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 .

Show/Hide Solution

(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)

Show/Hide Solution

(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.

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

Show/Hide Solution

(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}$

Show/Hide Solution

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

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

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

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

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

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 $-6$.

Show/Hide Solution

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

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

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

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

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.

Show/Hide Solution

(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?

Show/Hide Solution

$m = 0.05$, 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.

Show/Hide Solution

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

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 $8$?

Show/Hide Solution

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

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

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

$\therefore\ y= -5$.

Combination (Part - 1)

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

COMBINATION A combination is a selection of objects without regard to order or arrangement. The different groups or selections of a number of things taken some or all of them at a time are called combinations.

Combination နှင့် Permutation မတူညီသော အချက်မှာ

• အစုတစ်ခုအတွင်းမှာ အစုပိုင်းတစ်ခုကို ရွေးချယ်လိုက်သည် ဆိုပါစို့…။
• Permutation သည် ရွေးချယ်လိုက်သော အစုပိုင်းအတွင်းရှိ အစုဝင်တစ်ခုချင်းစီ၏ အစီအစဉ်ကို ထည့်သွင်း စဉ်းစားသည်။
• Combination သည် ရွေးချယ်လိုက်သော အစုပိုင်းအတွင်းရှိ အစုဝင်တစ်ခုချင်းစီ၏ အစီအစဉ်ကို ထည့်သွင်း စဉ်းစားခြင်း မရှိပါ။

အောက်ပါ ဥပမာကို လေ့လာကြည့်ကြမည်။

A, B, C, D နှင့် E ကျောင်းသား ၅ ယောက်ထဲမှ ကျောင်းသား ၂ ယောက်ပါသော ကိုယ်စားလှယ်အဖွဲ့ ဖွဲ့ရန်လိုအပ်သည်။

(a) ပ-ကိုယ်စားလှယ်၊ ဒု-ကိုယ်စားလှယ် ကျောင်းသား ၂ ယောက်ပါသော ကိုယ်စားလှယ် အဖွဲ့ပေါင်း မည်မျှ ဖွဲ့နိုင်သနည်း။

(b) ကျောင်းသား ၂ ယောက်ပါသော ကိုယ်စားလှယ် အဖွဲ့ပေါင်း မည်မျှ ဖွဲ့နိုင်သနည်း။

မေးခွန်း (a) တွင် ပ-ကိုယ်စားလှယ်၊ နှင့် ဒု-ကိုယ်စားလှယ် ဟူ၍ ခွဲခြားမေးထားသည်။ ထို့ကြောင့် မည်သူက ပ-ကိုယ်စားလှယ် ဖြစ်မည်၊ မည်သူက ဒု-ကိုယ်စားလှယ် ဖြစ်မည်၊ ဆိုသော အစီအစဉ်သည် အရေးပါသည်ကို တွေ့ရပါမည်။

မေးခွန်း (b) တွင် ကျောင်းသား ၂ ယောက်ပါသော ကိုယ်စားလှယ် အဖွဲ့ပေါင်းဟု မေးထားသည်။ ကိုယ်စားလှယ်အဖွဲ့၌ ကျောင်းသားနှစ်ယောက် ပါဝင်ရန်သာလိုအပ်သည်။ ပထမကျောင်းသား၊ ဒုတိယကျောင်းသား၊ မည်သူမည်ဝါ ဖြစ်ရမည်ဆိုသော သတ်မှတ်ချက်သည် အရေးမပါတော့ပါ။

ထို့ကြောင့် မေးခွန်း (a) အတွက် တွဲနိုင်သော အစီအစဉ်ပေါင်းမှာ $20$ ဖြစ်ပြီး၊ မေးခွန်း (B) အတွက် တွဲနိုင်သော အစီအစဉ်ပေါင်းမှာ $10$ ဖြစ်သည်။

အောက်ပါဇယားဖြင့် ယှဉ်တွဲလေ့လာကြည့်ပါ။

	A	B	C	D	E
A		AB	AC	AD	AE
B	BA		BC	BD	BE
C	CA	CB		CD	CE
D	DA	DB	DC		DE
E	EA	EB	EC	ED

မေးခွန်း (a) သည် အစီအစဉ် အရေးပါသောကြောင့် permutation ဖြစ်ပြီး မေးခွန်း (b) တွင် ကျောင်းသားနှစ်ယောက်၏ အစီအစဉ် အရေးမပါတော့ သောကြောင့် combination ဖြစ်သည်။

COMBINATION OF $n$ OBJECTS TAKEN $r$ AT A TIME The number of combinations of $n$ different things taken $r$ at a time is denoted by ${}^nC_r$ and is defined as                                   $\begin{array}{|c|} \hline ^{n}{{C}_{r}}=\displaystyle\frac{{n!}}{{r!}{(n-r)!}}\\ \hline\end{array}$

DIFFERENCE BETWEEN PERMUTATION AND COMBINATION

Permutation	Combination
Permutation is defined as arrangement of r things that can be done out of total n things.	Combination is defined as selection of r things that can be done out of total n things.
Represents arrangement.	Represents grouping or selection
Order of objects or arrangement matter	Order of grouping/selection does not matter
Denoted by $^{n}{{P}_{r}}=\displaystyle\frac{{n!}}{{(n-r)!}}$	Denoted by $^{n}{{C}_{r}}=\displaystyle\frac{{n!}}{{r!}{(n-r)!}}$
Many permutations can be derived from a single combination.	Only one combination can be derived with one permutation.

အောက်ပါ video ဖြင့် ယှဉ်တွဲလေ့လာကြည့်ပါ။

Video Credit : Steve Stein

Example (1) (a).          A local school board with 8 people needs to form a committee with three people. How many ways can this committee be formed? Order of 3 people doesn't matter. Thus, it is combination. Number of ways to constitute a committee $\begin{array}{l} ={\ }^{8}{{C}_{3}}\\\\ =\displaystyle\frac{{8!}}{{3!}{(8-3)!}}\\\\ =\displaystyle\frac{{8!}}{{3!}{5!}}\\\\ = \displaystyle\frac{{8\times7\times6\times5\times4\times3\times2\times1}}{{(3\times2\times1)}\cdot{(5\times4\times3\times2\times1)}}\\\\ =56 \end{array}$ (b).          A local school board with 8 people needs to form a committee with three different responsibilities. How many ways can this committee be formed? Order of 3 people matter for their responsibilities. Thus, it is permutation. Number of ways to constitute a committee. $\begin{array}{l} ={\ }^{8}{{P}_{3}}\\\\ =\displaystyle\frac{{8!}}{{(8-3)!}}\\\\ =\displaystyle\frac{{8!}}{{5!}}\\\\ = \displaystyle\frac{{8\times7\times6\times5\times4\times3\times2\times1}}{{5\times4\times3\times2\times1}}\\\\ =336 \end{array}$

PROPERTIES OF COMBINATIONS                                                     $\begin{array}{|c|}\hline\color{red}{ 1. {\ }^{n}{C}_{r}={\ }^{n}{C}_{n-r}}\\\hline\end{array}$ Proof: $\begin{aligned} \mathrm{LHS}&={ }^{n} C_{r}=\frac{n !}{r !(n-r) !} \\\\ \mathrm{RHS} &={ }^{n} C_{n-r}=\frac{n !}{(n-r) !(n-(n-r)) !} \\\\ &=\frac{n !}{(n-r) ! r !} \\\\ \therefore \mathrm{LHS} &=\mathrm{RHS} \\\\ \text { Note: }& 1. {\ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x= y \text { or } x+y=n\\\\ & 2. {\ }^{n} C_{0}={ }^{n} C_{n}=1\\\\ & 3. {\ }^{n} C_{1}={ }^{n} C_{n-1}=n\\\\ \end{aligned}$ $\begin{array}{|c|}\hline \color{red}{2. {\ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}}\\\hline\end{array}$ Proof: $\begin{aligned} \mathrm{LHS} &={ }^{n} C_{r}+{ }^{n} C_{r-1} \\\\ &=\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !} \\\\ &=\frac{n !}{r !(n-r) !} \times \frac{(n-r+1)}{(n-r+1)}+\frac{n !}{(r-1) !(n-r+1) !} \times \frac{r}{r} \\\\ &=\frac{n !(n-r+1)}{r !(n-r+1) !}+\frac{r \times n !}{r !(n-r+1) !}[n !=n(n-1) !] \\\\ &=\frac{n !(n+1)-r \times n !+r \times n !}{r !(n-r+1) !} \\\\ &=\frac{(n+1) !}{r !(n-r+1) !} \\\\ \operatorname{RHS} &={ }^{n+1} C_{r}\\\\ &=\frac{(n+1) !}{r !(n+1-r) !} \\\\ &=\frac{(n+1) !}{r !(n-r+1) !} \\\\ \therefore\ \mathrm{LHS} &=\mathrm{RHS} \end{aligned}$ $\begin{array}{|c|}\hline \color{red}{3. {\ }^{n} C_{r}=\displaystyle\frac{n_{n-1}}{r} C_{r-1}=\frac{n(n-1)_{n-2}}{r(r-1)} C_{r-2}=\ldots}\\\hline\end{array}$ Proof: $\begin{aligned} { }^{n} C_{r}&=\frac{n(n-1)(n-2) \ldots(n-r+1)}{1 \times 2 \times 3 \times \ldots(r-2)(r-1) r}\\\\ &=\frac{n}{r} \times \frac{(n-1)(n-2) \ldots(n-r+1)}{1 \times 2 \times 3 \times \ldots(r-2)(r-1)}\\\\ &=\frac{n}{r}\cdot{ }^{n-1} C_{r-1}\\\\ { }^{n} C_{r}&=\frac{n(n-1)(n-2) \ldots(n-r+1)}{1 \times 2 \times 3 \times \ldots(r-2)(r-1) r}\\\\ &=\frac{n(n-1)}{r(r-1)} \times \frac{(n-2) \ldots(n-r+1)}{1 \times 2 \times 3 \times \ldots(r-2)}\\\\ &=\frac{n(n-1)}{r(r-1)}\cdot{ }^{n-2} C_{r-2}\\\\ &\text{and so on.} \end{aligned}$

