form for the parallel line is $y-1=0$ .

Explanation of Solution

Given:

It is given in the question that the coordinates and equation are $(-5,1),y+2=0$ .

Concept Used:

In this , use the concept that the slope intercept form is $y=mx+c$ and the parallel line have the same slope.

Calculation: The equation is $y+2=0$ ,make this in slope intercept form,

  $\begin{array}{l}y+2=0\\ y=-2\end{array}$

This,is the given line and is a horizontal line.

So,a parallel line will also be a horizontal line having the general equation ,

  $y+k=0$

By,making this general equation satisfy the point $(-5,1),$ gives us the required parallel line,

  $\begin{array}{l}1+k=0\\ k=-1\end{array}$

The parallel line will be : $y-1=0$ .

Conclusion:

The equation is $y-1=0$ .

b.

To determine

Calculate the slope − intercept forms of the equations of the lines through the given points which are perpendicular to the given line.

Answer to Problem 75E

The slope intercept form for the perpendicular line is $x+5=0$ .

Explanation of Solution

Given:

It is given in the question that the coordinates and equation are $(-5,1),y+2=0$ .

Concept Used:

In this , use the concept that the slope intercept form is $y=mx+c$ and the perpendicular line has a slope that is the negative reciprocal of the line.

Calculation: The equation is $y+2=0$ ,make this in slope intercept form,

  $\begin{array}{l}y+2=0\\ y=-2\end{array}$

A perpendicular line to that in the question will be a vertical one with this general equation,

  $x+k=0$

Again,make it satisfy the point $(-5,1),$ to get the required perpendicular line.

  $\begin{array}{l}-5+k=0\\ k=5\end{array}$

Now,the perpendicular line will be $x+5=0$ .

Conclusion:

The equation is $x+5=0$ .