Chapter 1.5, Problem 28E

a.

To determine

To find: the standard form of equation for line.

Answer to Problem 28E

  $4x-5y+100=0$

Explanation of Solution

Given information:

Let the line whose standard equation needs to be evaluated be $p$ .

Line $p$ is $\parallel$ line $l$

Equation for line $l$ is $5y-4x=10$ .

Line $p$ passes through $\left(-15,8\right)$ .

Calculation:

Rearrange the equation of line $l$ in the slope intercept form and determine the slope as shown below.

  $\begin{array}{c}5y-4x=10\\ 5y=4x+10\\ y=\frac{4}{5}x+\frac{10}{5}\\ y=\frac{4}{5}x+2\\ \text{Compare with }y=mx+b\\ \therefore m=\frac{4}{5}\end{array}$

Since the slope of parallel lines is same. Thus, the slope of line $l\text{ and }p$ is $\frac{4}{5}$ .

Let’s put the value of m and $\left(-15,8\right)$ in the slope intercept formula and evaluate value of b as follows:

  $\begin{array}{c}y=mx+b\\ 8=\left(\frac{4}{5}\right)\left(-15\right)+b\\ 8=-12+b\\ b=8+12\\ =20\end{array}$

Now, substitute the value of m and b in the slope intercept formula and find the line equation of $p$ as follows:

  $\begin{array}{l}y=mx+b\\ y=\frac{4}{5}x+20\\ y=\frac{4}{5}x+20\left(\frac{5}{5}\right)\\ y=\frac{4}{5}x+\frac{100}{5}\left[\text{Taking common denominator}\right]\\ 5y=4x+100\\ 4x-5y+100=0\left[\begin{array}{l}\text{Rearranging in standard form}\\ \text{i}\text{.e}\text{. }Ax+By+b=0\end{array}\right]\end{array}$

Thus, the standard form equation of line $p$ is $4x-5y+100=0$ .

b.

To determine

To find: the standard form of equation for line.

Answer to Problem 28E

  $5x+4y+43=0$

Explanation of Solution

Given information:

Let the line whose standard equation needs to be evaluated be $p$ .

Line $p$ is $\perp$ line $l$

Equation for line $l$ is $5y-4x=10$ .

Line $p$ passes through $\left(-15,8\right)$ .

Slope of line $l$ = $\frac{4}{5}$

Calculation:

Since, line $p$ is $\perp$ line $l$ , so the slope of line $p$ can be evaluated as follows:

  $\begin{array}{l}\perp m=-\frac{1}{m}\\ =-\frac{1}{\left(\frac{4}{5}\right)}\left[\text{Substituting value of }m\text{ for line }l\right]\\ =-\frac{5}{4}\end{array}$

Thus, the slope of line $p$ is $-\frac{5}{4}$ .

Let’s put the value of m and $\left(-15,8\right)$ for line $p$ , in the slope intercept formula and evaluate value of b as follows:

  $\begin{array}{c}y=mx+b\\ 8=\left(-\frac{5}{4}\right)\left(-15\right)+b\\ 8=\frac{75}{4}+b\\ b=8-\frac{75}{4}\\ =\frac{32-75}{4}\\ =-\frac{43}{4}\end{array}$

Now, substitute the value of m and b in the slope intercept formula and find the line equation of $p$ as follows:

  $\begin{array}{l}y=mx+b\\ y=-\frac{5}{4}x+\left(-\frac{43}{4}\right)\\ y=-\frac{5}{4}x-\frac{43}{4}\\ y=\frac{-5x-43}{4}\\ 4y=-5x-43\\ 5x+4y+43=0\left[\begin{array}{l}\text{Rearranging in standard form}\\ \text{i}\text{.e}\text{. }Ax+By+b=0\end{array}\right]\end{array}$

Thus, the standard form equation of line $p$ is $5x+4y+43=0$ .