Introduction to Coordinate Geometry

Chapter 1

Introduction to Coordinate Geometry

Rene` Descarts' greatest contribution to mathematics was the discovery of coordinate systems and their applications to Problems of geometry. Coordinate system used in this book is referred to as Cartesian coordinate system. We have seen how coordinate systems work on a line. Coordinate system used in this book is referred to as Cartasian coordinate system.
သင်္ချာပညာရပ်ကိုအများဆုံးအထောက်အကူပေးခဲ့သူသင်္ချာပညာရှင် Rene` Descarts' သည် Coordinate System နှင့် Geometry ဆိုင်ရာပြဿနာများတွင်၎င်းတို့၏အသုံးပြုပုံများကိုရှာဖွေတွေ့ရှိခဲ့သူဖြစ်သည်။ ပြဌာန်းစာအုပ်တွင်အသုံးပြုသော Coordinate System ကို Cartasian Coordinate System အဖြစ်ရည်ညွှန်းဖော်ပြထားသည်။



Once we have set up a coordinate system on a line, every number corresponds to a point on the line and every point on the line corresponds to a number. We shall now extend this idea to the points in a plane, a point will correspond not to a single number, but to an ordered pair of numbers. The scheme works like this.
First we take two perpendicular lines, a horizontal line $X^{'}OX$ (the $X$- axis) and a vertical line $Y^{'}OY$ (the $Y$-axis), they intersect at zero point in the $XY$-plane. The zero point which is the intersection of these two lines is called the origin, normally labelled $O$. On the $X$-axis, values to the right are positive and those to the left are negative. On the $Y$-axis, values above the origin are positive and those below are negative.
ကိန်းစစ်တစ်ခုစီကိုအထက်ပါမျဉ်းပေါ်ရှိအမှတ်တစ်ခုစီဖြင့်သတ်မှတ်ပေးနိုင်ပြီးယင်းမျဉ်းပေါ်ရှိအမှတ်တစ်ခုစီကိုကိန်းစစ်တစ်ခုစီဖြင့်သတ်မှတ်ပေးနိုင်၏။ယင်းကဲ့သို့သတ်မှတ်ခြင်းကို coordinate system on a line ဟုခေါ်၏။ ၎င်းမျဉ်းကိုကိန်းစစ်မျဉ်းဟုခေါ်၏။အထက်ပါပုံတွင် $P$ အမှတ်နှင့်လိုက်ဖက်သည့်ကိန်းစစ်xကို $P$ အမှတ်၏ coordinate ဟုခေါ်၏။
ပြင်ညီတစ်ခုအတွင်းရှိအမှတ်တစ်ခုကိုကိန်းစစ်တစ်လုံးတည်းဖြင့်ဖော်ပြ၍မရပါ။ကိန်းစုံတွဲ ordered pair$(x , y)$တစ်ခုဖြင့်ဖော်ပြရသည်။
$XY$ ပြင်ညီတစ်ခုပေါ်တွင်သုညအမှတ်ကိုထောင့်မတ်ကျသောမျဉ်းဖြောင့်နှစ်ကြောင်းရေပြင်ညီမျဉ်း(horizontal line) $X$ ဝင်ရိုး ($X$- axis) $X^{'}OX$ နှင့်မတ်ရပ်မျဉ်း vertical line $Y$ ဝင်ရိုး ($Y$-axis) $Y^{'}OY$ ဆွဲလျှင်ထိုမျဉ်းနှစ်ကြောင်းဖြတ်သောသုညအမှတ်ကို origin (မူလမှတ်)ဟုခေါပြီး $O$ ဟုသတ်မှတ်သည်။
$X$ ဝင်ရိုးပေါ်တွင် $O$ ၏ညာဘက်တန်ဖိုးသည်အပေါင်း (positive)၊ဘယ်ဘက်တန်ဖိုးသည်အနုတ် (negative) ဖြစ်သည်။ $Y$ ဝင်ရိုးပေါ်တွင် $O$ ၏အပေါ်ဘက်တန်ဖိုးသည်အပေါင်း (positive)၊အောက်ဖက်တန်ဖိုးသည်အနုတ် (negative) ဖြစ်သည်။



We can now describe any point $P$ of the plane by an ordered pair of numbers, as follows. Draw $PM$ and $PN$, perpendicular to the $X$-axis and $Y$-axis. Let x be the coordinate of $M$ on the line $X^{'}OX$ and y be the coordinate of $N$ on the line $Y^{'}OY$.
ပြင်ညီတွင်းရှိမည်သည့်အမှတ် $P$ ကိုမဆိုကိန်းစုံတွဲတစ်ခုဖြင့်အောက်ပါအတိုင်းဖော်ပြနိုင်သည်။$PM$⊥$X^{'}OX$ ကိုဆွဲပါ။$x$ သည် $X^{'}OX$ ပေါ်ရှိ $M$ ၏ကိုဩဒိနိတ်ဖြစ်ပါစေ။$PN$⊥$Y^{'}OY$ ကိုဆွဲပါ။ $y$ သည် $Y^{'}OY$ ပေါ်ရှိ $M$ ၏ကိုဩဒိနိတ်ဖြစ်ပါစေ။



The number $x$ and $y$ are called $x$-coordinate and $y$-coordinate of $P$ respectively. In short, we indicate that $P$ has these coordinates by writing $P$(x,y). In particular, the origin $O$ has coordinates (0,0).
$x$ ကို $P$ ၏ $x$-coordinate ဟုခေါပြီး $y$ ကို $P$ ၏ $y$-coordinate ဟုခေါ်သည်။ $P(x, y)$ ဟုအတိုကောက်ရေးသည်။မူလမှတ် $O$ ၏ကိုဩဒိနိတ်မှာ $(0, 0)$ ဖြစ်သည်။
Just as single line separates the plane into two parts (each of which is a half-plane), so the two axes separate the $XY$-plane into four parts, called quadrants. The four quadrants are identified by the Roman numerals, I, II, III, IV.
မျဉ်းဖြောင့်တစ်ကြောင်းသည်ပြင်ညီတစ်ခုကိုနှစ်ပိုင်းပိုင်း၏။တစ်ပိုင်းစီကို 'a half-plane' ဟုခေါ်၏။ $X$ ဝင်ရိုးနှင့် $Y$ ဝင်ရိုးတို့သည်ပြင်ညီကိုလေးပိုင်းပိုင်း၍လေးပိုင်းစိတ်ဟုခေါ်သည်။ရောမဂဏန်း I, II, III, IV ဖြင့်သတ်မှတ်ဖော်ပြသည်။



We have shown that under the scheme, every point $P$ determines an ordered pair of real numbers. Does it work in reverse? That is, does every ordered pair (a,b) of real numbers determine a point? It is easy to see that the answer is "Yes".



Mark a point on the $X$-axis, so that the $x$-coordinate of that point is $a$. Draw a perpendicular line passes through that point. Then draw another perpendicular line at the point in which the $y$-coordinate is $b$. The point where these perpendiculars intersect is the point with the $y$-coordinates (a,b).
Thus we have a one-to-one correspondence between the points of the plane and the ordered pairs of real numbers. Such a correspondence is called a rectangular coordinate system.
$X$ ဝင်ရိုးပေါ်တွင်အမှတ်တစ်ခုကိုမှတ်သားပါ။ထိုအမှတ်၏ $x$ ကိုဩဒိနိတ်သည် $a$ ဖြစ်ပါစေ။ $a$ ကိုဖြတ်၍ထောင့်မတ်မျဉ်းတစ်ကြောင်းဆွဲပါ။ $y$ ကိုဩဒိနိတ် $b$ ဖြစ်သောအမှတ်တွင်ထောင့်မတ်မျဉ်းဆွဲပါ။ထောင့်မှတ်မျဉ်းနှစ်ကြောင်းဖြတ်သောအမှတ်သည်ကိုဩဒိနိတ် $(a, b)$ ရှိသောအမှတ်ဖြစ်သည်။
ထို့ကြောင့်ပြင်ညီ၏အမှတ်များအကြား one-one ဆက်သွယ်ချက်နှင့်ကိန်းစုံတွဲကိုရရှိသည်။ထိုကဲ့သို့သောဆက်သွယ်ချက်ကို rectangular coordinate system ဟုခေါ်သည်။

Midpoint and Length of a Line Segment

Every line segment has a midpoint. Both the midpoint and length of a line segment can be found by using the coordinates of the endpoints.
မျဉ်းပြတ်တိုင်းတွင်အလယ်မှတ်တစ်ခုရှိသည်။အဆုံးမှတ်များ၏ coordinate များကိုသုံး၍အလယ်မှတ်နှင့်မျဉ်းပြတ်အလျားတို့ကိုရှာနိုင်သည်။

1.1 Midpoint of a Line Segment in $XY$-plane

The midpoint $M$ of a line segment is the halfway point between the two endpoints. မျဉ်းပြတ်တစ်ခု၏အလယ်မှတ် $M$ ဆိုသည်မှာထိုမျဉ်းပြတ်၏အဆုံးမှတ်နှစ်ခုကြားတစ်ဝက်အကွာအဝေး၌ရှိသောအမှတ်ဖြစ်သည်။ $XY$ ပြင်ညီပေါ်တွင် ပေးရင်းအမှတ်နှစ်ခုကိုဆက်သွယ်ထားသောမျဉ်းပြတ်တစ်ကြောင်း(ရေပြင်ညီမျဉ်းမဟုတ်သော၊မတ်ရပ်မျဉ်းမဟုတ်သော)၏ အလယ်မှတ် coordinates များကိုရှာဖွေကြမည်။ Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$ be two given points. Let $M(x,y)$ be the midpoint of $AB$. Draw the perpendicular lines $AP$, $MQ$, $BR$ to the $X$-axis, and the perpendicular lines $AU$, $MT$, $BS$ to the $Y$-axis. It is clear that $AU$, $MT$, $BS$ are horizontal lines and $AP$, $MQ$, $BR$ are vertical lines. The point $M$ and $Q$ have the same x-coordinates and $M$ and $T$ have the same y-coordinates. $\begin{aligned} & Q \text{ is the midpoint of } PR,\\ &\text{ the } x-\text{ coordinate of } Q \text{ is } \dfrac{x_{1}+x_{2}}{2}.\\ & T \text{ is the midpoint of } SU,\\ & \text{ the } y-\text{ coordinate of } T \text{ is } \dfrac{y_{1}+y_{2}}{2}.\\ &\therefore x-\text{ coordinate of midpoint } M,\\ & x= \dfrac{x_{1}+x_{2}}{2}, \text{ and } \\ & y-\text{ coordinate of midpoint } M,\\ & y= \dfrac{y_{1}+y_{2}}{2}. \end{aligned}$ Midpoint Formula The coordinates of midpoint of $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ are $\begin{aligned} M(x,y)=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \end{aligned}$

