Chapter 1, Problem 11RE

a.

To determine

To find : the equation of the line.

Answer to Problem 11RE

the equation of line is $y=-\frac{3}{2}x-5$.

Explanation of Solution

Given information :

Point is $\left(0,-5\right)$.

Slope of the line is $m=-\frac{3}{2}$.

Formula used :

The equation of line is $y=mx+c$.

Calculation: first substitute the value of slope in the equation,

Since $m=-\frac{3}{2}$ , so substitute it in the equation,

  $\begin{array}{l}y=mx+c\\ y=-\frac{3}{2}x+c\text{ [substitute }m=-\frac{3}{2}]\end{array}$

Now find the value of $c$ , substitute $x=0,y=-5$

  $\begin{array}{l}y=-\frac{3}{2}x+c\\ -5=-\frac{3}{2}\left(0\right)+c\text{ [substitute }x=0,y=-5]\\ c=-5\text{ [simplify}]\end{array}$

Thus, the equation of the line is $y=-\frac{3}{2}x-5$.

b.

To determine

To find : the three additional points through the lines passes.

Answer to Problem 11RE

the points are $\left(1,-\frac{13}{2}\right)$ , $\left(-\frac{10}{3},0\right)$ , $\left(2,-8\right)$.

Explanation of Solution

Given information :

Point is $\left(-3,5\right)$.

Slope of the line is $m=-\frac{3}{2}$.

Calculation: from the above part it can be observed that,

the equation of the line is $y=-\frac{3}{2}x-5$.

To find the points on this line, first substitute $x=1$ ,

  $\begin{array}{l}y=-\frac{3}{2}x-5\\ y=-\frac{3}{2}\left(1\right)-\text{5 [substitute]}\\ y=-\frac{13}{2}\text{ [simplify]}\end{array}$

So the point is $\left(1,-\frac{13}{2}\right)$.

Now, substitute $y=0$ ,

  $\begin{array}{l}y=-\frac{3}{2}x-5\\ 0=-\frac{3}{2}x-5\text{ [substitute]}\\ y=-\frac{10}{3}\text{ [simplify]}\end{array}$

So the point is $\left(-\frac{10}{3},0\right)$.

Now , substitute $x=2$ ,

  $\begin{array}{l}y=-\frac{3}{2}x-5\\ y=-\frac{3}{2}\left(2\right)-5\text{ [substitute]}\\ x=-8\text{ [simplify]}\end{array}$

So the point is $\left(2,-8\right)$.

Thus, the points are $\left(1,-\frac{13}{2}\right)$ , $\left(-\frac{10}{3},0\right)$ , $\left(2,-8\right)$.