Chapter Closure
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Exercise 125 Page 722
a Use the Midpoint Formula.
b Start by using the Slope Formula to find the slope.
c The lines are perpendicular if the product of their slopes is $\N1.$
a Graph:
Is $\bm{D}$ the Midpoint of $\bm{\Seg{AC}}?$ Yes.
b $y=\N \dfrac{1}{3}x+8$
c Yes.
a We are given that $A(2,4),$ $B(9,5),$ and $C(4,10).$ Let's use these points to draw $\triangle ABC$ on a graph paper.
We want to check whether the point $D(3, 7)$ is a midpoint of $\Seg{AC}.$ To do so, let's find the midpoint of $\Seg{AC}$ using the Midpoint Formula. Then we will check whether its coordinates are the same as the coordinates of $D.$
\(M\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\)
SubstitutePoints
\SubstitutePoints{2, 4}{4, 10}
\(M\left(\dfrac{\col{2}+\colII{4}}{2},\dfrac{\col{4}+\colII{10}}{2}\right)\)
AddTerms
\AddTerms
\(M\left(\dfrac{6}{2},\dfrac{14}{2}\right)\)
CalcQuot
\CalcQuot
\(M\left(3, 7\right)\)
The midpoint is located at $(3, 7).$ Since $D$ has the same coordinates, point $D$ is the midpoint of $\Seg{AC}.$
b An equation in slope-intercept form follows a specific format.
\begin{aligned}
y=\colIII{m}x+\textcolor{darkorange}{b}
\end{aligned}
For an equation in this form, $\colIII{m}$ is the slope and $\textcolor{darkorange}{b}$ is the $y\text{-}$intercept.
Let's use the points $D(3, 7)$ and $B(9, 5)$ to calculate $\colIII{m}.$ We will start by substituting the points into the Slope Formula.
\(m=\dfrac{y_2-y_1}{x_2-x_1}\)
SubstitutePoints
\SubstitutePoints{3, 7}{9, 5}
\(m=\dfrac{\colII{5}-\col{7}}{\colII{9}-\col{3}}\)
▼
\MMSimpRHS
SubTerms
\SubTerms
\(m=\dfrac{\N 2}{6}\)
MoveNegNumToFrac
\MoveNegNumToFrac
\(m = \N \dfrac{2}{6}\)
ReduceFrac
\ReduceFrac{2}
\(m= \N \dfrac{1}{3}\)
A slope of $\colIII{\N\frac{1}{3}}$ means that for every $3$ horizontal step in the positive direction, we take $1$ vertical step in the negative direction. Now that we know the slope, we can write a partial version of the equation. \begin{aligned} \colII{y}=\colIII{\N\dfrac{1}{3}}\col{x}+\textcolor{darkorange}{b} \end{aligned} To complete the equation, we also need to determine the $y\text{-}$intercept, $\textcolor{darkorange}{b}.$ Since we know that the given points will satisfy the equation, we can substitute one of them into the equation to solve for $\textcolor{darkorange}{b}.$ Let's use $(\col{3},\colII{7}).$
\(y=\N \dfrac{1}{3}x + b\)
SubstituteII
\SubstituteII{x}{3}{y}{7}
\(\colII{7}=\N\dfrac{1}{3}(\col{3})+b\)
▼
\MMSolve{b}
MoveRightFacToNumOne
\MoveRightFacToNumOne
\(7 = \N\dfrac{3}{3}+b\)
CalcQuot
\CalcQuot
\(7=\N1+b\)
AddEqn
\AddEqn{1}
\(8 = b\)
RearrangeEqn
\RearrangeEqn
\(b = 8\)
A $y\text{-}$intercept of $\textcolor{darkorange}{8}$ means that the line crosses the $y\text{-}$axis at the point $(0,\textcolor{darkorange}{8}).$ We can now complete the equation. \begin{aligned} y=\colIII{\N \dfrac{1}{3}}x+\textcolor{darkorange}{8} \end{aligned}
c $\Seg{BD}$ is a height of $\triangle ABC$ if $\Seg{BD}$ is perpendicular to $\Seg{AC}.$ That is the case if the product of their slopes equals $\N1.$ Let's first find the slopes of these line segments. In Part B, we found that the slope of $\Seg{BD}$ is $\colIV{\N\frac{1}{3}}.$ To find the slope of $\Seg{AC},$ let's recall the Slope Formula.
\begin{gathered}
m = \dfrac{y_2 - y_1}{x_2 - x_1}
\end{gathered}
\(m=\dfrac{y_2-y_1}{x_2-x_1}\)
SubstitutePoints
\SubstitutePoints{2,4}{4,10}
\(m=\dfrac{\colII{10}-\col{4}}{\colII{4}-\col{2}}\)
SubTerms
\SubTerms
\(m=\dfrac{6}{2}\)
CalcQuot
\CalcQuot
\(m=\colVI{3}\)
Finally, let's find the product of $\colIV{\N\frac{1}{3}},$ the slope of $\Seg{BD},$ and $\colVI{3},$ the slope of $\Seg{AC}.$ If the result is $\N1,$ these line segments are perpendicular. \begin{gathered} \colIV{\N\dfrac{1}{3}} \t \colVI{3} = \N1 \end{gathered}
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