# 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.

(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)

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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.

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

Show/Hide Solution

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

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

$\therefore\ y= -5$.