Exercise 20 Page 229

An equation in slope-intercept form follows a specific format. For an equation in this form, $m$ is the slope and $b$ is the $y\text{-}$intercept. Let's use the given points to calculate $m.$ We will start by substituting the points into the Slope Formula.

$m=\frac{y_2-y_1}{x_2-x_1}$

SubstitutePoints

Substitute $\left(\text{-}1,\text{-}3\right)$ & $\left(\text{-}2,3\right)$

$m=\frac{3-(\text{-}3)}{\text{-}2-(\text{-}1)}$

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Simplify right-hand side

SubNeg

$a-(\text{-}b)=a+b$

$m=\frac{3+3}{\text{-}2+1}$

AddSubTerms

Add and subtract terms

$m=\frac{6}{\text{-}1}$

MoveNegDenomToFrac

Put minus sign in front of fraction

$m=\text{-}\frac{6}{1}$

DivByOne

$\frac{a}{1}=a$

$m=\text{-}6$

A slope of $\text{-}6$ means that for every $1$ horizontal step in the positive direction, we take $6$ vertical steps in the negative direction. Now that we know the slope, we can write a partial version of the equation.

$$\begin{array}{r}y=\text{-}6x+b\end{array}$$

To complete the equation, we also need to determine the $y\text{-}$intercept, $b.$ Since we know that the given points will satisfy the equation, we can substitute one of them into the equation to solve for $b.$ Let's use $(\text{-}1,\text{-}3).$

$y=\text{-}6x+b$

SubstituteII

$x=\text{-}1$, $y=\text{-}3$

$\text{-}3=\text{-}6(\text{-}1)+b$

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Solve for $b$

MultNegNegOnePar

$\text{-}a(\text{-}b)=a\cdot b$

$\text{-}3=6+b$

SubEqn

$\text{LHS}-6=\text{RHS}-6$

$\text{-}9=b$

RearrangeEqn

Rearrange equation

$b=\text{-}9$

A $y\text{-}$intercept of $\text{-}9$ means that the line crosses the $y\text{-}$axis at the point $(0,\text{-}9).$ We can now complete the equation.

$$\begin{array}{r}y=\text{-}6x+(\text{-}9)\iff y=\text{-}6x-9\end{array}$$