Chapter 1.1, Problem 76E

(a)

To determine

To Write: the slope intercept form of the equation passing through the point $\left(3,-1\right)$ and parallel to the line $y-2=0$ .

Answer to Problem 76E

  $y=-1$

Explanation of Solution

Given:

Given equation is $\left(3,-1\right)$ and point is $y-2=0$ .

Calculation:

Given line is

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

It is a straight line parallel to the x-axis having zero slope therefore line parallel to this will be vertical line passing through $y=-1$ having zero slope.

Substituting $x=3,y=-1$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ -1=0\times \left(3\right)+b\\ b=-1\end{array}$ .

Put $b=-1$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ y=0\times x-1\\ y=-1\end{array}$

Therefore, the answer is $y=-1$ .

(b)

To determine

To Write: the slope intercept form of the equation passing through the point $\left(3,-1\right)$ and perpendicular to the line $y-2=0$ ..

Answer to Problem 76E

  $x-3=0$

Explanation of Solution

Given; Given equation is $\left(3,-1\right)$ and point is $y-2=0$ .

Calculation;

Given line is

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

It is a straight line parallel to the y-axis having zero slope therefore line perpendicular to this will be vertical line parallel to x-axis passing through $x=3$ having undefined slope.

Required equation is.

  $\begin{array}{l}x=3\\ x-3=0\end{array}$

Therefore, the answer is $x-3=0$ .