Grade 10 - Chapter 1 - MCQ Questions

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Coordinate Geometry Quick Revision Note

Grade 10 သင်ရိုးသစ် Chapter (1) အတွက် Quick Revision Note ဖြစ်ပါတယ်...။ လိုအပ်သူတို့ အတွက် အဆင်ပြေပါစေကြောင်း ...

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.

Show/Hide Solution

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

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

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

$\therefore\ y= -5$.

Coordinate Geometry : Exercise (1.3) - Solutions

1.         Sketch the following lines.

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

Show/Hide Solution

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

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

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

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

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

2.         Graph each line in the same coordinate plane.

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

Show/Hide Solution

$\bullet\quad y=x$ 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.

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

Show/Hide Solution

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

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

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

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

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

Show/Hide Solution

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

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

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

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

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

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

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

Show/Hide Solution

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

$\therefore$ 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 $(9,4)$.

Show/Hide Solution

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.

Show/Hide Solution

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

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 $(-1,4)$.

Show/Hide Solution

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

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

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

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

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 $(-4, -2)$

Show/Hide Solution

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

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

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.