Logarithms : Exercise (3.5) - Solutions

1.          Given that $\log 2.345=0.3701 .$ What are the characteristics and the mantissas of each of the followings?

(a) $\log 234,500$

(b) $\log 0.0002345$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ \ \ \ \log 2.345=0.3701\\\\\text{(a)}\ \ \ \log 234,500\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{5}}} \right)\\\\\ \ \ =\log {{10}^{5}}+\log 2.345\\\\\ \ \ =5+0.3701\\\\\ \ \ =5.3701\\\\\therefore \ \text{characteristic}=5\\\\\ \ \ \text{mantissa}=.3701\\\\\\\text{(b)}\ \ \ \log 0.0002345\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{{-4}}}} \right)\\\\\ \ \ =\log {{10}^{{-4}}}+\log 2.345\\\\\ \ \ =-4+0.3701\\\\\therefore \ \text{characteristic}=-4\\\\\ \ \ \text{mantissa}=.3701\end{array}$

2.          Using $\log _{10} 2.74=0.4377, \log _{10} 2.83=0.4518, \log _{10} 5.97=0.7760$, $\log _{10} 6.21=0.7931, \log _{10} 8.18=0.9128$ and $\log _{10} 9.27=0.9671$, compute

(a) $\left(\displaystyle\frac{28.3}{597 \times 621}\right)^{2}$

(b) $\displaystyle\frac{274^{\frac{1}{3}}}{927 \times 818}$

(c) $\displaystyle\frac{28.3 \sqrt{0.621}}{597}$

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$\begin{array}{l}\ \ \ \ \ {{\log }_{{10}}}2.74=0.4377,\\\ \ \ \ \ {{\log }_{{10}}}2.83=0.4518,\\\ \ \ \ \ {{\log }_{{10}}}5.97=0.7760,\\\ \ \ \ \ {{\log }_{{10}}}6.21=0.7931,\\\ \ \ \ \ {{\log }_{{10}}}8.18=0.9128,\\\ \ \ \ \ {{\log }_{{10}}}9.27=0.9671.\\\text{(a)}\ \ \text{Let}\ x={{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}},\ \text{then}\\\ \ \ \ \ \log x=\log {{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}}\\\ \ \ \ \ \ \ \ \ \ \ \ =2\log \left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log (597\times 621)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {1.4518-2.7760-2.7931} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =-8.2346\\\ \ \ \ \ x={{10}^{{-8.2346}}}\\\\\text{(b)}\ \ \text{Let}\ x=\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\log \left( {927\times 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\left( {\log 927+\log 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\left( {2.4377} \right)-2.9671-2.9128\\\ \ \ \ \ \ \ \ \ \ \ \ =-5.0673\\\ \ \ \ \ x={{10}^{{-5.0673}}}\\\\\text{(c)}\ \ \text{Let}\ x=\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\log 28.3+\displaystyle\frac{1}{2}\log 0.621-\log 597\\\ \ \ \ \ \ \ \ \ \ \ \ =1.4518+\displaystyle\frac{1}{2}\left( {-1+0.7931} \right)-2.7760\\\ \ \ \ \ \ \ \ \ \ \ \ =-1.4277\\\ \ \ \ \ x={{10}^{{-1.4277}}}\end{array}$

3.          Express each to a natural logarithm in terms of $\ln 2$ and $\ln 5$.

(a) $\log_{8}25$

(b) $\log_{5}64$

(c) $\log_{25}100$

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$\begin{array}{l}\text{(a)}\ \ \ \ {{\log }_{8}}25\ \\\ \ \ \ =\displaystyle\frac{{\ln 25}}{{\ln 8}}\\\ \ \ \ =\displaystyle\frac{{\ln {{5}^{2}}}}{{\ln {{2}^{3}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5}}{{3\ln 2}}\\\\\text{(b)}\ \ \ \ {{\log }_{5}}64\ \\\ \ \ \ =\displaystyle\frac{{\ln 64}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{\ln {{2}^{6}}}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{6\ln 5}}{{\ln 2}}\\\\\text{(c)}\ \ \ \ {{\log }_{{25}}}100\ \\\ \ \ \ =\displaystyle\frac{{\ln 100}}{{\ln 25}}\\\ \ \ \ =\displaystyle\frac{{\ln ({{5}^{2}}\times {{2}^{2}})}}{{\ln {{5}^{2}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5+2\ln 2}}{{2\ln 5}}\\\ \ =1+\displaystyle\frac{{\ln 2}}{{\ln 5}}\ \ \ \end{array}$

4.          Write each expression as a single logarithm.

(a) $3\ln \left( {t + 5} \right) - 4\ln t - 2\ln \left( {s - 1} \right)$

(b) $2\ln x + 5\ln y - \displaystyle\frac{1}{2}\ln z$

(c) $\displaystyle \frac{1}{3}\ln a - 6\ln b + 2$

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$\begin{array}{l} \text{(a)}\ \ \ \ 3\ln \left( {t+5} \right)-4\ln t-2\ln \left( {s-1} \right)\\\\ \ \ \ \ =\ln {{\left( {t+5} \right)}^{3}}-\ln {{t}^{4}}-\ln {{\left( {s-1} \right)}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{{\left( {t+5} \right)}}^{3}}}}{{{{t}^{4}}{{{\left( {s-1} \right)}}^{2}}}}\\\\\\ \text{(b)}\ \ \ \ 2\ln x+5\ln y-\displaystyle\frac{1}{2}\ln z\\\\ \ \ \ \ =\ln {{x}^{2}}+\ln {{y}^{5}}-\ln \sqrt{z}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{x}^{2}}\cdot {{y}^{5}}}}{{\sqrt{z}}}\\\\\\ \text{(c)}\ \ \ \ \displaystyle\frac{1}{3}\ln a-6\ln b+2\\\\ \ \ \ \ =\ln {{a}^{{\frac{1}{3}}}}-\ln {{b}^{6}}+\ln {{e}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{\sqrt[3]{a}\cdot {{e}^{2}}}}{{{{b}^{6}}}} \end{array}$

5.          Solve the following logarithmic equations for $x$.

(a) $\ln \left( x \right) + \ln \left( {x + 3} \right) = \ln \left( {20 - 5x} \right)$

(b) $2\ln \left( {\sqrt x } \right) - \ln \left( {1 - x} \right) = 2$

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$\begin{array}{l}\text{(a)}\ \ \ln (x)+\ln \left( {x+3} \right)=\ln \left( {20-5x} \right)\\\ \\\ \ \ \ \ \ \ln \left( {x\left( {x+3} \right)} \right)=\ln \left( {20-5x} \right)\\\\\ \ \ \ \ \ x\left( {x+3} \right)=20-5x\\\\\ \ \ \ \ \ {{x}^{2}}+8x-20=0\\\\\ \ \ \ \ \ (x+10)(x-2)=0\\\\\ \ \ \ \ \ x=-10\ \text{or}\ x=2\\\\\ \ \ \ \ \ \text{Since }x>0,\ x=-10\ \text{is impossible}\text{.}\\\\\ \ \ \therefore \ \ x=2\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 2\ln \left( {\sqrt{x}} \right)-\ln \left( {1-x} \right)=2\\\ \\\ \ \ \ \ \ln \left( {\displaystyle\frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}}}{{1-x}}} \right)=2\\\\\ \ \ \ \ \displaystyle\frac{x}{{1-x}}={{e}^{2}}\\\ \\\ \ \ \ \ x={{e}^{2}}-{{e}^{2}}x\\\\\ \ \ \ \ {{e}^{2}}x+x={{e}^{2}}\\\\\ \ \ \ \ x\left( {{{e}^{2}}+1} \right)={{e}^{2}}\\\\\ \ \ \ \ x=\displaystyle\frac{{{{e}^{2}}}}{{{{e}^{2}}+1}}\end{array}$

6.          Solve the equations:

(a) $x - xe^{5x + 2} = 0$

(b) $7 + 15e^{1 - 3x} = 10$ using $\ln 5 = 1.6094$.

