# Slope of a Line : Exercise (1.2) - Solution

Ph မျက်နှာပြင်တွင် စာများအပြည့် မပေါ်လျှင် slider ကို ဆွဲ၍ လည်းကောင်း၊ ph ကို အလျားလိုက်ပုံစံ (landscape position) ပြောင်း၍ လည်းကောင်း ဖတ်ရှုနိုင်ပါသည်။

1.          Complete each sentence.

$\text{(a)}\quad$ The slope of the line passing through two points $(-6, 0)$ and $(2, 3)$ is __________.

$\text{(b)}\quad$ The slope of the line joining the point $(1, 2)$ and the origin is __________.

$\text{(c)}\quad$ A vertical line has __________ slope.

$\text{(d)}\quad$ A horizontal line has __________ slope .

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(a) $m=\displaystyle\frac{3-0}{2-(-6)}=\displaystyle\frac{3}{8}$

(b) $m=\displaystyle\frac{2-0}{1-0}=2$

(c) undefined

(d) zero

2.          For each graph state whether the slope is positive, negative, zero or undefined, then find the slope if possible.

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(a) undefined slope

(b) undefined slope

(c) zero slope

(d) zero slope

(e) positive slope, $m=1$

(f) positive slope, $m=\displaystyle\frac{2}{3}$

(g) negative slope, $m=-\displaystyle\frac{3}{4}$

(h) negative slope, $m=-\displaystyle\frac{2}{3}$

3.          Which pairs of points given below will determine horizontal lines? Which ones vertical lines? Determine the slope of each line without calculation.

$\displaystyle \begin{array}{l} \text{(a)}\quad (5,2)\ \text{and}\ (-3,2) \\\\ \text{(b)}\quad (0,5)\ \text{and}\ (-1,5)\\\\ \text{(c)}\quad (2,3)\ \text{and}\ (2,6) \\\\ \text{(d)}\quad (0,0)\ \text{and}\ (0,-2)\\\\ \text{(e)}\quad (1,-2)\ \text{and}\ (-3,-2)\\\\ \text{(f)}\quad (a,b)\ \text{and}\ (a,c) \end{array}$

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(a) horizontal line, slope = 0 ($\because$ same y-coordinate)

(b) horizontal line, slope = 0 ($\because$ same y-coordinate)

(c) vertical line, slope = undefined ($\because$ same x-coordinate)

(d) vertical line, slope = undefined ($\because$ same x-coordinate)

(e) horizontal line, slope = 0 ($\because$ same y-coordinate)

(f) vertical line, slope = undefined ($\because$ same x-coordinate)

4.          Find the slope of each line which contains each pair of points listed below.

$\displaystyle \begin{array}{l} \text{(a) }\quad A(0,0)\ \text{ and }\ B(8,4)\\\\ \text{(b) }\quad C(10,5)\ \text{ and }\ D(6,8)\\\\ \text{(c) }\quad E(-5,7)\ \text{ and }\ F(-2,-4)\\\\ \text{(d) }\quad G(23,15)\ \text{ and }\ H(18,5)\\\\ \text{(e) }\quad I(-2,0)\ \text{and }\ J(0,\ 6)\\\\ \text{(f) }\quad K(15,6)\ \text{ and }\ L(-2,23) \end{array}$

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The slope of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $m=\displaystyle\frac{y_2-y_1}{x_2-x_1}.$

\begin{aligned} \text{(a)}\quad m_{AB}&=\displaystyle\frac{4-0}{8-0}\\\\ &= \displaystyle\frac{1}{2} \end{aligned}

\begin{aligned} \text{(b)}\quad m_{CD}&=\displaystyle\frac{8-5}{6-10}\\\\ &= -\displaystyle\frac{3}{4} \end{aligned}

\begin{aligned} \text{(c)}\quad m_{EF}&=\displaystyle\frac{-4-7}{-2-(-5)}\\\\ &= -\displaystyle\frac{11}{3} \end{aligned}

\begin{aligned} \text{(d)}\quad m_{GH}&=\displaystyle\frac{5-15}{18-23}\\\\ &= \displaystyle\frac{-10}{-5}\\\\ &=2 \end{aligned}

\begin{aligned} \text{(f)}\quad m_{IJ}&=\displaystyle\frac{6-0}{0-(-2)}\\\\ &= \displaystyle\frac{6}{2}\\\\ &=3 \end{aligned} \begin{aligned} \text{(f)}\quad m_{KL}&=\displaystyle\frac{23-6}{-2-15}\\\\ &= \displaystyle\frac{17}{-17}\\\\ &=-1 \end{aligned}

5.          Find the slope of each line which contains each pair of points listed below.

$\displaystyle \begin{array}{l} \text{(a) }\quad E\left( {\displaystyle\frac{3}{4},\frac{4}{5}\text{ }} \right)\ \text{and }\ F\left( {-\displaystyle\frac{1}{2},\frac{7}{5}} \right)\\\\ \text{(b) }\quad G(-a,b)\ \text{ and }\ H(3a,2b)\\\\ \text{(c) }\quad L\left( {\sqrt{{12}},\sqrt{{18}}} \right)\ \text{ and }\ M\left( {\sqrt{{27}},\sqrt{8}} \right)\\\\ \text{(d) }\quad P(0,a)\ \text{ and }\ Q(a,0) \end{array}$