The Length of a line Segment in $XY$-plane

Let $AB$ be a non-horizontal, non-vertical line segment with $A(x_{1},y_{1})$ and $(x_{2},y_{2})$. $\begin{aligned} \text{ Horizontal distance } AC &= x_{2}-x_{1}\\\\ \text{ Vertical distance } BC&= y_{2}-y_{1}\\\\ \end{aligned}$ In right $\triangle ACB$, By Pythagoras Theorem, $\begin{aligned} AB^{2} &= AC^{2}+BC^{2}\\\\ AB &= \sqrt {AC^{2}+BC^{2}}\\\\ AB &= \sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \end{aligned}$ Therefore the length of the line segment $AB$ is $\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$. Distance Formula The distance beteween two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

1.2 Slope of a Straight Line

The slope of the line passihg through any two points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is a measure of the steepness of the line $AB$. Simply it is the ratio of the vertical change (rise) divided by the horizontal change (run) . To determine rise and run, select any two points on the line. The horizontal destance between the two points is called the run and the vertical distance is called the rise. $\begin{aligned} \text{ slope m }& = \dfrac{\text{ vertical change }}{\text{ horizontal change }}\\\\ &= \dfrac{\text{ rise }}{\text{ run }}\\\\ &= \dfrac{BC}{AC}\\\\ \text{ slope m }& = \left|\dfrac{(y_{2}-y_{1})}{(x_{2}-x_{1})} \text{ (or) } \dfrac{(y_{1}-y_{2})}{(x_{1}-x_{2})}\right| \end{aligned}$ Slope Formula The slope of the line passing through any two points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is Slope m = $\dfrac{(y_{2}-y_{1})}{(x_{2}-x_{1})}$

To find the slope of a slanting line, select any two points on a line to determine the rise and run.
Create a right-angled triangle to determine rise and run.

(i)Positive Slope


$\begin{aligned} \text{ increase in } x, \text{ run } &= 3\\ \text{ increase in } y, \text{ rise } &= 3\\ \text{ slope } = \dfrac{ \text{ rise }}{\text{ run }} =\dfrac{3}{3} &= 1 \end{aligned}$

(ii)Negative Slope


$\begin{aligned} \text{ increase in } x, \text{ run } &= 3\\ \text{ decrease in } y, \text{ rise } &= -2\\ \text{ slope } = \dfrac{ \text{ rise }}{\text{ run }} =\dfrac{-2}{3} &= -\dfrac{2}{3} \end{aligned}$

(iii)Zero Slope


$\begin{aligned} \text{ increase in } x, \text{ run } &= 3\\ \text{ decrease in } y, \text{ rise } &= 0\\ \text{ slope } = \dfrac{ \text{ rise }}{\text{ run }} =\dfrac{3-3}{4-1} &= -\dfrac{0}{3}=0 \end{aligned}$
Note that the slope of the horizonal line is zero.

(iii)Undefined Slope


$\begin{aligned} \text{ increase in } x, \text{ run } &= 0\\ \text{ decrease in } y, \text{ rise } &= 3\\ \text{ slope } = \dfrac{ \text{ rise }}{\text{ run }} =\dfrac{4-1}{3-3} &= -\dfrac{3}{0}\quad(\text{ undefined }) \end{aligned}$
Note that the vertical line has undefined slope.

"The graph of a line " ကိုကြည့်၍ slope သည် positive ဖြစ်နိုင်သည်၊ negative ဖြစ်နိုင်သည်၊ zero ဖြစ်နိုင်သည်၊ undefined ဖြစ်နိုင်သည်စသည်ဖြင့် slope ၏ဖြစ်နိုင်မည့်ပုံစံများကိုတွက်ချက်ခြင်းမပြုပဲ အောက်ပါဇယားအတိုင်းလွယ်ကူစွာသိနိုင်သည်။

1.3 Lines in the Coordinate Plane

Graph of a Linear Equation Linear equation ဆိုသည်မှာအက္ခရာသင်္ချာဆိုင်ရာညီမျှခြင်းတစ်ခုဖြစ်ပြီး၎င်းညီမျှခြင်းတွင်ပါဝင်သောကိန်းရှင် (variable) တို့၏ထပ်ညွှန်း $1$ ဖြစ်သောညီမျှခြင်းဖြစ်သည်။ Linear equation y=mx+c ၏ graph ကိုဆွဲရန် $x$ နှင့် $y$ တန်ဖိုးများရှာရန် table တည်ဆောက်ပါ။ Linear equation တွင်ပေးထားသောကြားပိုင်းအတွင်းမှ $x$ တန်ဖိုးနှစ်ခု(သို့မဟုတ်)နှစ်ခုထက်ပို၍ရွေးချယ်ပြီးအစားသွင်းခြင်းဖြင့် $y$ တန်ဖိုးများရရှိမည်။ ကြားပိုင်းတန်ဖိုးသတ်မှတ်မထားပါကကြိုက်နှစ်သက်ရာ $x$ တန်ဖိုးကိုရွေးချယ်အစားသွင်း၍ $y$ တန်ဖိုးရှာနိုင်သည်။ ရရှိလာသော $x$ နှင့် $y$ ကိန်းစုံတွဲများ (coordinates) ကိုပြင်ညီ ($XY$-plane) ပေါ်တွင်နေရာချပေးခြင်းဖြင့်ပေးထားသောညီမျှခြင်းကိုပြေလည်စေမည့် graph ကိုရရှိမည်။ Linear equation ၏ graph များသည်မျဉ်းဖြောင့်များ (straight line) ဖြစ်သည်ကိုတွေ့ရမည်။ Consider the graph of y = 2x+1. $\begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -3 & -1 & 1 & 3 & 5 \\ \hline \end{array}$ Let $(x_{1},y_{1})=(1,3)$ and $(x_{2},y_{2})=(2,5)$, then $\begin{aligned} \text{ slope } m &= \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ &=\dfrac{5-3}{2-1}\\ &=\dfrac{2}{1}\\ &=2 \end{aligned}$ Let $(x_{1},y_{1})=(-2,-3)$ and $(x_{2},y_{2})=(1,3)$, then $\begin{aligned} \text{ slope } m &= \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ &=\dfrac{3-(-3)}{1-(-2)}\\ &=\dfrac{6}{3}\\ &=2 \end{aligned}$ Table မှမည်သည့် coordinates နှစ်စုံကိုမဆိုရွေးချယ်၍ slope တန်ဖိုးရှာပါက slope တန်ဖိုးများတူညီနေကြောင်းတွေ့ရမည်။ Slope တန်ဖိုးသည်ပေးထားသောမျဉ်းဖြောင့်ညီမျှခြင်းမှx၏မြှောက်ဖော်ကိန်းနှင့်တူညီနေကြောင်းလည်းတွေ့ရသည်။ Graph line ဖြတ်သွားသော $y$ဝင်ရိုးပေါ်ရှိအမှတ် (y-intercept) သည်ပေးထားသောမျဉ်းဖြောင့်ညီမျှခြင်းတွင်ပါဝင်သောကိန်းသေ(constant)တန်ဖိုးပင်ဖြစ်သည်။

Slope-Intercept Form

The equation of the form y = mx + c is the equation of a straight line with slope m and y-intercept c, which is called the slope-intercept form. $y$ ဖြတ်မှတ်နှင့် slope တန်ဖိုးပေးထားလျှင် "The equation of the line" ကိုရှာနိုင်သည်။

Point-Slope Form

အမှတ်တစ်ခုနှင့် slope တန်ဖိုးပေးထားလျှင် "The equation of the line" ကိုရှာနိုင်သည်။ Consider a line with slope m, and passes through the known point $A(x_{1},y_{1})$. Let $P(x,y)$ be any point on the line, then m = $\dfrac{(y-y_{1})}{(x-x_{1})}$ $(y-y_{1}) = m (x-x_{1})$ This equation is called the point-slope form.

Collinear

On a line, all segments have the same slope. All the points on a line are said to be collinear.
မျဉ်းဖြောင့်တစ်ကြောင်းပေါ်ရှိမျဉ်းပြတ်များအားလုံးတွင်တူညီသော slope များရှိကြသည်။ထိုမျဉ်းပေါ်ရှိအမှတ်များအားလုံးကိုတစ်မျဉ်းမှတ်များဟုခေါ်သည်။

မျဉ်းဖြောင့်တစ်ကြောင်းပေါ်ရှိမျဉ်းပြတ်များအားလုံး၏ slopeအားလုံးတူညီကြသည်။
$m_{AB} = m_{BC} = m_{CD} = m_{AD}$

A, B , C are collinear ⇔ slope of AB = slope of BC.