(c) $4e^{1 + 3x} - 9e^{5 - 2x} = 0$ using $\ln 2=0.6931$ and $\ln 3=1.0986$.

No(3) မှ No(6) အထိ မေးခွန်းများသည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ Natural Logarithm ဆိုင်ရာ ပုစ္တာများကို တိုးချဲ့လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

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$\begin{array}{l}\text{(a)}\ \ \ x-x{{e}^{{5x+2}}}=0\\\ \\\ \ \ \ \ \ x\left( {1-{{e}^{{5x+2}}}} \right)=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ 1-{{e}^{{5x+2}}}=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ {{e}^{{5x+2}}}=1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=\ln 1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=0\\\\\ \ \therefore \ \ \ x=0\ \text{or}\ x=-\displaystyle\frac{2}{5}\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 7+15{{e}^{{1-3x}}}=10\\\ \\\ \ \ \ \ \ 15{{e}^{{1-3x}}}=3\\\\\ \ \ \ \ \ {{e}^{{1-3x}}}=\displaystyle\frac{1}{5}\\\ \\\ \ \ \ \ \ 1-3x=\ln \left( {\displaystyle\frac{1}{5}} \right)\\\\\ \ \ \ \ \ 1-3x=\ln \left( {{{5}^{{-1}}}} \right)\\\\\ \ \ \ \ \ 1-3x=-\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\text{1}\text{.6094}\\\\\ \ \ \ \ \ 3x=2.\text{6094}\\\\\ \ \ \ \ \ x=\text{0}\text{.8698}\\\\\\\text{(c)}\ \ \ 4{{e}^{{1+3x}}}-9{{e}^{{5-2x}}}=0\\\\\ \ \ \ \ 4{{e}^{{1+3x}}}=9{{e}^{{5-2x}}}\\\\\ \ \ \ \ \displaystyle\frac{{{{e}^{{1+3x}}}}}{{{{e}^{{5-2x}}}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ {{e}^{{5x-4}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ 5x-4=\ln \displaystyle\frac{{{{3}^{2}}}}{{{{2}^{2}}}}\\\\\ \ \ \ \ 5x=2\left( {\ln 3-\ln 2} \right)+4\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\ln 3-\ln 2+2} \right)\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\text{1}\text{.0986}-\text{0}\text{.6931+2}} \right)\\\\\ \ \ \ \ x=\text{0}\text{.9622}\end{array}$

Logarithms : Exercise (3.4) - Solutions

1.          If $\log _{a} b+\log _{b} a^{2}=3,$ find $b$ in terms of $a$.

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$\begin{array}{l}{{\log }_{a}}\,b+{{\log }_{b}}\,{{a}^{2}}=3\\\\{{\log }_{a}}\,b+2{{\log }_{b}}a=3\\\\{{\log }_{a}}\,b+\displaystyle\frac{2}{{{{{\log }}_{a}}\,b}}=3\\\\\text{Let}\ {{\log }_{a}}\,b=u,\ \text{then}\\\\u+\displaystyle\frac{2}{u}=3\\\\\therefore \ \ {{u}^{2}}+2=3u\\\\\therefore \ \ {{u}^{2}}-3u+2=0\\\\\therefore \ \ (u-1)(u-2)=0\\\\\therefore \ \ {{\log }_{a}}\,b=1\ \text{or}\ {{\log }_{a}}\,b=2\\\\\therefore \ \ \,b=a\ \text{or}\ \,b={{a}^{2}}\end{array}$

2.          Show that

(a) $\log _{4} x=2 \log _{16} x$.

(b) $\log _{b} x=3 \log _{b^{3}} x \quad$.

(c) $\log _{2} x=\left(1+\log _{2} 3\right) \log _{6} x$.

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$\begin{array}{l}(\text{a})\ \ \ \text{LHS}={{\log }_{4}}x\\\\\ \ \ \ \ \ \text{RHS}=2{{\log }_{{16}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{{{4}^{2}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{2}{2}{{\log }_{4}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{4}}x\\\\\ \ \therefore \ \ {{\log }_{4}}x=2{{\log }_{{16}}}x\\\\\\(\text{b})\ \ \text{LHS}={{\log }_{b}}x\\\\\ \ \ \ \ \text{RHS}=3{{\log }_{{{{b}^{3}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{3}{3}{{\log }_{b}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}x\\\\\ \ \therefore \ \ {{\log }_{b}}x=3{{\log }_{{{{b}^{3}}}}}x\\\\\\(\text{c})\ \ \text{LHS}={{\log }_{2}}x\\\\\ \ \ \ \ \text{RHS}=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {{{{\log }}_{2}}2+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}6\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{{{{{\log }}_{6}}2}}\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{{{{{\log }}_{6}}x}}{{{{{\log }}_{6}}2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}x\\\\\ \ \therefore \ \ {{\log }_{2}}x=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\end{array}$

3.          If $a=\log _{b} c, b=\log _{c} a$ and $c=\log _{a} b,$ prove that $a b c=1$.

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$ \begin{array}{l}\ \ \ a={{\log }_{b}}c,\ b={{\log }_{c}}a,\ c={{\log }_{a}}b\\\\\therefore \ \ abc={{\log }_{b}}c\cdot \ {{\log }_{c}}a\cdot {{\log }_{a}}b\\\\\therefore \ \ abc=\displaystyle\frac{{\log c}}{{\log b}}\cdot \ \displaystyle\frac{{\log a}}{{\log c}}\cdot \displaystyle\frac{{\log b}}{{\log a}}\ \left[ {\because \ \text{by L8}} \right]\\\\\therefore \ \ abc=1\end{array}$

4.          Show that

(a) $\left(\log _{10} 4-\log _{10} 2\right) \log _{2} 10=1$.

(b) $2 \log _{2} 3\left(\log _{9} 2+\log _{9} 4\right)=3$.

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$\begin{array}{l}\text{(a)}\ \ \ \left( {{{{\log }}_{{10}}}4-{{{\log }}_{{10}}}2} \right)\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}\displaystyle\frac{4}{2}\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot \displaystyle\frac{1}{{{{{\log }}_{{10}}}2}}\\\\\ \ \ =\ 1\\\\\\\text{(b)}\ \ \ 2\cdot {{\log }_{2}}3\cdot \left( {{{{\log }}_{9}}2+{{{\log }}_{9}}4} \right)\\\\\ \ \ ={{\log }_{2}}{{3}^{2}}\cdot {{\log }_{9}}(2\times 4)\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}8\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}{{2}^{3}}\\\\\ \ \ ={{\log }_{2}}9\cdot 3{{\log }_{9}}2\\\\\ \ \ =3\cdot \displaystyle\frac{1}{{{{{\log }}_{9}}2}}\cdot {{\log }_{9}}2\\\\\ \ \ =3\end{array}$

5.          Compute

(a) $3^{\log _{2} 5}-5^{\log _{2} 3}$

(b) $4^{\log _{2} 3}$

(c) $2^{\log _{2 \sqrt{2}} 27}$

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$\begin{array}{l}\text{(a)}\ \ \ {{3}^{{{{{\log }}_{2}}\,5}}}-{{5}^{{{{{\log }}_{2}}\,3}}}\\\\\ \ \ \ ={{3}^{{{{{\log }}_{2}}\,5}}}-{{3}^{{{{{\log }}_{2}}\,5}}}\\\\\ \ \ \ =0\\\\\text{(b)}\ \ \ {{4}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{2}}} \right)}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\\\\\\\text{(c)}\ \ \ {{2}^{{{{{\log }}_{{2\sqrt{2}}}}27}}}\\\\\ \ \ \ ={{2}^{{{{{\log }}_{{{{2}^{{3/2}}}}}}{{3}^{3}}}}}\\\\\ \ \ \ ={{2}^{{\frac{2}{3}\times 3{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{2}^{{2{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\end{array}$

6.          Using properties of logarithm, solve the following equations.

(a) $ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2$

(b) $25^{\log _{10} x}=5+4 x^{\log _{10} 5}$

(c) $9^{\log _{3}\left(\log _{2} x\right)}=\log _{2} x-\left(\log _{2} x\right)^{2}+1$