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The slope of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $m=\displaystyle\frac{y_2-y_1}{x_2-x_1}.$

\begin{aligned} \text{(a)}\quad m_{EF}&=\displaystyle\frac{\frac{7}{5}-\frac{4}{5}}{-\frac{1}{2}-\frac{3}{4}}\\\\ &= \displaystyle\frac{\frac{3}{5}}{-\frac{5}{4}}\\\\ &= -\displaystyle\frac{12}{25} \end{aligned}

\begin{aligned} \text{(b)}\quad m_{CD}&=\displaystyle\frac{2b-b}{3a-(-a)}\\\\ &= \displaystyle\frac{b}{4a} \end{aligned}

\begin{aligned} \text{(c)}\quad m_{EF}&=\displaystyle\frac{\sqrt{8}-\sqrt{18}}{\sqrt{27}-\sqrt{12}}\\\\ &= \displaystyle\frac{2\sqrt{2}-3\sqrt{2}}{3\sqrt{3}-2\sqrt{3}}\\\\ &= \displaystyle\frac{-\sqrt{2}}{\sqrt{3}}\\\\ &= -\displaystyle\frac{\sqrt{6}}{3} \end{aligned}

\begin{aligned} \text{(d)}\quad m_{GH}&=\displaystyle\frac{0-a}{a-0}\\\\ &= \displaystyle\frac{-a}{a}\\\\ &=-1 \end{aligned}

6.          Find $p, q, r$ in the followings:

$\text{(a) }\quad$ The slope joining the points $(0,3)$ and $(1,p)$ is $5$.

$\text{(b) }\quad$ The slope joining the points $(-2, q)$ and $(0,1)$ is $-1$.

$\text{(c) }\quad$ The slope joining the points $(-4, -2)$ and $(r, -6)$ is $-6$.

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The slope of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $m=\displaystyle\frac{y_2-y_1}{x_2-x_1}.$

\begin{aligned} \text{(a)}\quad \displaystyle\frac{p-3}{1-0}&=5\\\\ p-3 &= 5\\\\ p &= 8 \end{aligned}

\begin{aligned} \text{(b)}\quad \displaystyle\frac{1-q}{0-(-2)}&=-1\\\\ \displaystyle\frac{1-q}{2}&=-1\\\\ 1-q&=-2\\\\ q & = 3 \end{aligned}

\begin{aligned} \text{(c)}\quad \displaystyle\frac{-6-(-2)}{r-(-4)}&=-6\\\\ \displaystyle\frac{-4}{r+4}&=-6\\\\ r+4&=\displaystyle\frac{-4}{-6}\\\\ r+4&=\displaystyle\frac{2}{3}\\\\ r&=-\displaystyle\frac{10}{3} \end{aligned}

7.          Find the slope corresponding to the following events.

$\text{(a) }\quad$ A man climbs $10$ m for every $200$ meters horizontally.

$\text{(b) }\quad$ A motorbike rises $20$ km for every $100$ kilometers horizontally.

$\text{(c) }\quad$ A plane takes off $35$ km for every $5$ kilometers horizontally.

$\text{(d) }\quad$ A submarine descends $120$ m for every $15$ meters horizontally.

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(a) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{10}{200}=\displaystyle\frac{1}{20}$

(b) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{20}{100}=\displaystyle\frac{1}{5}$

(c) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{35}{5}=7$

(d) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{-120}{15}=-8$

8.          A train climbs a hill with slope $0.05$. How far horizontally has the train travelled after rising $15$ meters?

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$m = 0.05$, rise $= 15 m$
Since $m=\displaystyle\frac{\text{ rise}}{\text{ run}}$,
$0.05=\displaystyle\frac{15}{\text{ run}} \Rightarrow\text{ run} = 300.$

9.          The vertices of a triangle are the points $A(-2,3)$, $B(5,-4)$ and $C(1,8)$. Find the slope of each side and perimeter of a triangle.

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$A=(-2,3),\ B=(5,-4),\ C=(1,8)$

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

$m_{AB}=\displaystyle\frac{-4-3}{5+2}=-1$

$m_{BC}=\displaystyle\frac{8+4}{1-5}= -3$

$m_{AC}=\displaystyle\frac{8-3}{1+2}=\frac{5}{3}$

length of a segment $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\therefore\ AB=\sqrt{(5+2)^2+(-4-3)^2}=\sqrt{98}=9.9$

$BC=\sqrt{(1-5)^2+(8+4)^2}=\sqrt{160}=12.6$

$AC=\sqrt{(1+2)^2+(8-3)^2}=\sqrt{34}=5.8$

$\therefore\ \text{ the perimeter of}\ \triangle ABC = AB + BC + AC = 9.9+12.7+5.8=28.4$

10.          The vertices of a parallelogram are the points $P(1,4)$, $Q(3,2)$, $R(4,6)$ and $S(2,8)$. Find the slope of each side.

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$P=(1,4)$, $Q=(3,2)$, $R=(4,6), S=(2,8)$

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

$m_{PQ}=\displaystyle\frac{2-4}{3-1}=-1$

$m_{QR}=\displaystyle\frac{6-2}{4-3}= 4$

$m_{RS}=\displaystyle\frac{8-6}{2-4}=-1$

$m_{PS}=\displaystyle\frac{8-4}{2-1}=4$

11.          A line having a slope of $-1$ contains the point $(-2,5)$. What is the $y$-coordinate of the point on that line whose $x$-coordinate is $8$?

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Let the required point be $(8, y).$

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

$-1=\displaystyle\frac{y-5}{8+2}$

$\therefore\ y= -5$.