Parallel and Perpendicular Lines

Any two horizontal lines are parallel. Any two vertical lines are parallel. Vertical line and horizontal line are perpendicular. Two non-vertical lines are parallel if and only if they have the same slope. Two non-vertical lines are perpendicular if and only if the product of their slopes is -1 (i.e, one is the negative reciprocal of the other). မည်သည့်ရေညီမျဉ်းနှစ်ကြောင်းမဆိုပြိုင်ကြသည်။ မည်သည့်မတ်ရပ်မျဉ်းနှစ်ကြောင်းမဆိုပြိုင်ကြသည်။ မည်သည့်မတ်ရပ်မျဉ်းနှင့်ရေညီမျဉ်းမဆိုတစ်ခုနှင့်တစ်ခုထောင့်မတ်ကျသည်။ vertical line မဟုတ်သောမျဉ်းနှစ်ကြောင်းပြိုင်လျှင်၎င်းတို့၏slope များတူကြသည်။အပြန်အလှန်အားဖြင့် slope တူသောမျဉ်းနှစ်ကြောင်းပြိုင်ကြသည်။ vertical line မဟုတ်သောမျဉ်းနှစ်ကြောင်းထောင့်မတ်ကျလျှင်ထိုမျဉ်းတို့၏ slope မြှောက်လဒ် $-1$ ဖြစ်သည်။အပြန်အလှန်အားဖြင့်မျဉ်းနှစ်ကြောင်းတို့၏ slope မြှောက်လဒ် $-1$ ဖြစ်လျှင် ထိုမျဉ်းနှစ်ကြောင်းတို့တစ်ခုကိုတစ်ခုထောင့်မတ်ကျသည်။ In the figure, lines $l_{1}$ ∥ $l_{2}$, $l_{1}$ ⊥ $l_{3}$ and $l_{2}$ ⊥ $l_{3}$ $\begin{aligned} \text{ slope of line } l_{1} &= m_{1} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ &= \dfrac{3-1}{0-(-1)}\\ &= \dfrac{2}{1}\\ &=2 \end{aligned}$ $\begin{aligned} \text{ slope of line } l_{2} &= m_{2} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ &= \dfrac{4-2}{3-2}\\ &= \dfrac{2}{1}\\ &=2 \end{aligned}$ $\begin{aligned} \text{ slope of line } l_{3} &= m_{3} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ &= \dfrac{-1-2}{3-(-3)}\\ &= -\dfrac{-3}{6}\\ &= -\dfrac{1}{2}\\ \end{aligned}$ ∴ $m_{1} = m_{2} = 2$, $m_{1} × m_{3} = 2 × -\dfrac{1}{2} = -1$ $m_{2} × m_{3} = 2 × -\dfrac{1}{2} = -1$ $l_{1} ∥ l_{2} ⇔ m_{1} = m_{2}$ $l_{1} ⊥ l_{3} ⇔ m_{1} × m_{3} = -1$        ($m_{1} = -\dfrac{1}{m_{3}}$) $l_{2} ⊥ l_{3} ⇔ m_{2} × m_{3} = -1$        ($m_{2} = -\dfrac{1}{m_{3}}$)

Problems

Example 1

Find the coordinates of the midpoint of $PQ$ with endpoints $P(2,5)$ and $Q(6,1)$.

$\begin{aligned} \text{ Let } (2,5)&=\left(x_{1}, y_{1}\right) \text{ and } (6,1)=\left(x_{2}, y_{2}\right)\\\\ \end{aligned}$ Let $M$ be the midpoint of PQ. $\begin{aligned} M &=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \\\\ &=\left(\dfrac{2+6}{2}, \dfrac{5+1}{2}\right) \\\\ &=(4,3) \end{aligned}$

Example 2

Find the length of $PQ$ with endpoints $P(1,5)$ and $Q(4,1)$.

$\begin{aligned} \text { Let } (1,5) &=\left(x_{1}, y_{1}\right) \text{ and } (4,1) =\left(x_{2}, y_{2}\right) \\\\ \text { Length of } P Q &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(4-1)^{2}+(1-5)^{2}} \\\\ &=\sqrt{3^{2}+(-4)^{2}} \\\\ &=\sqrt{25}\\\\ &=5 \end{aligned}$

Example 3

$A$ is the point ($5,-3$) and $B$ is the point ($-2,1$).
(a) Find the midpoint of $AB$.
(b) Find the length of $AB$.

(a) Let $(5,-3)=\left(x_{1}, y_{1}\right),(-2,1)=\left(x_{2}, y_{2}\right)$ Let $M$ be the midpoint of $A B$. $\begin{aligned} M &=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \\\\ M &=\left(\dfrac{5+(-2)}{2}, \dfrac{-3+1}{2}\right) \\\\ &=\left(\dfrac{3}{2},-1\right) \end{aligned}$ $\begin{aligned} (b) \text{ Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(-2-5)^{2}+(1-(-3))^{2}} \\\\ &=\sqrt{49+16} \\\\ &=\sqrt{65} \end{aligned}$

Example 4

The point $M$($a,4$) is the midpoint of the line segment with endpoints at $A$($1,3$) and $B$($5,b$). Find the value of a and b.

$M$($a, 4$) is the midpoint of $A$($1,3$) and $B$($5, b$). Let $\left(x_{1}, y_{1}\right)=(1,3),\left(x_{2}, y_{2}\right)=(5, b)$ $\begin{aligned} \text{ Midpoint of } A B &=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ (a, 4)&=\left(\dfrac{1+5}{2}, \dfrac{3+b}{2}\right) \\\\ \therefore \quad a&=\dfrac{1+5}{2},\quad 4=\dfrac{3+b}{2} \\\\ 2a&=6,\quad\quad 8=3+b\\\\ a&=3,\quad b=5 \end{aligned}$

Example 5

The distance between two points $R$($9,a$) and $S$($a+1,2$) is $6$. Find the two possible values of a.

The distance between $R$ ($9, a$) and $S$ ($a+1, 2$) is $6$. Let $\left(x_{1}, y_{1}\right)=(a, a)$ and $\left(x_{2}, y_{2}\right)=(a+1, 2)$ Distance between $R$ and $S$ = $6$ $\sqrt{(a+1-9)^{2}+(2-a)^{2}}=6$ squaring both sides, $\begin{aligned} (a-8)^{2}+(2-a)^{2} &=36 \\\\ a^{2}-16 a+64+4-4 a+a^{2} &=36 \\\\ 2 a^{2}-20 a+32 &=0 \\\\ a^{2}-10 a+16 &=0 \\\\ (a-8)(a-2) &=0 \\\\ a-8=0 \text { (or) } a-2 &=0 \\\\ a=8 \text { (or) } a &=2 \end{aligned}$

Example 6

In a parallelogram $A B C D$, three vertices are $A$ ($-3, 1$), $B$ ($2, 4$) and $C$ ($3, 1$).
(a) Find the midpoint of the diagonal $A C$.
(b) Find the coordinates of $D$.

$\begin{aligned} (a)\quad \text{ Midpoint of } A C &= \left(\dfrac{-3+3}{2}, \dfrac{1+1}{2}\right)\\\\ &=(0,1)\\\\ (b)\quad \text{ Let the coordinates of }& D \text{ be } (x, y).\\\\ \text{ Midpoint of } A C&= \text{ Midpoint of } B D \quad ( \because A B C D \text{ is a parallelogram })\\\\ (0,1)&=\left(\dfrac{2+x}{2}, \dfrac{4+y}{2}\right) \\\\ \dfrac{2+x}{2}&=0 \text { and } \dfrac{4+y}{2}=1 \\\\ 2+x &= 0 \text { and } 4+y =2\\\\ x&=-2 \text { and } y=-2 \\\\ \therefore D(x, y)&=(-2,-2) \end{aligned}$

Example 7

Find the slope of the line $A B$ that passes through the points $A$ ($4, -2$) and $B$ ($-1, 2$).

Let $(4,-2)=\left(x_{1}, y_{1}\right),(-1,2)=\left(x_{2}, y_{2}\right)$. $\begin{aligned} \text{ slope } m &= \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &=\dfrac{2-(-2)}{-1-4} \\\\ &=-\dfrac{4}{5} \end{aligned}$ The slope of the line $A B=-\dfrac{4}{5}$.

Example 8

Find the slope of the given line.

Let $(3,3)=\left(x_{1}, y_{1}\right)$ and $(-2,-1)=\left(x_{2}, y_{2}\right)$ $\begin{aligned} \text { slope } m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{-1-3}{-2-3} \\\\ &=\dfrac{-4}{-5} \\\\ &=\dfrac{4}{5} \end{aligned}$

Example 9

Find the slope of the line that passes through the following points.

$\begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline y & 5 & 5 & 5 & 5 \\ \hline \end{array}$

Let us choose any two points on the line from the table, let $(0,5)=\left(x_{1}, y_{1}\right)$ and $(2,5)=\left(x_{2}, y_{2}\right)$. $\begin{aligned} \text { slope } m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{5-5}{2-0} \\\\ &=\dfrac{0}{2} \\\\ &=0 \end{aligned}$

Example 10

Find the y-intercept and the slope of each line.
(a) y-3x+4 = 0
(b) y+5x = 1
(c) x+y = 8

(a) $y-3 x+4=0$ $y=3 x-4$ Compare this equation with $y=m x+c$. slope $m=3$ y-intercept $c=-4$ (b) $y+5 x=1$ $y=-5 x+1$ Compare this equation with $y=m x+c$. $\therefore$ slope $m=-5$ $y$-intercept $c=1$ (c) $x+y=8$ $y=-x+8$ Compare this equation with $y=m x+c$. $\therefore$ slope $m=-1$ $y$-intercept $c=8$ $\begin{array}{|c|c|c|c|} \hline \text{ equation } &y = m x + c & \text{ slope } & y-\text{ intercept } \\ \hline y - 3x + 4 = 0 & y = 3 x - 4 & 3 & -4 \\ \hline y + 5x = 1 & y = -5 x + 1 & -5 & 1 \\ \hline x + y = 8 & y = - x + 8 & -1 & 8 \\ \hline \end{array}$

Example 11

Find the equation of the line with slope $-5$ and passes through the point ($2, 3$). Find the equation of the line.

slope $m=-5$ Let point $=(x, y)=(2,3)$ Using slope-intercept form, $y=-5 x+c$ since the point $(2,3)$ is on the line, $\begin{aligned} &3=-5(2)+c \\\\ &c=13 \end{aligned}$ Thus the equation of the line is $y=-5 x+13$ Alternative method v slope $m=-5$ Let point $=(2,3)=\left(x_{1}, y_{1}\right)$ Using point_slope form, $\begin{gathered} y-y_{1}=m(x-x, 1) \\\\ y-3=-5(x-2) \\\\ y-3=-5 x+10 \end{gathered}$ Thus the equation of the line is $y=-5 x+13$

Example 12

Find the equation of the line passing through the points $A$ ($1, 3$) and $B$ ($4, 9$).Find also the y-intercept.

$A(1,3), B(4,9)$ Let $\left(x_{1}, y_{1}\right)$ be $(1,3)$ and $\left(x_{2}, y_{2}\right)$ be $(4,9)$. $\begin{aligned} \text { slope }m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{9-3}{4-1} \\\\ &=2 \end{aligned}$ Using slope-intercept form $y=m x+c$, we get $y=2 x+c$ Since the line passing through the points $(1,3)$ and $(4,9)$, we can use one of these two poirs of coordinates, $3 = 2(1))+ c$ we get $c=1$ Thus the equation of the line is, $y=2 x+1$ and the $y$-intercept is 1 .