(d) $\log _{2} x+\log _{4} x+\log _{16} x=7$

(e) $\log _{7} 2+\log _{49} x=\log _{\frac{1}{7}} \sqrt{3}$

(f) $\log _{3} x-\log _{\frac{1}{3}} x^{2}=6$

(g) $ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4$

No (6) သည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ။ Change of Base နှင့် သက်ဆိုင်သော ဉာဏ်စမ်းမေးခွန်း များကို ထပ်ဆင့်လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

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$ \begin{array}{l} \text{(a)}\ \ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10+\displaystyle\frac{1}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ \displaystyle\frac{3}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10=\displaystyle\frac{4}{3}\\\\ \ \ \ \ \ {{x}^{{\frac{4}{3}}}}=10\\\\ \ \ \ \ \ x={{10}^{{\frac{3}{4}}}}\\\\\\ \text{(b)}\ \ {{25}^{{{{{\log }}_{{10}}}x}}}=5+4{{x}^{{{{{\log }}_{{10}}}5}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{2}}} \right)}^{{{{{\log }}_{{10}}}x}}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{5}^{{2\cdot }}}^{{{{{\log }}_{{10}}}x}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{{{{{\log }}_{{10}}}x}}}} \right)}^{2}}-4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}-5=0\\\\ \ \ \ \ \ \text{Let}\ {{5}^{{{{{\log }}_{{10}}}x}}}=u,\ \text{then}\\\\ \ \ \ \ \ {{u}^{2}}-4u-5=0\\\\ \ \ \ \ \ (u-5)(u+1)=0\\\\ \ \therefore \ \ \ u=5\ \text{or}\ u=-1\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\ \text{or}\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\\\\ \ \ \ \ \ \text{Since}\ {{5}^{{{{{\log }}_{{10}}}x}}}>0,\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\ \text{is}\ \text{impossible}\text{.}\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\\\\ \ \therefore \ \ \ {{\log }_{{10}}}x=1\\\\ \ \therefore \ \ \ x=10\\\\\\ \text{(c)}\ \ {{9}^{{{{{\log }}_{3}}\left( {{{{\log }}_{2}}x} \right)}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}9}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}{{3}^{2}}}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{2}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\ \ \ \ \ \ 2{{\left( {{{{\log }}_{2}}x} \right)}^{2}}-{{\log }_{2}}x-1=0\\\\ \ \ \ \ \ \left( {2\cdot {{{\log }}_{2}}x+1} \right)\left( {{{{\log }}_{2}}x-1} \right)=0\\\\ \ \ \ \ \ {{\log }_{2}}x=-\displaystyle\frac{1}{2}\ \text{or}\ {{\log }_{2}}x=1\\\\ \ \therefore \ \ \ x={{2}^{{-\frac{1}{2}}}}\ \text{or}\ x=2\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{{\sqrt{2}}}\ \text{or}\ x=2\\\\\\ \text{(d)}\ \ {{\log }_{2}}x+{{\log }_{4}}x+{{\log }_{{16}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+{{\log }_{{{{2}^{2}}}}}x+{{\log }_{{{{2}^{4}}}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+\displaystyle\frac{1}{2}{{\log }_{2}}x+\displaystyle\frac{1}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ \displaystyle\frac{7}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x=4\\\ \therefore \ \ \ x={{2}^{4}}=16\\\\\\ \text{(e)}\ \ {{\log }_{7}}2+{{\log }_{{49}}}x={{\log }_{{\displaystyle\frac{1}{7}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{{{{7}^{2}}}}}x={{\log }_{{{{7}^{{-1}}}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+\displaystyle\frac{1}{2}{{\log }_{7}}x=-{{\log }_{7}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{7}}{{x}^{{\frac{1}{2}}}}={{\log }_{7}}{{3}^{{-\frac{1}{2}}}}\\\\ \ \ \ \ \ {{\log }_{7}}2{{x}^{{\frac{1}{2}}}}={{\log }_{7}}\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 2{{x}^{{\frac{1}{2}}}}=\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 4x=\displaystyle\frac{1}{3}\\\ \ \therefore \ \ x=\displaystyle\frac{1}{{12}}\\\\\\ \text{(f)}\ \ {{\log }_{3}}x-{{\log }_{{\displaystyle\frac{1}{3}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x-{{\log }_{{{{3}^{{-1}}}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x+{{\log }_{3}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}{{x}^{3}}=6\\\\ \ \ \ \ \ {{x}^{3}}={{3}^{6}}\\\\ \ \therefore \ \ \ x={{3}^{2}}=9\\\\\\ \text{(g)}\ \ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ \displaystyle\frac{{{{{\log }}_{3}}(9{{x}^{2}})}}{{{{{\log }}_{3}}x}}\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ {{\log }_{3}}(9{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}9+{{\log }_{3}}{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}{{3}^{2}}+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ (2+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \therefore \ \ (1+{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=2\\\\ \ \therefore \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}+\left( {{{{\log }}_{3}}x} \right)-2=0\\\\ \ \therefore \ \ ({{\log }_{3}}x+2)({{\log }_{3}}x-1)=0\\\\ \ \therefore \ \ \ {{\log }_{3}}x=-2\ \text{or}\ {{\log }_{3}}x=1\\\\ \ \therefore \ \ \ x={{3}^{{-2}}}\ \text{or}\ x=3\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{9}\ \text{or}\ x=3 \end{array}$

Exponents and Radicals : Exercise (2.3) - Solutions

1.         Write the following in radical form.

$\begin{array}{ll} \text{(a)}\quad (5)^{^{\tfrac{1}{2}}} & \text{(b)}\quad (-9)^{^{\tfrac{1}{3}}} \\\\ \text{(c)}\quad (2)^{^{-\frac{1}{2}}} & \text{(d)}\quad \left(-\displaystyle\frac{3}{4}\right)^{^{\tfrac{2}{5}}}\\\\ \text{(e)}\quad \left(\displaystyle\frac{2}{7}\right)^{^{\tfrac{5}{2}}} \end{array}$

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$\begin{array}{l} \text{(a)}\ \ {{\left( 5 \right)}^{{\tfrac{{1\,}}{{2\,}}}}}=\sqrt{5}\\\\ \text{(b)}\ \ {{\left( {-9} \right)}^{{\tfrac{1}{3}}}}=\sqrt[3]{{-9}}\\\\ \text{(c)}\ \ {{\left( 2 \right)}^{{-\tfrac{{1\,}}{{2\,}}}}}=\displaystyle \frac{1}{{{{{\left( 2 \right)}}^{{\tfrac{{1\,}}{{2\,}}}}}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\\ \text{(d)}\ \ {{\left( {-\displaystyle \frac{3}{4}} \right)}^{{\tfrac{2}{5}}}}=\sqrt[5]{{{{{\left( {-\displaystyle \frac{3}{4}} \right)}}^{2}}}}=\sqrt[5]{{\displaystyle \frac{9}{{16}}}}\\\\ \text{(e)}\ \ {{\left( {\displaystyle \frac{2}{7}} \right)}^{{\tfrac{5}{2}}}}=\sqrt{{{{{\left( {\displaystyle \frac{2}{7}} \right)}}^{5}}}}=\sqrt{{\displaystyle \frac{{32}}{{16807}}}}\\ \end{array}$

2.         Write the following in fractional exponent form.

$\begin{array}{ll} \text{(a)}\quad \sqrt[6]{c^{^{5}}} & \text{(b)}\quad \sqrt[3]{-2} \\\\ \text{(c)}\quad \sqrt[5]{a^{^{4}} \sqrt[3]{b^{^{5}}}} & \text{(d)}\quad \sqrt[4]{\left(\displaystyle\frac{3}{7}\right)^{^{3}}} \end{array}$

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$\begin{array}{l} (\text{a})\ \ \sqrt[6]{{{{c}^{5}}}}={{c}^{{\frac{5}{6}}}}\\\\ (\text{b})\ \ \sqrt[3]{{-2}}={{(-2)}^{{\frac{1}{3}}}}\\\\ (\text{c})\ \ \sqrt[5]{{{{a}^{4}}\sqrt[3]{{{{b}^{5}}}}}}=\sqrt[5]{{{{a}^{4}}{{b}^{{\frac{5}{3}}}}}}={{\left( {{{a}^{4}}{{b}^{{\frac{5}{3}}}}} \right)}^{{\frac{1}{5}}}}={{a}^{{\frac{4}{5}}}}{{b}^{{\frac{1}{3}}}}\\\\ (\text{d})\ \ \sqrt[4]{{{{{\left( {\displaystyle \frac{3}{7}} \right)}}^{3}}}}={{\left( {\displaystyle \frac{3}{7}} \right)}^{{\frac{3}{4}}}} \end{array}$

