Chapter 1.3, Problem 78E

a .

To determine

To find: the equation of line that passes through the point $\left(3,-2\right)$ and is parallel to the line $x-4=0$ .

Answer to Problem 78E

  $x=3$

Explanation of Solution

Concept Used:

Slope intercept form of equation of line is $y=mx+b$ , where m is the slope of the line and $\left(0,b\right)$ is the y -intercept of the line.

Point slope form of equation of line is $\left(y-y_1\right)=m\left(x-x_1\right)$ , where m is the slope of the line and line passes through the point $\left(x_1,y_1\right)$ .

Consider,

  $\begin{array}{c}x-4=0\\ x=4\end{array}$

The lines of the form $x=a$ are vertical lines and have undefined slope.

Now, two lines are parallel to each other only if they have the same slope.

So, the required line also have undefined slope, so it implies that the line is a vertical line of the form $x=a$

Also it is given that is passes through the point $\left(3,-2\right)$ .

Substitute $\left(3,-2\right)$ in the equation $x=a$ to find value of a ,

  $\begin{array}{c}x=a\\ 3=a\end{array}$

Thus, the equation of required line is,

  $x=3$

Conclusion:

The equation of line that passes through the point $\left(3,-2\right)$ and is parallel to the line $x-4=0$ is $x=3$ .

b .

To determine

To find: the equation of line that passes through the point $\left(3,-2\right)$ and is perpendicular to the line $x-4=0$ .

Answer to Problem 78E

  $y=-2$

Explanation of Solution

Concept Used:

Slope intercept form of equation of line is $y=mx+b$ , where m is the slope of the line and $\left(0,b\right)$ is the y -intercept of the line.

Point slope form of equation of line is $\left(y-y_1\right)=m\left(x-x_1\right)$ , where m is the slope of the line and line passes through the point $\left(x_1,y_1\right)$ .

Consider,

  $\begin{array}{c}x-4=0\\ x=4\end{array}$

The lines of the form $x=a$ are vertical lines and have undefined slope.

Now, two lines are perpendicular to each other only if their slopes are negative reciprocals of each other.

So, the slope of the required line is $m=0$ .

This implies that the line is a horizontal line of the form $y=b$

Also it is given that is passes through the point

  $\left(3,-2\right)$ .

Substitute $\left(3,-2\right)$ in the equation $y=b$ to find value of b ,

  $\begin{array}{c}y=b\\ -2=b\end{array}$

Thus, the equation of required line is,

  $y=-2$

Conclusion:

The equation of line that passes through the point $\left(3,-2\right)$ and is perpendicular to the line $x-4=0$ is $y=-2$ .