Example 13

The coordinates of the points $A$ and $B$ are ($2, 5$) and ($-1, 5$) respectively. Find the equation of the line $A B$.

$A(2,5), B(-1,5)$ Since $y$ coordinates of the two points are same, the slope $m=0$. It is a horizontal line. Let $(2,5)$ be $\left(x_{1}, y_{1}\right)$. Using point-slope form, $\begin{aligned} y-y_{1} &=m\left(x-x_{1}\right) \\\\ y-5 &=0(x-2) \\\\ y &=5 \end{aligned}$ Thus the equation of the line is,$y=5$.

Exercise 1.1

1. Draw a set of coordinate axes. Locate the points, $A$ ($2, 3$), $B$($2, -4$) and $C$ ($-4, 3$). Lable each point with its coordinates. Determine whether each of the line segments $A B$, $B C$, $C A$ is horizontal or vertical.



segment $A B$ is a vertical line. segment $B C$ is a non-horizontal and non-vertical line. Segment CA is a horizontal line.



2. Find the missing coordinates in the following table if $M$ is the midpoint of points $P$ and $Q$.

$\begin{array}{|c|c|c|} \hline \text{ P } &\text{ Q } & \text{ R } \\ \hline (2, 6) & & (3, 3) \\ \hline (3, 2) & (-3, -1) & \\ \hline & (0, -1) & (-3, 2) \\ \hline (1, 5) & & (2.5, 3.5) \\ \hline \end{array}$



(i) Let $P=\left(x_{1}, y_{1}\right)=(2,6), Q=\left(x_{2}, y_{2}\right)$ and $M=(x, y)=(3,3)$. $M$ is the midpoint of $P Q$. $\begin{aligned} M=(x, y) &=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \\\\ (3,3) &=\left(\dfrac{2+x_{2}}{2}, \dfrac{6+y_{2}}{2}\right) \\\\ \dfrac{2+x_{2}}{2} &=3 \text { and } \dfrac{6+y_{2}}{2}=3 \\\\ 2+x_{2} &=6 \text { and } 6+y_{2}=6 \\\\ x_{2} &=4 \text { and } y=0 \\\\ \therefore Q &=(x, y)=(4,0) \end{aligned}$ (ii) Let $P=\left(x_{1}, y_{1}\right)=(3,2), Q=\left(x_{2}, y_{2}\right)=(-3,-1)$, and $M=(a, b)$. $M$ is the midpoint of $P$ and $Q$. $\begin{aligned} M=(a, b) &=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \\\\ &=\left(\dfrac{3+(-3)}{2}, \dfrac{2+(-1)}{2}\right) \\\\ &=\left(0, \dfrac{1}{2}\right) \\\\ \therefore M=(a, b) &=\left(0, \dfrac{1}{2}\right) \end{aligned}$ (iii) Let $P=\left(x_{1}, y_{1}\right), Q=\left(x_{2}, y_{2}\right)=(0,-1)$ and $M=(p, q)=(-3,2)$. $M$ is the midpoint of $P$ and $Q$. $M=(p, q)=\left(\dfrac{x_{1}+0}{2}, \dfrac{y_{1}+(-1)}{2}\right)$ $\dfrac{x_{1}}{2}=-3$ and $\dfrac{y_{1}-1}{2}=2$ $x_{1}=-6$ and $y_{1}-1=4$ $x_{1}=-6$ and $y_{1}=5$ $\therefore P=\left(x_{1}, y_{1}\right)=(-6,5)$ (iv) Let $P=\left(x_{1}, y_{1}\right)=(1,5), Q=\left(x_{2}, y_{2}\right)$ and $M=(x, y)=(2.5,3.5)$ $M$ is the midpoint of $P$ and $Q$. $\begin{gathered} M=(x, y)=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \\\\ (2.5,3.5)=\left(\dfrac{1+x_{2}}{2}, \dfrac{5+y_{2}}{2}\right) \\\\ \dfrac{1+x_{2}}{2}=2.5 \text { and } \dfrac{5+y_{2}}{2}=3.5 \\\\ 1+x_{2}=5 \quad \text { and } \quad 5+y_{2}=7 \\\\ x_{2}=4 \quad \text { and } \quad y_{2}=2 \\\\ \therefore Q=\left(x_{2}, y_{2}\right)=(4,2) \end{gathered}$ $\begin{array}{|c|c|c|} \hline \text{ P } &\text{ Q } & \text{ R } \\ \hline (2, 6) & (4, 0) & (3, 3) \\ \hline (3, 2) & (-3, -1) & (0, 0.5) \\ \hline (-6, 5) & (0, -1) & (-3, 2) \\ \hline (1, 5) & (4, 2) & (2.5, 3.5) \\ \hline \end{array}$



3. Find the coordinates of the midpoint and the length of the line segment joining these pairs of points.



(a) $(0 , 0)$ and $(4, -4)$	(b) $(1 , 5)$ and $(3, 1)$	(c) $(-3, -3)$ and $(0, 0)$
(d) $(-1, 3)$ and $(5, 1)$	(e) $(-1, 6)$ and $(2, -2)$	(f) $(-3, -4)$ and $(3, -1)$




( a ) $A=\left(x_{1}, y_{1}\right)=(0,0), B=\left(x_{2}, y_{2}\right)=(4,-4)$ $\begin{aligned} Midpoint of A B&=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ &=\left(\dfrac{0+4}{2}, \dfrac{0+(-4)}{2}\right)\\\\ &=(2,-2)\\\\ Length of A B&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\\\\ &=\sqrt{(4-0)^{2}+(-4-0)^{2}} \\\\ &=\sqrt{16+16} \\\\ &=\sqrt{32} \\\\ &=4 \sqrt{2} \end{aligned}$ $\begin{aligned} (\text{ b })A=\left(x_{1}, y_{1}\right)&=(1,5), B=\left(x_{2}, y_{2}\right)=(3,1) \\\\ \text { Midpoint of } A B &=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right) \\\\ &=\left(\dfrac{1+3}{2}, \dfrac{5+1}{2}\right) \\\\ &=(2,3) \\ \\ \text { Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(3-1)^{2}+(1-5)^{2}} \\\\ &=\sqrt{4+16} \\ \\ &=\sqrt{20} \\\\ &=2 \sqrt{5} \end{aligned}$ ( c ) $A=\left(x_{1}, y_{1}\right)=(-3,-3), B=\left(x_{2}, y_{2}\right)=(0,0)$ $\begin{aligned} \text{ Midpoint of } A B&=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ &=\left(\dfrac{-3+0}{2}, \dfrac{-3+0}{2}\right) \\\\ &=(-1.5,-1.5)\\\\ \text { Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{\left(0-(-3)^{2}+(0-(-3))^{2}\right.} \\\\ &=\sqrt{9+9} \\\\ &=\sqrt{18} \\\\ &=3 \sqrt{2} \end{aligned}$ ( d ) $A=\left(x_{1}, y_{1}\right)=(-1,3), B=\left(x_{2}, y_{2}\right)=(5,1)$ $\begin{aligned} \text{ Midpoint of } A B&=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ &=\left(\dfrac{-1+5}{2}, \dfrac{3+1}{2}\right) \\\\ &=(2,2) \\\\ \text { Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(5-(-1))^{2}+(1-3)^{2}} \\\\ &=\sqrt{36+4} \\\\ &=\sqrt{40} \\\\ &=2 \sqrt{10} \end{aligned}$ ( e ) $A=\left(x_{1}, y_{1}\right)=(-1,6), B=\left(x_{2}, y_{2}\right)=(2,-2)$ $\begin{aligned} \text{ Midpoint of } A B&=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ &=\left(\dfrac{-1+2}{2}, \dfrac{6+(-2)}{2}\right) \\\\ &=(0.5,2)\\\\ \text { Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(2-(-1))^{2}+(-2-6)^{2}} \\\\ &=\sqrt{9+64} \\\\ &=\sqrt{73} \end{aligned}$ ( f ) $A=\left(x_{1}, y_{1}\right)=(-3,-4), B=\left(x_{2}, y_{2}\right)=(3,-1)$ $\begin{aligned} \text{ Midpoint of } A B&=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ &=\left(\dfrac{-3+3}{2}, \dfrac{-4+(-1)}{2}\right) \\\\ &=(0,-2.5) \\\\ \text { Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(3-(-3))^{2}+(-1-(-4))^{2}} \\\\ &=\sqrt{36+9} \\\\ &=\sqrt{45} \\\\ &=3 \sqrt{5} \end{aligned}$



4. If ($1, 0$) is the midpoint of the line passing through the points $A$ ($-5, 2$) and $B$ ($x, y$), find the value of $x$ and $y$.



$A=\left(x_{1}, y_{1}\right)=(-5,2), B=(x, y), M=(1,0)$ $M$ is the midpoint of $A B$. $\begin{aligned} \text{ Midpoint of } A B&=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)\\\\ (1,0)&=\left(\dfrac{-5+x}{2}, \dfrac{2+y}{2}\right)\\\\ \dfrac{-5+x}{2}&=1 and \dfrac{2+y}{2}=0\\\\ -5+x&=2 and 2+y=0\\\\ x&=7 and y=-2 \end{aligned}$



5. Calculate the perimeter of given polygons correct to one decimal place.
(a) A triangle with vertices $P (-2, 3)$, $Q (5, -4)$ and $R (1, 8)$.
(b) A parallelogram with vertices $A (-10, 1)$, $B (6, -2)$, $C (14, 4)$ and $D (-2, 7)$.
(c) A trapezium with vertices $E (-6, -2)$, $F (1, -2)$, $G (0, 4)$ and $H (-5, 4)$.