3.         Change the expression with the same radical and simplify the radicands.

$\begin{array}{ll} \text{(a)}\quad 6 \sqrt{2} \quad& \text{(b)}\quad 3 a \sqrt[3]{x} \\\\ \text{(c)}\quad 2 \sqrt[5]{2} \quad& \text{(d)}\quad \sqrt[4]{\displaystyle\frac{1}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{^{3}}} \end{array}$

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$\begin{array}{l} \text{(a)}\quad 6 \sqrt{2}=\sqrt{6^{2} \cdot 2}=\sqrt{72}\\\\ \text{(b)}\quad 3 a \sqrt[3]{x}=\sqrt[3]{3^{3} \cdot a^{3} x}=\sqrt[3]{27 a^{3} x}\\\\ \text{(c)}\quad 2 \sqrt[5]{2}=\sqrt[5]{2^{5} \cdot 2}=\sqrt[5]{64}\\\\ \text{(d)}\quad 3 \sqrt[4]{\displaystyle\frac{1}{2}}=\sqrt[4]{3^{4} \cdot \displaystyle\frac{1}{2}}=\sqrt[4]{\displaystyle\frac{81}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{3}}=\sqrt{3^{2} \cdot x^{3}}=\sqrt{9 x^{3}} \end{array}$

4.         Simplify.

$\begin{array}{ll} \text{(a)}\quad \sqrt{32} & \text{(b)}\quad \sqrt[5]{-32} \\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{^{16}}}{16 y^{^{4}}}} & \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{^{2}}}{4 y}} \\\\ \text{(e)}\ \displaystyle\frac{9^{^{\tfrac{1}{2}}}}{\sqrt[3]{27}}& \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{^{3}}}} & \text{(h)}\quad \sqrt[n]{\displaystyle\frac{32}{2^{^{5+n}}}} \end{array}$

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$\begin{array}{l} \text{(a)}\quad \sqrt{32}=\sqrt{16 \cdot 2}=4 \sqrt{2}\\\\ \text{(b)}\quad \sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2\\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{16}}{16 y^{4}}}=\sqrt[4]{\displaystyle\frac{3^{4}\left(x^{4}\right)^{4}}{2^{4} \cdot y^{4}}}=\displaystyle\frac{3 x^{4}}{2 y}\\\\ \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{2}}{4 y}}=\sqrt[3]{\displaystyle\frac{3^{3} \cdot 3 x^{2}}{4 y}}=3 \sqrt[3]{\displaystyle\frac{3 x^{2}}{4 y}}\\\\ \text{(e)}\quad \displaystyle\frac{9^{\displaystyle\frac{1}{2}}}{\sqrt[3]{27}}=\displaystyle\frac{\left(3^{2}\right)^{\frac{1}{2}}}{\sqrt[3]{3^{3}}}=\displaystyle\frac{3}{3}=1\\\\ \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{2}{3} \times \displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{25}{49}}=\displaystyle\frac{5}{7}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{3}}}=\sqrt[3]{\displaystyle\frac{(-6)^{3}}{2^{3} \times 10^{3}}}=\displaystyle\frac{-6}{2 \times 10}=-\displaystyle\frac{3}{10}\\\\ \text{(h)}\quad \sqrt[4]{\displaystyle\frac{32}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{2^{5}}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{1}{2^{n}}}=\displaystyle\frac{1}{2} \end{array}$

5.         Rationalize the denominators.

$\begin{array}{ll} \text{(a)}\quad \displaystyle\frac{4 \sqrt{35}}{3 \sqrt{7}}\quad & \text{(b)}\quad \displaystyle\frac{20}{\sqrt{5}} \\\\ \text{(c)}\quad \displaystyle\frac{18}{\sqrt[3]{2}}\quad& \text{(d)}\quad \displaystyle\frac{\sqrt[3]{32}}{\sqrt[4]{27}} \\\\ \text{(e)}\quad \displaystyle\frac{\sqrt[3]{36 a^{^{2}}}}{\sqrt[3]{9 a}}\quad & \text{(f)}\quad \displaystyle\frac{\sqrt[3]{2}}{\sqrt[6]{12}} \\\\ \text{(g)}\quad \displaystyle\frac{1}{\sqrt[3]{x y^{^{2}}}}\quad & \text{(h)}\quad \sqrt[m]{{\displaystyle\frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}} \end{array}$

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$\begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \frac{{4\sqrt{{35}}}}{{3\sqrt{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{35}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{7\times 5}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{{4\sqrt{5}}}{3}\\\\\\\text{(b)}\ \ \ \ \displaystyle \frac{{20}}{{\sqrt{5}}}\\\\\ \ \ \ \ =\displaystyle \frac{{20}}{{\sqrt{5}}}\times \displaystyle \frac{{\sqrt{5}}}{{\sqrt{5}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{20\sqrt{5}}}{5}\\\\\ \ \ \ \ =4\sqrt{5}\\\\\\\text{(c)}\ \ \ \ \displaystyle \frac{{18}}{{\sqrt[3]{2}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18}}{{\sqrt[3]{2}}}\times \displaystyle \frac{{\sqrt[3]{{{{2}^{2}}}}}}{{\sqrt[3]{{{{2}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{{\sqrt[3]{{{{2}^{3}}}}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{2}\\\ \ \ \ \ =9\sqrt[3]{4}\\\\\text{(d)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{32}}}}{{\sqrt[4]{{27}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{2}^{3}}\cdot 4}}}}{{\sqrt[4]{{{{3}^{3}}}}}}\times \displaystyle \frac{{\sqrt[4]{3}}}{{\sqrt[4]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{2\sqrt[3]{4}\sqrt[4]{3}}}{3}\\\\\\\text{(e)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{36{{a}^{2}}}}}}{{\sqrt[3]{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{\displaystyle \frac{{36{{a}^{2}}}}{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{4a}}\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{12}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{{{2}^{2}}}}\cdot \sqrt[6]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[3]{2}\cdot \sqrt[6]{3}}}\times \displaystyle \frac{{\sqrt[6]{{{{3}^{5}}}}}}{{\sqrt[6]{{{{3}^{5}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[6]{{243}}}}{3}\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\times \displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{2}}y}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{3}}{{y}^{3}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{xy}}\\\\\text{(h)}\ \ \ \ \ \sqrt[m]{{\displaystyle \frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}}\\\\\ \ \ \ \ =\sqrt[m]{{\displaystyle \frac{{2y}}{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[m]{{2y}}}}{{\sqrt[m]{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2y}}}}{{y\sqrt[m]{{{{3}^{2}}{{x}^{3}}}}}}\times \displaystyle \frac{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2\cdot {{3}^{{m-2}}}{{x}^{{m-3}}}y}}}}{{3xy}}\end{array}$

6.         Reduce the order as far as possible.

$\begin{array}{ll} \text{(a)}\quad \sqrt[4]{25} \quad& \text{(b)}\quad \sqrt[6]{4} \\\\ \text{(c)}\quad \sqrt[6]{8} \quad& \text{(d)}\quad \sqrt[9]{8 y^{^{3}}} \\\\ \text{(e)}\quad \sqrt[6]{27^{^{3}}} \quad& \text{(f)}\quad \sqrt[8]{a^{2} b^{^{4}}} \\\\ \text{(g)}\quad \sqrt[12]{64 a^{^{2}} b^{^{6}}} & \text{(h)}\quad (72)^{^{\tfrac{3}{5}}} \\\\ \text{(i)}\quad \sqrt[3]{768} \end{array}$

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$\begin{array}{l} \text{(a)}\quad \sqrt[4]{25}=\sqrt[4]{5^{2}}=\sqrt{5}\\\\ \text{(b)}\quad \sqrt[6]{4}=\sqrt[6]{2^{2}}=\sqrt[3]{2}\\\\ \text{(c)}\quad \sqrt[6]{8}=\sqrt[6]{2^{3}}=\sqrt{2}\\\\ \text{(d)}\quad \sqrt[9]{8 y^{3}}=\sqrt[9]{2^{3} \cdot y^{3}}=\sqrt[3]{2 y}\\\\ \text{(e)}\quad \sqrt[6]{27^{3}}=\sqrt{27}=3 \sqrt{3}\\\\ \text{(f)}\quad \sqrt[8]{a^{2} b^{4}}=\sqrt[4]{a b^{2}}\\\\ \text{(g)}\quad \sqrt[12]{64 a^{2} b^{6}}=\sqrt[12]{8^{2} \cdot a^{2} b^{6}}=\sqrt[6]{8 a b^{3}}\\\\ \text{(h)}\quad (72)^{\frac{3}{5}}=\sqrt[5]{72^{3}}=\sqrt[5]{\left(3^{2} \cdot 2^{3}\right)^{3}}=\sqrt[5]{3^{6} \cdot 2^{9}}=6 \sqrt[5]{3 \cdot 16}=6 \sqrt[5]{48}\\\\ \text{(i)}\quad \sqrt[3]{768}=\sqrt[3]{4^{3} \cdot 12}=4 \sqrt[3]{12} \end{array}$

