Chapter 1.3, Problem 79E

a .

To determine

To find: the equation of line that passes through the point $\left(2.5,6.8\right)$ and is parallel to the line $x-y=4$ .

Answer to Problem 79E

  $y=x+4.3$

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 $\left(x_1,y_1\right)$ .

First write the given line in slope intercept form to find its slope,

  $\begin{array}{c}x-y=4\\ y=x-4\\ y=1\cdot x+\left(-4\right)\end{array}$

So, the slope of this line is $m=1$ .

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

So, the required line have slope $m=1$ and it is given that is passes through the point $\left(2.5,6.8\right)$ .

Then by point slope form, the equation of line is,

  $\begin{array}{c}\left(y-y_1\right)=m\left(x-x_1\right)\\ y-6.8=1\left(x-2.5\right)\\ y-6.8=x-2.5\end{array}$

  $\begin{array}{c}y=x-2.5+6.8\\ y=x+4.3\end{array}$

Conclusion:

The equation of line that passes through the point $\left(2.5,6.8\right)$ and is parallel to the line $x-y=4$ is $y=x+4.3$ .

b .

To determine

To find: the equation of line that passes through the point $\left(2.5,6.8\right)$ and is perpendicular to the line $x-y=4$ .

Answer to Problem 79E

  $y=-x+9.3$

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 $\left(x_1,y_1\right)$ .

First write the given line in slope intercept form to find its slope,

  $\begin{array}{c}x-y=4\\ y=x-4\\ y=1\cdot x+\left(-4\right)\end{array}$

So, the slope of this line is $m=1$ .

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

So, the required line have slope $m=-\frac{1}{1}=-1$ and it is given that is passes through the point $\left(2.5,6.8\right)$ .

Then by point slope form, the equation of line is,

  $\begin{array}{c}\left(y-y_1\right)=m\left(x-x_1\right)\\ y-6.8=-1\left(x-2.5\right)\\ y-6.8=-x+2.5\end{array}$

  $\begin{array}{c}y=-x+2.5+6.8\\ y=-x+9.3\end{array}$

Conclusion:

The equation of line that passes through the point $\left(2.5,6.8\right)$ and is perpendicular to the line $x-y=4$ is $y=-x+9.3$ .