(a) $P(-2,3), Q(5,-4), R(1,8)$ $\begin{aligned} \text{ Length of } P Q&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\\\\ &=\sqrt{\left(5-(-2)^{2}+(-4-3)^{2}\right.}\\\\ &=\sqrt{49+49}\\\\ &=\sqrt{98}\\\\ &=9.9\\\\ \text { Length of } Q R &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(1-5)^{2}+(8-(-4))^{2}}\\\\ &=\sqrt{16+144}\\\\ &=\sqrt{160}\\\\ &=12.7\\\\ \end{aligned}$ $\begin{aligned} \text { Length of } P R &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(1-(-2))^{2}+(8-3)^{2}} \\\\ &=\sqrt{9+25} \\\\ &=\sqrt{34} \\\\ &=5.8 \end{aligned}$ $\begin{aligned} \text { The perimeter of } \triangle P Q R &=P Q+Q R+P R \\ &=9.9+12.7+5.8 \\ &=28.4 \end{aligned}$ (b) $A(-10,1), B(6,-2), C(14,4), D(-2,7)$ $\begin{aligned} \text { Length of } A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(6-(-10))^{2}+(-2-1)^{2}} \\\\ &=\sqrt{256+9} \\\\ &=\sqrt{265} \\\\ &=16.3 \\\\ \text { Length of } B C &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(14-6)^{2}+(4-(-2))^{2}} \\\\ &=\sqrt{64+36} \\\\ &=\sqrt{100} \\\\ &=10 \end{aligned}$ $\begin{aligned} \text { Length of } C D &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(-2-14)^{2}+(7-4)^{2}} \\\\ &=\sqrt{256+9} \\\\ &=\sqrt{265} \\\\ &=16.3 \\\\ \text { Length of } D A &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(-2-(-10))^{2}+(7-1)^{2}} \\\\ &=\sqrt{64+36} \\\\ &=\sqrt{100} \\\\ &=10 \end{aligned}$ $\begin{aligned} \text{ The perimeter of parallelogram } A B C D&=A B+B C+C D+D A\\\\ &=16.3+10+16.3+10 \\\\ &=52.6\\\\ \end{aligned}$ (c) $E(-6,-2), F(1,-2), G(0,4), H(-5,4)$ $\begin{aligned} \text{ Length of } E F&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\\\\ &=\sqrt{(1-(-6))^{2}+(-2-(-2))^{2}} \\\\ &=\sqrt{49+0}\\\\ &=7 \end{aligned}$ $\begin{aligned} \text {Length of } F G &=\sqrt{(0-1)^{2}+(4-(-2))^{2}} \\\\ &=\sqrt{1+36} \\\\ &=\sqrt{37} \\\\ &=6.1 \end{aligned}$ $\begin{aligned} \text{ Length of }G H&=\sqrt{(-5-0)^{2}+(4-4)^{2}}\\\\ &=\sqrt{25+0} \\\\ &=\sqrt{25} \\\\ &=5 \end{aligned}$ $\begin{aligned} \text { Length of } E H &=\sqrt{(-5-(-6))^{2}+(4-(-2))^{2}} \\\\ &=\sqrt{1+36} \\\\ &=\sqrt{37} \\\\ &=6.1 \end{aligned}$ $\begin{aligned} \text{ The perimeter of trapezium } E F G H&=E F+F G+G H+E H\\\\ &=7+6.1+5+6.1 \\\\ &=24.2 \end{aligned}$



6. A circle has centre ($2, 1$). Find the coordinates of the endpoint of a diameter if one endpoint is ($7, 1$).



The centre of a circle $=(2,1)$ (i.e) the midpoint of the diameter. Let $A B$ be the diameter. $\begin{aligned} \text{ Let } A=(x, y) and B&=(7,1)\\\\ \text{ Midpoint of } A B&=\left(\dfrac{x+7}{2}, \dfrac{y+1}{2}\right)\\\\ (2,1)&=\left(\dfrac{x+7}{2}, \dfrac{y+1}{2}\right)\\\\ \dfrac{x+7}{2}&=2 \text{ and } \dfrac{y+1}{2}=1\\\\ x+7&=4 \quad y+1=2\\\\ x&=-3 \quad y=1\\\\ \end{aligned}$ $\therefore$ The coordinates of the endpoint of a diameter $=A=(x, y)=(-3,1)$.



7. △$KLM$ has vertices $K$ ($-5, 18$), $L$ ($10, 14$) and $M$ ($-5, -10$).
(a) Find the length of each side.
(b) Find the perimeter of △KLM.
(c)Find the area of △KLM.




$K(-5,18)$, $L(10,-2)$ and $M(-5,-10)$ $\begin{aligned} \text{ (a) Length of } K L&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\\\\ &=\sqrt{(10-(-5))^{2}+(-2-18)^{2}} \\\\ &=\sqrt{225+400} \\\\ &=\sqrt{625} \\\\ &=25 \end{aligned}$ $\begin{aligned} \text{ Length of } L M&=\sqrt{(-5-10)^{2}+(-10-(-2))^{2}}\\\\ &=\sqrt{225+64}\\\\ &=\sqrt{289} \\\\ &=17 \end{aligned}$ $\begin{aligned} \text{ Length of } KM&= \text{ vertical distance }\\\\ &=18-(-10) \\\\ &=28 \end{aligned}$ $\begin{aligned} \text{ (b) The perimeter of } \triangle K L M&=K L+L M+K M\\\\ &=25+17+28 \\\\ &=70 \end{aligned}$ (c) $\begin{aligned} \text{ Draw } LN &\perp KM.\\\\ \text{ Length of } L N&= \text{ horizontal distance }\\\\ & =10-(-5)\\\\ &=15\\\\ \text{ The area of } \triangle K L M&=\dfrac{1}{2} \times K M \times L N\\\\ &=\dfrac{1}{2} \times 28 \times 15\\\\ &=210 \text{ square units } \end{aligned}$



8. Prove that the triangle whose vertices are $P$ ($2, 3$), $Q$ ($-1, -1$), $R$ ($3,-4$) is isosceles.



$\begin{aligned} P(2,3) &, Q(-1,-1) \text { and } R(3,-4) \\\\ P Q &=\sqrt{\left(x_{2-x_{1}}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(-1-2)^{2}+(-1-3)^{2}} \\\\ &=\sqrt{9+16} \\ &=\sqrt{25} \\\\ &=5 \\\\ Q R &=\sqrt{\left(3-(-1)^{2}+(-4-(-1))^{2}\right.} \\\\ &=\sqrt{16+9} \\\\ &=\sqrt{25} \\\\ &=5 \\\\ P R &=\sqrt{(3-2)^{2}+(-4-3)^{2}} \\\\ &=\sqrt{1+49} \\\\ &=5 \sqrt{50}\\ P Q&=Q R \end{aligned}$ $\therefore \triangle P Q R$ is isosceles.



9. A triangle has vertices $E$ ($0, 7$), $F$ ($5, -5$), and $G$ ($10, 7$). Find the lingth of the altitute to the shortest side.



$\begin{aligned} E(0,7)&, F(5,-5) and G=(10,7).\\\\ E F &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(5-0)^{2}+(-5-7)^{2}} \\\\ &=\sqrt{25+144} \\\\ &=\sqrt{169} \\\\ &=13 \\\\ F G &=\sqrt{(10-5)^{2}+(7-(-5))^{2}} \\\\ &=\sqrt{25+144} \\\\ &=\sqrt{169} \\\\ &=13\\\\ E G&= \text{ horizontal distance }\\\\ &=10-0 \\\\ &=10 \end{aligned}$ $E G$ is the shortest side. $\therefore F H \perp E G$. The length of the altitute to the shortest side =$F A=7-(-5)=12$



10. The vertices of a quadrilateral are $A$ ($4, -3$), $B$ ($7, 10$), $C$ ($-8, 2$) and $D$ ($-1, -5$). Find the length of each diagonal.



$A(4,-3), B(7,10), C(-8,2)$ and $D(-1,-5)$. $A C$ and $B D$ are diagonals. $\begin{aligned} A C &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(-8-4)^{2}+(2-(-3))^{2}} \\\\ &=\sqrt{144+25} \\\\ &=\sqrt{169} \\\\ &=13 \\\\ B D &=\sqrt{(-1-7)^{2}+(-5-10)^{2}} \\\\ &=\sqrt{64+225} \\\\ &=\sqrt{289} \\\\ &=17 \end{aligned}$



11. The distance between the two points $M$ ($15, a$) and $N$ ($a, -5$) is $20$. Find the value of $a$.



$M(15, a)$ and $N(a,-5)$. Distance between $M N=20$ $\begin{aligned} M N=\sqrt{(a-15)^{2}+(-5-a)^{2}}&=20 \\\\ (a-15)^{2}+(-5-a)^{2} &=400 \\\\ a^{2}-30 a+225+25+10 a+a^{2}&=400 \\\\ 2 a^{2}-20 a+250&=400 \\\\ 2 a^{2}-20 a-150&=0 \\\\ a^{2}-10 a-75&=0 \\\\ (a-15)(a+5)&=0 \\\\ a-15=0 \text { (or) } a+5&=0 \\\\ a=15 \text { (or) } a&=-5 \end{aligned}$



12. In a parallelogram $P Q R S$, three of vertices are $P$ ($1, 1$), $Q$ ($2, 6$) and $R$ ($5, 3$). Find the midpoint of $P R$ and use it to find the fourth vertex $S$. Find also the lengths of the diagonals.



$P(1,1)$, $Q(2,6)$ and $R(5,3)$. $\begin{aligned} \text{ Midpoint of } P R&=\left(\dfrac{1+5}{2}, \dfrac{1+3}{2}\right)\\\\ &=(3,2) \end{aligned}$ Let the coordinates of $S=(x, y)$. $P Q R S$ is a parallelogram, $P R$ and $Q S$ are diagonals. $\begin{aligned} \therefore \text{ Midpoint of } P R&= \text{ Midpoint of } Q S\\\\ (3,2)&=\left(\dfrac{2+x}{2}, \dfrac{6+y}{2}\right)\\\\ \dfrac{2+x}{2}&=3 \text{ and } \dfrac{6+y}{2}=2\\\\ 2+x&=6 \text{ and } 6+y=4\\\\ x&=4 \text{ and } y=-2 \end{aligned}$ $\therefore$ The coordinates of $S=(4,-2)$. $\begin{aligned} P R &=\sqrt{\left(x_{2-x_{1}}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\\\ &=\sqrt{(5-1)^{2}+(3-1)^{2}} \\\\ &=\sqrt{16+4} \\\\ &=\sqrt{20} \\\\ &=2 \sqrt{5} \\\\ Q S &=\sqrt{(4-2)^{2}+(-2-6)^{2}} \\\\ &=\sqrt{4+64} \\ &=\sqrt{68} \\\\ &=2 \sqrt{17} \end{aligned}$

Exercise 1.2

1. Complete each sentence.
1. The slope of the line passing through two points ($-6, 0$) and ($2, 3$) is -----.
2. The slope of the line joining the point ($1, 2$) and the origin is -----.
3. A vertical line has ----- slope.
4. A horizontal line has ----- slope.