7.         Find the simplified forms.

$\begin{array}{ll} \text{(a)}\quad \sqrt{\displaystyle\frac{9}{50}} \quad& \text{(b)}\quad \sqrt[3]{\displaystyle\frac{-192}{49}} \\\\ \text{(c)}\quad \sqrt[4]{16}\quad& \text{(d)}\quad 2 \sqrt[3]{56} \end{array}$

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$\begin{array}{l} \text{(a)}\quad\sqrt{\displaystyle \frac{9}{50}}=\sqrt{\displaystyle \frac{3^{2}}{5^{2} \cdot 2}}=\displaystyle \frac{3}{5 \sqrt{2}} \times \displaystyle \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{3 \sqrt{2}}{10}\\\\ \text{(b)}\quad\sqrt[3]{\displaystyle \frac{-192}{49}}=\sqrt[3]{\displaystyle \frac{(-4)^{3} \cdot 3}{7^{2}} \times \displaystyle \frac{7}{7}}=-\displaystyle \frac{4 \sqrt[3]{21}}{7}\\\\ \text{(c)}\quad\sqrt[4]{16}=\sqrt[4]{2^{4}}=2\\\\ \text{(d)}\quad 2 \sqrt[3]{56}=2 \sqrt[3]{2^{3} \cdot 7}=4 \sqrt[3]{7} \end{array}$

Coordinate Geometry : Exercise (1.3) - Solutions

1.         Sketch the following lines.

$\begin{array} \text{(a)}\ y=3\\\\ \text{(b)}\ x=-2\\\\ \text{(c)}\ y=5x\\\\ \text{(d)}\ y=-3x\\\\ \text{(e)}\ y=\displaystyle\frac{1}{2}x \end{array}$

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$\text{(a)}\ y=3$

$\text{(b)}\ x=-2$

$\begin{array}{l} \text{(c)}\ y=5x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=5x & 0 & 5\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(d)}\ y=-3x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x & 0 & -3\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(e)}\ y=\displaystyle\frac{1}{2}x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x & 0 & -3\\ \hline \end{array}\end{array}$

2.         Graph each line in the same coordinate plane.

$\begin{array}{lll} \text{(a)}\ y=x+2 \\\\ \text{(b)}\ y=x+3 \\\\ \text{(c)}\ y=x+5 \\\\ \text{(d)}\ y=x-1 \\\\ \text{(e)}\ y=x-2 \\\\ \text{(f)}\ y=x-4 \end{array}$

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$\bullet\quad y=x$ is a line with slope equal to 1 and passing through origin.

$\bullet\quad y=x + 2$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 2).

$\bullet\quad y=x + 3$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 3).

$\bullet\quad y=x + 5$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 5).

$\bullet\quad y=x -1 $ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -1).

$\bullet\quad y=x -2$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -2).

$\bullet\quad y=x -4$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -4).

3.         Graph each line.

$\begin{array} \text{(a)}\ y=2x+2\\\\ \text{(b)}\ y=\displaystyle\frac{1}{2}x+2\\\\ \text{(c)}\ y=-\displaystyle\frac{1}{2}x+2 \end{array}$

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$\begin{array}{l} \text{(a)}\ y=2x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & -1\\ \hline y=2x+2 & 2 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(b)}\ y=\displaystyle\frac{1}{2}x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & -4\\ \hline y=\displaystyle\frac{1}{2}x+2 & 2 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(b)}\ y=-\displaystyle\frac{1}{2}x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & 4\\ \hline -\displaystyle\frac{1}{2}x+2 & 2 & 0\\ \hline \end{array}\end{array}$

4.         Find the slope and $y$-intercept for the following equations and sketch their graphs.

$\begin{array}{l} \text{(a)}\ y=x-2 \\\\ \text{(b)}\ y=-3x-3 \\\\ \text{(c)}\ y=\displaystyle\frac{1}{2}x+1\\\\ \text{(d)}\ y=-\displaystyle\frac{1}{2}x+1 \\\\ \text{(e)}\ y+3x=3 \\\\ \text{(f)}\ x-y=3 \end{array}$

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$\begin{array}{l} \text{(a)}\ y=x-2\\\\ \text{slope}= 1\\\\ y\text{-intercept}=-2\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 2\\ \hline y=x-2 & -2 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(b)}\ y=-3x-3\\\\ \text{slope}= -3\\\\ y\text{-intercept}=-3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x-3 & -3 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(c)}\ y=\displaystyle\frac{1}{2}x+1\\\\ \text{slope}= \displaystyle\frac{1}{2}\\\\ y\text{-intercept}=1\\\\ \begin{array}{|l|l|l|} \hline x & 0 & -2\\ \hline y=\displaystyle\frac{1}{2}x+1 & 1 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(d)}\ y=-\displaystyle\frac{1}{2}x+1\\\\ \text{slope}= -\displaystyle\frac{1}{2}\\\\ y\text{-intercept}=1\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 2\\ \hline y=-\displaystyle\frac{1}{2}x+1 & 1 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(e)}\ y+3x=3\\\\ \quad y=-3x+3\\\\ \text{slope}=-3\\\\ y\text{-intercept}=3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x+3 & 3 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(f)}\ x-y=3\\\\ \quad y=x-3\\\\ \text{slope}=1\\\\ y\text{-intercept}=-3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 3\\ \hline y=x-3 & -3 & 0\\ \hline \end{array}\end{array}$

5.         Find the equation of the straight line with the given slope and $y$-intercept.

(a) slope 3, $y$-intercept 4

(b) slope 2, $y$-intercept 0

(c) slope 0, $y$-intercept 2

(d) slope 0, $y$-intercept 0

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$\begin{array}{l} \text{(a)}\quad m=3,\ c=4\\\\ \quad \quad \ \ y=m x+c\\\\ \quad \quad \ \ \ \ =3 x+4 \end{array}$

$\therefore$ The equation of the line is $y=3 x+4$.

$\begin{array}{l} \text{(b)}\quad m=2,\ c=0\\\\ \quad \quad \ \ y=m x+c\\\\ \quad \quad \ \ \ \ =2 x+0\\\\ \quad \quad \ \ \ \ =2 x \end{array}$

$\therefore$ The equation of the line is $y=2x$.

$\begin{array}{l} \text{(c)}\quad m=0,\ c=2\\\\ \quad \quad \ \ y=(0) x+2\\\\ \quad \quad \ \ \ \ =2 \end{array}$

$\therefore$ The equation of the line is $y=2$.

$\begin{array}{l} \text{(d)}\quad m=0,\ c=0\\\\ \quad \quad \ \ y=(0) x+0\\\\ \quad \quad \ \ \ \ =0 \end{array}$

$\therefore$ The equation of the line is $y=0$.

6.         Find the equation of the line which has a slope $m$ of $\displaystyle -\frac{2}{3}$ and passes through the point $(9,4)$.

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Let $\left(x_{1}, y_{1}\right)=(9,4). m=-\displaystyle \frac{2}{3}$

$y-y_{1}=m\left(x-x_{1}\right)$

$y-4=-\displaystyle \frac{2}{3}(x-9)=-\displaystyle \frac{2}{3} x+6$

$y=-\displaystyle \frac{2}{3} x+10$

$\therefore 2x+3y=30$

$\therefore$ The equation of the line is $2x+3y=30$.

7.         A line has slope $-2$ and $y$-intercept $6$, find its $x$-intercept.

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$$\begin{array}{l} m=-2,\ c=6 \\\\ \text{The equation of the line having slope} \\\\ m\ \text{and}\ y\text{-intercept}\ c\ \text{is}\ y=m x+c.\\\\ \therefore\ \text{The equation of given line is}\ y=-2x + 6.\\\\ \text { When }\ y=0,-2 x+6=0\Rightarrow x=3 \\\\ \therefore x \text {-intercept }=3 \end{array}$$

8.         Find the equation of the line which:

(a) has a slope of $5$ and passes through the point $(2,9)$.

(b) has a slope of $1$ and passes through the point $(1,-2)$.

(c) has a slope of $-3$ and passes through the point $(-1,6)$.

(d) has a slope of $-2$ and passes through the point $(-1,4)$.