$\dfrac{3}{8}$ $2$ undefined zero



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




(a) The slope is undefined. (b) The slope is undefined. (c) The slope is zero. (d) The slope is zero. (e) The slope is positive. $\begin{aligned} \text { slope } &=\dfrac{\text { rise }}{\text { run }} \\ &=\dfrac{4-0}{4-0} \\ &=1 \end{aligned}$ (f) The slope is positive. $\begin{aligned} \text { Slope } &=\dfrac{\text { rise }}{\text { run }} \\ &=\dfrac{2-0}{3-0} \\ &=\dfrac{2}{3} \end{aligned}$ (g) The slope is negative. $\begin{aligned} \text { slope } &=\dfrac{\text { rise }}{\text { ran }} \\ &=\dfrac{0-3}{4-0} \\ &=-\dfrac{3}{4} \end{aligned}$ (h) The slope is negative. $\begin{aligned} \text { Slope } &=\dfrac{\text { rise }}{\text { run }} \\ &=\dfrac{-2}{3} \end{aligned}$



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

(a) $(5 , 2)$ and $(-3 , 2)$	(b) $(0 , 5)$ and $(-1 , 5)$	(c) $(2 , 3)$ and $(2 , 6)$
(d) $(0 , 0)$ and $(0 , -2)$	(e) $(1 , -2)$ and $(-3 , -2)$	(f) $(a , b)$ and $(a , c)$




(a) The point $(5,2)$ and $(-3,2)$ have the same y-coordinates 2 . So it is a horizontal line. The slope of the line is zero. (b) The point $(0,5)$ and $(-1,5)$ have the same y-coordinates 5 . So it is a horizontal line. The slope of the line is zero. (c) The point $(2,3)$ and $(2,6)$ have the same x-coordinates 2 . So it is a vertical line. The slope of the line is undefined. (d) The point $(0,0)$ and $(0,-2)$ have the same x-coordinates 0 . so it is a vertical line. The slope of the line is undefined. (e) The point $(1,-2)$ and $(-3,-2)$ have the same y-coordinates $-2$. So it is a horizontal line. The slope of the line is zero. (f) The point $(a, b)$ and $(a, c)$ have the same x-coordinates $a$. So it is a vertical line. The slope of the line is undefined.



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

(a) $A(0 , 0)$ and $B(8 , 4)$	(b) $C(10 , 5)$ and $D(6 , 8)$	(c) $E(-5 , 7)$ and $F(-2 , -4)$
(d) $G(23 , 15)$ and $H(18 , 5)$	(e) $I(-2 , 0)$ and $J(0 , 6)$	(f) $K(15 , 6)$ and $L(-2 , 23)$




(a) $A(0,0)$ and $B(8,4)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{4-0}{8-0} \\\\ &=\dfrac{1}{2} \end{aligned}$ $\therefore$ Slope of $A B=\dfrac{1}{2}$ (b) $C(10,5)$ and $D(6,8)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{8-5}{6-10} \\\\ &=\dfrac{3}{-4} \end{aligned}$ $\therefore$ Slope of $C D=-\dfrac{3}{4}$ (c) $E(-5,7)$ and $F(-2,-4)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{-4-7}{-2-(-5)} \\\\ &=\dfrac{-11}{3} \end{aligned}$ Slope of $E F=-\dfrac{11}{3}$ (d) $G(23,15)$ and $H(18,5)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{5-15}{18-23} \\\\ &=\dfrac{-10}{-5} \\\\ &=2 \end{aligned}$ $\therefore$ Slope of $G H=2$ (e) $I(-2,0)$ and $J(0,6)$ $\begin{aligned} \text { slope }=m &=\dfrac{x_{2}-x_{1}}{} \\\\ &=\dfrac{6-0}{0-(-2)} \\\\ &=3 \end{aligned}$ $\therefore $ Slope of $IJ =3 $. (f) $K(15,6)$ and $L(-2,23)$. $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ &=\dfrac{23-6}{-2-15} \\ &=\dfrac{17}{-17} \\ &=-1 \end{aligned}$ $\therefore$ Slope of $K L=-1$



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

(a) $E(\dfrac{3}{4}, \dfrac{4}{5})$ and $F(-\dfrac{1}{2}, \dfrac{7}{5})$	(b) $G(-a, b)$ and $H(3a, 2b)$
(c) $L(\sqrt{12}, \sqrt{18})$ and $M(\sqrt{27}, \sqrt{8})$	(d) $P(0, a)$ and $Q(a, 0)$




(a) $E\left(\dfrac{3}{4}, \dfrac{4}{5}\right)$ and $F\left(-\frac{1}{2}, \dfrac{7}{5}\right)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{\dfrac{7}{5}-\dfrac{4}{5}}{-\dfrac{1}{2}-\dfrac{3}{4}} \\\\ &=\dfrac{\dfrac{3}{5}}{-\dfrac{5}{4}}\\\\ \text{ Slope of } E F&=-\dfrac{12}{25}\\\\ \end{aligned}$ (b) $G(-a, b)$ and $H(3 a, 2 b)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{2 b-b}{3 a-(-a)}\\\\ \text{ Slope of } GH &=\dfrac{b}{4 a} \quad(a \neq 0) \end{aligned}$ (c) $L(\sqrt{12}, \sqrt{18})$ and $M(\sqrt{27}, \sqrt{8})$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{\sqrt{8}-\sqrt{18}}{\sqrt{27}-\sqrt{12}} \\\\ &=\dfrac{2 \sqrt{2}-3 \sqrt{2}}{3 \sqrt{3}-2 \sqrt{3}} \\\\ &=\dfrac{-\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}\\\\ \text{ Slope of } L M&=\dfrac{-\sqrt{6}}{3} \end{aligned}$ (d) $P(0, a)$ and $Q(a, 0)$. $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{0-a}{a-0}\\\\ \text{ slope of } P Q&=-1 \end{aligned}$



6. Find $p$, $q$, $r$ in the followings.
(a) The slope joining the points $(0, 3)$ and $(1, p)$ is $5$.
(b) The slope joining the points $(-2, q)$ and $(0, 1)$ is $-1$.
(c) The slope joining the points $(-4, -2)$ and $(r, -6)$ is $-6$.




(a) The points $(0,3)$ and $(1, p)$. $\begin{aligned} m&=5 \\\\ \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ 5 &=\dfrac{p-3}{1-0} \\\\ p-3 &=5 \\\\ p &=8 \end{aligned}$ (b) The points $(-2, q)$ and $(0,1)$. $\begin{aligned} m&=-1 \\\\ \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ -1 &=\dfrac{1-q}{0-(-2)} \\\\ 1-q &=-2 \\\\ q &=3 \end{aligned}$ (c) The points $(-4,-2)$ and $(r,-6)$. $\begin{aligned} m&=-6\\\\ \text{ slope } =m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ -6&=\dfrac{-6-(-2)}{r-(-4)}\\\\ -6 r-24&=-4\\\\ 6 \gamma&=-20\\\\ r&=-\dfrac{20}{6}\\\\ r&=-\dfrac{10}{3} \end{aligned}$



7. Find the slope corresponding to the following events.
(a) A man climbs $10$ m for every $200$ meters horizontally.
(b) A motorbike rises $20$ km for every $100$ kilometers horizontally.
(c) A plane takes off $35$ km for every $5$ kilometers horizontally.
(d) A submarine descends $120$ m for every $15$ meters horizontally.




$\begin{aligned} (a) \text { rise } &=10 \mathrm{~m}, \text { run }=200 \mathrm{~m} \\\\ \text { slope }=m &=\dfrac{\text { rise }}{\text { run }} \\\\ &=\dfrac{10}{200} \\\\ &=\dfrac{1}{20} \end{aligned}$ $\begin{aligned} (b) \text{ rise } &=20 \mathrm{~km}, \text{ run }=100 \mathrm{~km}\\\\ \text { slope }=m &=\dfrac{\text { rise }}{\text { run }} \\\\ &=\dfrac{20}{100} \\\\ &=\dfrac{1}{5} \end{aligned}$ $\begin{aligned} (c) \text{ rise }&=-35 \mathrm{~km}, \text{ run } =5 \mathrm{~km}\\\\ \text { slope }=m &=\dfrac{\text { rise }}{\text{ run }} \\\\ &=\dfrac{-35}{5} \\\\ &=-7 \end{aligned}$ $\begin{aligned} (d) \text { rise }&=-120 \mathrm{~m}, \text { run }=15 \mathrm{~m}\\\\ \text { slope }=m&=\dfrac{\text { rise }}{\text { run }}\\\\ &=\dfrac{-120}{15}\\\\ &=-8 \end{aligned}$



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



$\begin{aligned} m=0.05, \text { rise }&=15 m, \text { run }=\text { ? } \\\\ \text { slope }=m&=\dfrac{\text { rise }}{\text{ run }} \\\\ 0.05&=\dfrac{15}{\text{ run }} \\\\ \text { run }&=\dfrac{15}{0.05} \\\\ &=\dfrac{1500}{5} \\\\ &=300 \mathrm{~m} \end{aligned}$ The trains travelled $300 \mathrm{~m}$ horizontally.



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.



$A(-2,3), B(5,-4), C(1,8)$. $\begin{aligned} \text{ slope of } A B=m_{A B}&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &= \dfrac{-4-3}{5-(-2)} \\\\ &=-\dfrac{7}{7} \\\\ &=-1 \\\\ \text { Slope of } B C=m_{B C}& =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{8-(-4)}{1-5} \\\\ &=\dfrac{12}{-4} \\\\ &=-3 \\\\ \text { Slope of } A C=m_{A C}& =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{8-3}{1-(-2)} \\\\ &=\dfrac{5}{3} \end{aligned}$



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.