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$$\begin{array}{l} \text { (a) } \text{Let}\ \left(x_{1}, y_{1}\right)=(2,9).\ m=5\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-9 =5(x-2)\\\\ \quad \quad y-9 =5 x-10 \\\\ \quad \quad y=5 x-1 \end{array}$$

$$\begin{array}{l} \text { (b) } \text{Let}\ \left(x_{1}, y_{1}\right)=(1,-2). \ m=1\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-(-2)=1(x-1)\\\\ \quad \quad y+2 =x-1 \\\\ \quad \quad y=x-3 \end{array}$$

$$\begin{array}{l} \text { (c) } \text{Let}\ \left(x_{1}, y_{1}\right)=(-1,6).\ m=-3\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-6=-3\left(x-(-1)\right)\\\\ \quad \quad y-6 =-3 x-3 \\\\ \quad \quad y=-3 x+3 \end{array}$$

$$\begin{array}{l} \text { (d) } \text{Let}\ \left(x_{1}, y_{1}\right)=(-1,4).\ m=-2\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-4=-2\left(x-(-1)\right) \\\\ \quad \quad y-4 =-2 x-2 \\\\ \quad \quad y=-2 x+2 \end{array}$$

9.         Find the slope and equation of the line joining the following pairs of points.

(a) $(2,4)$ and $(6,8)$

(b) $(-3,5)$ and $(6,-1)$

(c) $(-2, 1)$ and $(-4, -2)$

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(a) Let $\left(x_{1}, y_{1}\right)=(2,4)$ and $\left(x_{2}, y_{2}\right)=(6,8)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{8-4}{6-2}=\frac{4}{4}=1$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-4=1(x-2)$

$y-4=x-2$

$y=x+2$

(b) Let $\left(x_{1}, y_{1}\right)=(-3,5)$ and $\left(x_{2}, y_{2}\right)=(6,-1)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{-1-5}{6-(-3)}=-\frac{6}{9}=-\frac{2}{3}$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-5=-\displaystyle\frac{2}{3}\left(x-(-3)\right)$

$2x+3y = 9$

(c) Let $\left(x_{1}, y_{1}\right)=(-2,1)$ and $\left(x_{2}, y_{2}\right)=(-4,-2)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{-2-1}{-4-(-2)}=\frac{3}{2}$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-1=\displaystyle\frac{3}{2}\left(x-(-2)\right)$

$3x-2y = -8$

10.        Determine which of the pairs of lines in each case with given equations are parallel or perpendicular or neither.

(a) $y=3x-2$ and $y=3x+9$

(b) $y=\displaystyle \frac{2}{3}x-5$ and $y=\displaystyle \frac{3}{2}x-5$

(c) $y=3x-2$ and $y=-\displaystyle \frac{1}{3}x+9$

(d) $y=\displaystyle \frac{2}{3}x-5$ and $y=-\displaystyle \frac{3}{2}x-5$

Show/Hide Solution

$\begin{array}{l}(\text{a})\ \ \ \ {{l}_{1}}:y=3x-2\ \text{and}\ {{l}_{2}}:y=3x+9\\\ \ \ \therefore \ \ {{m}_{1}}=3\ \text{and}\ {{m}_{2}}=3\\\ \ \ \therefore \ \ {{m}_{1}}={{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\parallel \ {{l}_{2}}\\\\(\text{b})\ \ \ \ {{l}_{1}}:y=\displaystyle \frac{2}{3}x-5\ \text{and}\ {{l}_{2}}:y=\displaystyle \frac{3}{2}x-5\\\ \ \ \therefore \ \ {{m}_{1}}=\displaystyle \frac{2}{3}\ \text{and}\ {{m}_{2}}=\displaystyle \frac{3}{2}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \ \ \ \ \text{Again,}\ {{m}_{1}}\cdot {{m}_{2}}=\displaystyle \frac{2}{3}\times \displaystyle \frac{3}{2}=1\ne -1\\\ \ \ \therefore \ \ {{l}_{1}}\ \text{and}\ {{l}_{2}}\ \text{are neither parallel nor }\\\ \ \ \ \ \ \text{perpendicular to each other}\text{.}\\\\(\text{c})\ \ \ \ {{l}_{1}}:y=3x-2\ \text{and}\ {{l}_{2}}:y=-\displaystyle \frac{1}{3}x+9\\\ \ \ \therefore \ \ {{m}_{1}}=3\ \text{and}\ {{m}_{2}}=-\displaystyle \frac{1}{3}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\nparallel \ {{l}_{2}}\\\ \ \ \ \ \ \text{But,}\ {{m}_{1}}\cdot {{m}_{2}}=3\left( {-\displaystyle \frac{1}{3}} \right)=-1\\\ \ \ \therefore \ \ {{l}_{1}}\bot \ {{l}_{2}}\\\\(\text{d})\ \ \ \ {{l}_{1}}:y=\displaystyle \frac{2}{3}x-5\ \text{and}\ {{l}_{2}}:y=-\displaystyle \frac{3}{2}x-5\\\ \ \ \therefore \ \ {{m}_{1}}=\displaystyle \frac{2}{3}\ \text{and}\ {{m}_{2}}=-\displaystyle \frac{3}{2}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\nparallel \ {{l}_{2}}\\\ \ \ \ \ \ \text{But,}\ {{m}_{1}}\cdot {{m}_{2}}=\displaystyle \frac{2}{3}\left( {-\displaystyle \frac{3}{2}} \right)=-1\\\ \ \ \therefore \ \ {{l}_{1}}\bot \ {{l}_{2}}\end{array}$

11.         Find the equation of the line which is parallel to the line:

(a) with equation $y=4x+2$ and passes through $(0,8)$.

(b) with equation $y=-x+3$ and passes through ($0,5)$.

(c) with equation $y=-2x-3$ and passes through $(0,-7)$.

(d) with equation $y=-\displaystyle \frac{4}{5}x-3$ and passes through $ \displaystyle \left(0,\frac{1}{2}\right)$.

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$\begin{array}{l}(\text{a})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=4x+2\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=4\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=4\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,8)}\ \text{is}\\\\\ \ \ \ y-8=4(x-0)\\\\\ \ \ \ y=4x+8\\\\(\text{b})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-x+3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=-1.\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,5)}\ \text{is}\\\\\ \ \ \ y-5=-(x-0)\\\\\ \ \ \ y=-x+5\\\\(\text{c})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-2x-3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-2\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}},\ {{m}_{2}}=-2.\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}-\text{7)}\ \text{is}\\\\\ \ \ \ y-(-7)=-2(x-0)\\\\\ \ \ \ y=-2x-7\\\\(\text{d})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-\displaystyle \frac{4}{5}x-3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-\displaystyle \frac{4}{5}\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{4}{5}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}\displaystyle \frac{1}{2}\text{)}\ \text{is}\\\\\ \ \ \ y-\displaystyle \frac{1}{2}=-\displaystyle \frac{4}{5}(x-0)\\\\\ \ \ \ y=-\displaystyle \frac{4}{5}x+\displaystyle \frac{1}{2}\end{array}$

12.         Find the equation of the line which is perpendicular to the line:

(a) with equation $y=5x-4$ and passes through $(0,7)$

(b) with equation $y=-x+7$ and passes through $(0,4)$

(c) with equation $y=-2x+3$ and passes through $(0,-4)$

(d) with equation $ \displaystyle y=x-\frac{3}{2}$ and passes through $ \displaystyle \left( {0,\frac{5}{4}} \right)$

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$\begin{array}{l}(\text{a})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=5x-4\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=5\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=-\displaystyle \frac{1}{5}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,7)}\ \text{is}\\\\\ \ \ \ y-7=-\displaystyle \frac{1}{5}(x-0)\\\\\ \ \ \ y=-\displaystyle \frac{1}{5}x+7\\\\(\text{b})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-x+7\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=1\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,4)}\ \text{is}\\\\\ \ \ \ y-4=1(x-0)\\\\\ \ \ \ y=x+4\\\\(\text{c})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-2x+3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-2\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=\displaystyle \frac{1}{2}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}-\text{4)}\ \text{is}\\\\\ \ \ \ y-(-4)=\displaystyle \frac{1}{2}(x-0)\\\\\ \ \ \ y=\displaystyle \frac{1}{2}x-4\\\\(\text{d})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=x-\displaystyle \frac{3}{2}\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=-1\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point }\left( {\text{0,}\displaystyle \frac{5}{4}} \right)\ \text{is}\\\\\ \ \ \ y-\displaystyle \frac{5}{4}=-1(x-0)\\\\\ \ \ \ y=-x+\displaystyle \frac{5}{4}\end{array}$

13.         Show that the line through $(3n,0)$ and $(0,7n)$ is parallel to the line through $(0,21n)$ and $(9n,0)$.

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$\begin{array}{l}\text{Let }{{l}_{1}}\ \text{contain }(3n,\ 0)\ \text{and}\ (0,\ 7n)\ \\\text{and}\ {{l}_{2}}\ \text{contain }(0,\ 21n)\ \text{and}\ (9n,\ 0).\\\\\therefore \ {{m}_{1}}=\displaystyle \frac{{7n-0}}{{0-3n}}=\displaystyle \frac{{7n}}{{-3n}}=-\displaystyle \frac{7}{3}\\\\\therefore \ {{m}_{2}}=\displaystyle \frac{{0-21n}}{{9n-0}}=\displaystyle \frac{{-21n}}{{9n}}=-\displaystyle \frac{7}{3}\\\\\therefore \ {{m}_{1}}={{m}_{2}}\\\\\therefore \ {{l}_{1}}\parallel {{l}_{2}}\end{array}$