$P(1,4), Q(3,2), R(4,6)$ and $S(2,8)$ $\begin{aligned} \text { Slope of } P Q=m_{P Q}& =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{2-4}{3-1} \\\\ &=-1 \end{aligned}$ $\begin{aligned} \text{ Slope of } Q R=m_{P R}&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &=\dfrac{6-2}{4-3} \\\\ &=4 \end{aligned}$ $\begin{aligned} \text{ Slope of } R S=m_{R S}&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &=\dfrac{8-6}{2-4} \\\\ &=-1 \end{aligned}$ $\begin{aligned} \text { Slope of } P S=m_{P S}&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{8-4}{2-1} \\\\ &=4 \end{aligned}$



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 $8$?



Let $\left(x_{1}, y_{1}\right)=(-2,5)$ and $\left(x_{2}, y_{2}\right)=\left(8, y_{2}\right)$ $\begin{aligned} m&=-1\\\\ \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ -1 &=\dfrac{y_{2}-5}{8-(-2)} \\\\ y_{2}-5 &=-10 \\\\ y_{2} &=-5 \end{aligned}$ The $y$-coordinate of the point is $-5$.

Exercise 1.3

1. Sketch the following lines.

(a) $y = 3$

(b) $x = -2$

(c) $y = 5x$

(d) $y = -3x$

(e) $y = \dfrac{1}{2}x$




(a) $y=3$ (b) $x=-2$ (c) $y=5 x$ $\begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -10 & -5 & 0 & 5 & 10 \\ \hline \end{array}$ (d) $y=-3 x$ $\begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 6 & 3 & 0 & -3 & -6 \\ \hline \end{array}$ (e) $y=\dfrac{1}{2} x$ $\begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -1 & -\dfrac{1}{2} & 0 & \dfrac{1}{2} & 1 \\ \hline \end{array}$



2. Graph each line in the same coordinate plane.

(a) $y = x + 2$	(b) $y = x + 3$	(c) $y = x + 5$
(d) $y = x - 1$	(e) $y = x - 2$	(f) $y = x - 4$




(a) $y=x+2$ $\begin{array}{|c|c|c|} \hline x & -2 & 0 \\ \hline y & 0 & 2 \\ \hline \end{array}$ (b) $y=x+3$ $\begin{array}{|c|c|c|} \hline x & -3 & 0 \\ \hline y & 0 & 3 \\ \hline \end{array}$ (c) $y=x+5$ $\begin{array}{|c|c|c|} \hline x & -5 & 0 \\ \hline y & 0 & 5 \\ \hline \end{array}$ (d) $y=x-1$ $\begin{array}{|c|c|c|} \hline x & 1 & 0 \\ \hline y & 0 & -1 \\ \hline \end{array}$ (e) $y=x-2$ $\begin{array}{|c|c|c|} \hline x & 2 & 0 \\ \hline y & 0 & -2 \\ \hline \end{array}$ (f) $y=x-4$ $\begin{array}{|c|c|c|} \hline x & 4 & 0 \\ \hline y & 0 & -4 \\ \hline \end{array}$



3. Graph each line.

(a) $y = 2x + 2$

(b) $y = \dfrac{1}{2}x + 2$

(c) $y = -\dfrac{1}{2}x + 2$




(a) $y=2 x+2$ $\begin{array}{|c|c|c|} \hline x & -1 & 0 \\ \hline y & 0 & 2 \\ \hline \end{array}$ (b) $y=\dfrac{1}{2} x+2$ $\begin{array}{|r|r|r|} \hline x & -4 & 0 \\ \hline y & 0 & 2 \\ \hline \end{array}$ (c) $y=-\dfrac{1}{2} x+2$ $\begin{array}{|l|l|l|} \hline x & 4 & 0 \\ \hline y & 0 & 2 \\ \hline \end{array}$



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

(a) $y = x - 2$	(b) $y = -3x - 3$	(c) $y =\dfrac{1}{2} x + 1$
(d) $y =-\dfrac{1}{2} x + 1$	(e) $y + 3 x = 3$	(f) $x - y = 3$




(a) $y=x-2$ Compare this equation with $y=m x+c$, $m=1, c=-2$ slope $=1, y$-intercept $=-2$ rise $=2$, run $=2$ (b) $y=-3 x-3$ Compare this equation with $y=m x+c$, $m=-3, c=-3$ rise $=3$, run $=-1$ (c) $y=\dfrac{1}{2} x+1$ Compare this equation with $y=m x+c$, $m=\dfrac{1}{2}, c=1$ rise $=1$, run $=2$ (d) $y=-\dfrac{1}{2} x+1$ Compare this equation with $y=m x+c$, $m=-\dfrac{1}{2}, c=1$ $\text { rise }=1, \text { run }=2$ $\begin{aligned} \text { (e) }y+3 x &=3 \\\\ y &=-3 x+3 \end{aligned}$ Compare this equation with $y=m x+c$, $m=-3$, $c=3$ . $\text { rise }=3, \text { run }=-1$ (f) $x-y=3$ $y =x-3 $ Compare this equation with $y=m x+c$, $\begin{aligned} m&=1, c=-3 \\\\ \text { rise }&=2, \text { run }=2 \end{aligned}$



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 $0$




(a) $m=3, c=4$ The equation of the line is $y=m x+c$ $y=3 x+4$ (b) $m=2, c=0$ The equation of the line is $y=m x+c$ $y=2 x+0$ $y=2 x$ (c) $m=0, c=2$ The equation of the line is $y=m x+c$ $y=0 \times x+2$ $y=2$ (d) $m=0, c=0$ The equation of the line is $y=m x+c$ $y=0 \times x+0$ $y=0 $



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



slope $=m=-\dfrac{2}{3}$, Let $\left(x_{1}, y_{1}\right)=(9,4)$. The equation of the line through the point $(9,4)$ is $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right) \\\\ y-4&=-\dfrac{2}{3}(x-9) \\\\ 3 y-12&=-2 x+18 \\\\ 2 x+3 y&=30 \end{aligned}$



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



slope $=m=-2, y$-intercept $=c=6$ The equation of the line is $\begin{aligned} y&=m x+c \\\\ y&=-2 x+6 \end{aligned}$ When $y=0$, $\begin{aligned} -2 x+6 &=0 \\\\ 2 x &=6 \\\\ x &=3 \\\\ \therefore x \text {-intercept }&=3 \end{aligned}$



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 $(-1, 4)$.




(a) slope $=m=5$, Let the point $(2,9)=\left(x_{1}, y_{1}\right)$. The equation of the line is $\begin{gathered} y-y_{1}&=m\left(x-x_{1}\right) \\\\ y-9&=5(x-2) \\\\ y-9&=5 x-10 \\\\ y&=5 x-1 \\\\ 5 x&-y=1 \end{gathered}$ (b) slope $=m=1$, Let the point $(1,-2)=\left(x, y_{1}\right)$. The equation of the line is $\begin{aligned} y-y_{1} &=m\left(x-x_{1}\right) \\\\ y-(-2) &=1(x-1) \\\\ y+2 &=x-1 \\\\ x-y &=3 \end{aligned}$ (c) slope $=m=-3$, let the point $(-1,6)=(x, y)$. The equation of the line is $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right)\\\\ y-6&=-3(x-(-1)) \\\\ y-6&=-3 x-3 \\\\ 3 x+y&=3 \end{aligned}$ (d) slope $=m=-2$, let the point $(-1,4)=(x, y),$, The equation of the line is $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right) \\\\ y-4&=-2(x-(-1)) \\\\ y-4&=-2 x-2 \\\\ 2 x+y&=2 \end{aligned}$



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 $(-4, -2)$




(a) Let $(2,4)=\left(x_{1}, y_{1}\right)$, $(6,8)=\left(x_{2}, y_{2}\right)$ $\begin{aligned} \text { slope }=m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &=\dfrac{8-4}{6-2}\\\\ &=\dfrac{4}{4}\\\\ &=1 \end{aligned}$ The equation of the line is $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right)\\\\ y-4&=1(x-2)\\\\ x-y&=-2 \end{aligned}$ (b) Let $(-3,5)=\left(x_{1}, y_{1}\right),(6,-1)=\left(x_{2}, y_{2}\right)$ $\begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{-1-5}{6-(-3)} \\\\ &=\dfrac{-6}{9} \\\\ &=-\dfrac{2}{3} \end{aligned}$ The equation of the line is $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right) \\\\ y-5&=-\dfrac{2}{3}(x-(-3)) \\\\ 3 y-15&=-2 x-6 \\\\ 2 x+3 y&=9 \end{aligned}$ (c) Let$(-2,1)=\left(x_{1}, y_{1}\right),(-4,-2)=\left(x_{2}, y_{2}\right)$ $ \begin{aligned} \text { slope }=m &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\ &=\dfrac{-2-1}{-4-(-2)} \\\\ &=\dfrac{-3}{-2} \\\\ &=\dfrac{3}{2} \end{aligned}$ The equation of the line is $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right) \\\\ y-1&=\dfrac{3}{2}(x-(-2)) \\\\ y-1&=\dfrac{3}{2}(x+2) \\\\ 2 y-2&=3 x+6 \\\\ 3 x-2 y&=-8 \end{aligned}$



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) $y = \dfrac{2}{3}x - 5$ and $y = \dfrac{3}{2}x - 5$
(c) $y = 3x - 2$ and $y = -\dfrac{1}{3}x + 9$	(d) $y = \dfrac{2}{3}x - 5$ and $y = -\dfrac{3}{2}x - 5$




$\begin{aligned} (a)\quad y &=3 x-2 \\\\ m_{1} &=3 \\\\ y &=3 x+9 \\\\ m_{2} &=3 \\\\ m_{1} &=m_{2} \end{aligned}$ $\therefore$ Two lines are parallel. $\begin{aligned} (b)\quad y&=\dfrac{2}{3} x-5 \\\\ m_{1}&=\dfrac{2}{3} \\\\ y&=\dfrac{3}{2} x-5 \\\\ m_{2}&=\dfrac{3}{2} \\\\ m_{1} &\neq m_{2} \end{aligned}$ $\therefore$ Two lines are not parallel. $\begin{aligned} (c)\quad y&=3 x-2 \\\\ m_{1}&=3 \\\\ y&=-\dfrac{1}{3} x+9 \\\\ m_{2}&=-\dfrac{1}{3} \\\\ m_{1} \times m_{2}&=3 \times-\dfrac{1}{3}=-1 \end{aligned}$ $\therefore$ Two lines are perpendicular. $\begin{aligned} (d)\quad y&=\dfrac{2}{3} x-5 \\\\ m_{1}&=\dfrac{2}{3} \\\\ y&=-\dfrac{3}{2} x-5 \\\\ m_{2}&=-\dfrac{3}{2} \\\\ m_{1} \times m_{2}&=\dfrac{2}{3} \times-\dfrac{3}{2}=-1 \end{aligned}$ $\therefore$ Tro lines are perpendicular.