14.         Prove that the triangle whose vertices are $H(-12,1)$, $K(9,3)$ and $M(11,-18)$ is a right triangle.

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$ \begin{array}{l}H=(-12,1),\ K=(9,3),\ M=(11,-18)\\\\{{m}_{{HK}}}=\displaystyle \frac{{3-1}}{{9-(-12)}}=\displaystyle \frac{2}{{21}}\\\\{{m}_{{KM}}}=\displaystyle \frac{{-18-3}}{{11-9}}=-\displaystyle \frac{{21}}{2}\\\\\therefore {{m}_{{HK}}}\cdot {{m}_{{KM}}}=\displaystyle \frac{2}{{21}}\left( {-\displaystyle \frac{{21}}{2}} \right)=-1\\\\\therefore \ HK\bot KM\\\\\therefore \ \triangle HKM\ \text{is a right triangle}\text{.}\end{array}$

15.        Given the points $P(1,2)$, $Q(5,-6)$ and $R(b,b)$, determine the value of $b$ so that angle $PQR$ is a right angle.

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$ \begin{array}{l}P=(1,2),\ Q=(5,-6),\ R=(b,b)\\\\{{m}_{{PQ}}}=\displaystyle \frac{{-6-2}}{{5-1}}=-2\\\\{{m}_{{QR}}}=\displaystyle \frac{{b-(-6)}}{{b-5}}=\displaystyle \frac{{b+6}}{{b-5}}\\\\\text{Since}\ \angle PQR\text{ is a right angle,}\\\\\therefore {{m}_{{PQ}}}\cdot {{m}_{{QR}}}=-1\\\\\therefore \ -2\left( {\displaystyle \frac{{b+6}}{{b-5}}} \right)=-1\\\\\therefore \ 2b+12=b-5\\\\\therefore b=17\end{array}$

16.        A right-angled isosceles triangle has vertices at $(0,5)$, $(5,0)$ and $(-5,0)$. Find the equation of each of the three sides.

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$\begin{array}{l}\ \ \ \ \text{Let}\ A=(0,\ 5),\ B=(-5,\ 0)\ \text{and}\ C=(5,\ 0).\\\\\therefore \ \ {{m}_{{AB}}}=\displaystyle \frac{{0-5}}{{-5-0}}\ =1\\\\\therefore \ \ \text{The equation of the line through }A\ \text{and}\ B\ \text{is}\\\ \ \text{ }\ y-5=1(x-0)\Rightarrow y=x+5\\\\\,\ \ \ \ {{m}_{{AC}}}=\displaystyle \frac{{0-5}}{{5-0}}\ =-1\\\\\therefore \ \ \text{The equation of the line through }A\ \text{and}\ C\ \text{is}\\\ \ \text{ }\ y-5=-1(x-0)\Rightarrow y=-x+5\\\\\ \ \ \ \text{Since }B\ \text{and}\ C\ \text{lie on }x\text{-axis, the equation }\\\ \ \ \ \text{of the line through }B\ \text{and}\ C\ \text{is}\ y=0.\end{array}$

17.         Determine the slope of each side of the quadrilateral whose vertices are $A(5,6)$, $B(13,6)$, $C(11,2)$ and $D(1,2)$. Can you tell what kind of a quadrilateral it is?

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$\begin{array}{l}A=(5,6),\ B=(13,6),\ C=(11,2),\ D=(1,2)\\\\\text{The slope of the line passing through }\\({{x}_{1}},\ {{y}_{1}})\ \text{and}\ ({{x}_{2}},\ {{y}_{2}})\ \text{is}\ m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}.\\\\\therefore \ \ {{m}_{{AB}}}=\displaystyle \frac{{6-6}}{{13-5}}=0.\\\\\,\ \ \ \ {{m}_{{BC}}}=\displaystyle \frac{{2-6}}{{11-13}}=2.\\\\\,\ \ \ \ {{m}_{{CD}}}=\displaystyle \frac{{2-2}}{{1-13}}=0.\\\\\,\ \ \ \ {{m}_{{AD}}}=\displaystyle \frac{{2-6}}{{1-5}}=1.\\\\\therefore \ \ {{m}_{{AB}}}={{m}_{{CD}}}\\\\\therefore \ \ AB\parallel CD\\\\\therefore \ \ ABCD\ \text{is a trapezium}\text{.}\end{array}$

18.         Given the points $D(-4,6)$, $E(1,1)$, and $F(4,6)$, find the slopes of $DE$ and $EF$. Are the points $D$, $E$ and $F$ collinear, explain why?

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$\begin{array}{l}D=(-4,6),\ E=(1,1),\ F=(4,6)\\\\\text{The slope of the line passing through }\\({{x}_{1}},\ {{y}_{1}})\ \text{and}\ ({{x}_{2}},\ {{y}_{2}})\ \text{is}\ m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}.\\\\\therefore \ \ {{m}_{{DE}}}=\displaystyle \frac{{1-6}}{{1-(-4)}}=-\displaystyle \frac{5}{5}=-1\\\\\,\ \ \ \ {{m}_{{EF}}}=\displaystyle \frac{{6-1}}{{4-1}}=\displaystyle \frac{5}{3}\\\\\therefore \ \ {{m}_{{DE}}}\ne {{m}_{{EF}}}\\\\\therefore \ \ D,\ E\ \text{and}\ F\ \text{are not collinear}\text{.}\end{array}$

19.        Prove that the quadrilateral with vertices $A(-2,2)$, $B(2,-2)$, $C(4,2)$ and $D(2,4)$ is a trapezoid with perpendicular diagonals.

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$ D=(-4,6),\ E=(1,1),\ F=(4,6)$

The slope of the line passing through $({{x}_{1}},\ {{y}_{1}})$ and $({{x}_{2}},\ {{y}_{2}})$ is $m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$.

$\begin{aligned} \therefore m_{AB}&=\displaystyle \frac{-2-2}{2+2} = -1\\\\ m_{BC}&=\displaystyle \frac{2+2}{4-2} =2\\\\ m_{CD}&=\displaystyle \frac{4-2}{2-4} = -1\\\\ m_{AD}&=\displaystyle \frac{4-2}{2+2} = \displaystyle \frac{1}{2}\\\\ m_{AC}&=\displaystyle \frac{2-2}{4+2} = 0\\\\ m_{BD}&=\displaystyle \frac{4+2}{2-2} =\text{undefined} \end{aligned}$

Since $m_{AB}=m_{CD},\ AB\parallel CD.$

$m_{AC} = 0\Rightarrow AC$ is a horizontal line.

$m_{BD} =\text{undefined}\Rightarrow BD$ is a vertical line.

$\therefore AC\bot BD$.

Hence, the quadrilateral $ABCD$ is a trapezoid with perpendicular diagonals.

20.         Find the slopes of the six lines determined by the points $A(-5,4)$, $B(3,5)$, $C(7,-2)$, $D(-1,-3)$. Prove that $ABCD$ is a rhombus.

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$A=(-5,4),\ B=(3,5),\ C=(7,-2),\ D=(-1,-3)$

The slope of the line passing through $({{x}_{1}},\ {{y}_{1}})$ and $({{x}_{2}},\ {{y}_{2}})$ is $m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$.

$\begin{aligned} m_{AB}&=\displaystyle \frac{5-4}{3+5}=\displaystyle \frac{1}{8}\\\\ m_{BC}&=\displaystyle \frac{-2-5}{7-3} =-\displaystyle \frac{7}{4} \\\\ m_{CD}&=\displaystyle \frac{-3+2}{-1-7} = \displaystyle \frac{1}{8}\\\\ m_{AD}&=\displaystyle \frac{-3-4}{-1+5} = -\displaystyle \frac{7}{4}\\\\ m_{AC}&=\displaystyle \frac{-2-4}{7+5} = -\displaystyle \frac{1}{2}\\\\\ m_{BD}&=\displaystyle \frac{-3-5}{-1-3}=2 \end{aligned}$

$\therefore\ m_{AB}=m_{CD}\Rightarrow AB\parallel CD$

$\quad\ m_{BC}=m_{AD}\Rightarrow BC\parallel AD$

$\quad\ m_{AC}\cdot m_{BD}=-\displaystyle \frac{1}{2}\times 2 = -1\Rightarrow AC\bot BD$

Hence $ABCD$ is a rhombus.