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 $y = -\dfrac{4}{5}x - 3$ and passes through $(0, \dfrac{1}{2})$




(a) Given line $y=4 x+2$ Compare this equation with $y=m x+c$, $m=4$ $\therefore$ The slope of the parallel line to the given line, $m=m_{1}=4$ The equation of the line passes through $(0,8)$ is $\begin{aligned} y-y_{1} &=m_{1}\left(x-x_{1}\right) \\\\ y-8 &=4(x-0) \\\\ y-8 &=4 x \\\\ 4 x-y &=-8 \end{aligned}$ (b) Given line $y=-x+3$ Compare this equation with $y=m x+c$, $m=-1$ $\therefore$ The slope of the parallel line to the given line, $m=m_{1}=-1$ The equation of the line passes through $(0,5)$ is $\begin{aligned} y-y_{1} &=m_{1}\left(x-x_{1}\right) \\\\ y-5 &=-1(x-0) \\\\ y-5 &=-x \\\\ x+y &=5 \end{aligned}$ (c) Given line $y=-2 x-3$ Compare this equation with $y=m x+c$, $m=-2$ $\therefore$ The slope of the parallel line to the given line, $m=m_{1}=-2$ The equation of the line passes through $(0,-7)$ is $\begin{aligned} y-y_{1} &=m_{1}\left(x-x_{1}\right) \\\\ y-(-7) &=-2(x-0) \\\\ y+7 &=-2 x \\\\ 2 x+y &=-7 \end{aligned}$ (d) Given line $y=-\dfrac{4}{5} x-3$ Compare this equation with $y=m x+c$, $m=-\dfrac{4}{5}$ $\therefore$ The slope of the parallel line to the given line, $m=m_{1}=-\dfrac{4}{5}$ The equation of the line passes through $\left(0, \dfrac{1}{2}\right)$ is $\begin{aligned} y-y_{1}&=m_{1}\left(x-x_{1}\right) \\\\ y-\dfrac{1}{2}&=-\dfrac{4}{5}(x-0) \\\\ \dfrac{2 y-1}{2}&=-\dfrac{4 x}{5} \\\\ 10 y-5&=-8 x \\\\ 8 x+10 y&=5 \end{aligned}$



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 $y = x -\dfrac{3}{2}$ and passes through $(0, \dfrac{5}{4})$




(a) Given line $y=5 x-4$ Compare this equation with $y=m x+c$, $m=5$ $\therefore$ The slope of the perpendicular line to the given line, $\begin{aligned} m \times m_{1} &=-1 \\\\ 5 \times m_{1} &=-1 \\\\ m_{1} &=-\dfrac{1}{5} \end{aligned}$ The equation of the line passes through $(0,7)$ is $\begin{aligned} y-y_{1} &=m_{1}\left(x-x_{1}\right) \\\\ y-7 &=-\dfrac{1}{5}(x-0) \\\\ y-7 &=-\dfrac{1}{5} x \\\\ 5 y-35 &=-x \\\\ x+5 y &=35 \end{aligned}$ (b) Given line $y=-x+7$ Compare this equation with $y=m x+c$, $m=-1$ $\therefore$ The slope of the perpendicular line to the given line, $\begin{aligned} m \times m_{1} &=-1 \\\\ -1 \times m_{1} &=-1 \\\\ m_{1} &=1 \end{aligned}$ The equation of the line passes through $(0,4)$ is $\begin{aligned} y-y_{1} &=m_{1}\left(x-x_{1}\right) \\\\ y-4 &=1(x-0) \\\\ y-4 &=x \\\\ x-y &=-4 \end{aligned}$ (c) Given line $y=-2 x+3$ Compare this equation with $y=m x+c$, $m=-2$ $\therefore$ The slope of the perpendicular line to the given line, $\begin{aligned} m \times m_{1} &=-1 \\\\ -2 \times m_{1} &=-1 \\\\ m_{1} &=\dfrac{1}{2} \end{aligned}$ The equation of the line passes through $(0,-4)$ is $\begin{aligned} y-y_{1}&=m_{1}\left(x-x_{1}\right) \\\\ y-(-4)&=\dfrac{1}{2}(x-0) \\\\ 2y+8&= x \\\\ x-2y&=8 \end{aligned}$ (d) Given line $y=x-\dfrac{3}{2}$ with $y=m x+c$, $m=1$ $\therefore$ The slope of the perpendicular line to the given line, $\begin{aligned} m \times m_{1} &=-1 \\\\ 1 \times m_{1} &=-1 \\\\ m_{1} &=-1 \end{aligned}$ The equation of the line parses through $\left(0, \dfrac{5}{4}\right)$ is $\begin{aligned} y-y_{1}=m_{1}\left(x-x_{1}\right) \\\\ y-\dfrac{5}{4}=1 \quad(x-0) \\\\ y-\dfrac{5}{4}=x \\\\ x-y=\dfrac{5}{4} \\\\ 4 x-4 y=5 \end{aligned}$



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



Let $P=(3 n, 0), Q=(0,7 n)$ $R=(0,21 n), S=(9 n, 0)$ $\begin{aligned} \text{ Slope } =m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text{ Slope of } P Q=m_{P Q}&=\dfrac{7 n-0}{0-3 n}\\\\ &=-\dfrac{7}{3}\\\\ \text{ Slope of } R S=m_{R S}&=\dfrac{0-21 n}{9 n-0}\\\\ &=-\dfrac{7}{3} \\\\ m_{ P Q }&=m_{R S} \\\\ \therefore \quad P Q & \| R S \end{aligned}$



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



$H(-12,1), K(9,3), M(11,-18)$ $\begin{aligned} \text { slope }=m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text { Slope of } H K=m_{H K}&=\dfrac{3-1}{9-(-12)}\\\\ &=\dfrac{2}{21}\\\\ \text { Slope of } K M=m_{K M}&=\dfrac{-18-3}{11-9}\\\\ &=\dfrac{-21}{2}\\\\ m_{H K} \times m_{K M}&=\dfrac{2}{21} \times-\dfrac{21}{2}\\\\ &=-1\\\\ \therefore H K &\perp K M \end{aligned}$ $\therefore \triangle H K M$ is aright triangle.



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.



$P(1,2), Q(5,-6)$ and $R(6, b)$ $\begin{aligned} \text{ Slope } =m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text{ Slope of } P Q=m_{P Q}&=\dfrac{-6-2}{5-1}\\\\ &=-\dfrac{8}{4}\\\\ &=-2\\\\ \text{ Slope of } Q R=m_{Q R}&=\dfrac{b-(-6)}{b-5}\\\\ &=\dfrac{b+6}{b-5} \end{aligned}$ Since angle $P Q R$ is a right angle, $\begin{aligned} P Q & \perp Q R \\\\ m_{P Q} \times m_{Q R} &=-1 \\\\ -2 \times \frac{b+6}{b-5} &=-1 \\\\ -2 b-12 &=-b+5 \\\\ b &=-17 \end{aligned}$



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.



Let $P(0,5), Q(5,0)$ and $R(-5,0)$. $\begin{aligned} \text{ Slope }=m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text{ Slope of } P Q=m_{P Q}&=\dfrac{0-5}{5-0}=-1 \end{aligned}$ The equation of the line passes through the point $P$, $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right)\\\\ y-5&=-1(x-0)\\\\ y-5&=-x \\\\ x+y&=5 \end{aligned}$ slope of $P R=m_{P R}=\dfrac{0-5}{-5-0}=1$ The equation of the line passes through the point $R$, $\begin{aligned} y-y_{1} &=m\left(x-x_{1}\right) \\\\ y-0 &=1(x-(-5)) \\\\ y &=x+5 \\\\ x-y &=5 \end{aligned}$ Since the $y$-coordinates of $Q$ and $R$ are $0$, $O R$ is a horizonal line. $\therefore$ Slope of $Q R=m_{Q R}=0$. The equation of the line pares through the point $Q$, $\begin{aligned} y-y_{1}&=m\left(x-x_{1}\right)\\\\ y-0 &=0(x-5) \\\\ y &=0 \end{aligned}$



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?



$A(5,6), B(13,6), C(11,2), D(1,2)$ $\begin{aligned} \text { slope }=m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text { Slope of } A B=m_{A B}&=\dfrac{6-6}{13-5}\\\\ &=\dfrac{13-5}{8}\\\\ &=0\\\\ \text { Slope of } B C=m_{B C}&=\dfrac{2-6}{11-13}\\\\ &=\dfrac{-4}{-2}\\\\ &=2\\\\ \text { Slope of } C D=m_{ C D }&=\dfrac{2-2}{1-11}\\\\ &=\dfrac{0}{-10}\\\\ &=0\\\\ \text { slope of } A D=m_{ AD }&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ &=\dfrac{2-6}{1-5}\\\\ &=\dfrac{-4}{-4}\\\\ &=1 \end{aligned}$ $m_{A B}=m_{ C D }=0$ $\therefore A B$ and $C D$ are horizontal line and $A B \| C D$. Quadrilateral $A B C D$ is a trapezium.



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?



$D(-4,6)$, $E(1,1)$ and $F(4,6)$ $\begin{aligned} \text{ Slope } =m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text{ Slope of } D E=m_{D E}&=\dfrac{1-6}{1-(-4)}\\\\ &=\dfrac{-5}{5} \\\\ &=-1\\\\ \text{ Slope of } E F=m_{E F}&=\dfrac{6-1}{4-1}\\\\ &=\dfrac{5}{3} \\\\ m_{D E} & \neq m_{E F} \end{aligned}$ $\therefore D, E$ and $F$ are not collinear.



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.



$A(-2,2), B(2,-2), C(4,2)$ and $D(2,4)$ $\begin{aligned} \text{ slope } =m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text{ slope of } A B=m_{A B}&=\dfrac{-2-2}{2-(-2)}\\\\ &=\dfrac{-4}{4} \\\\ &=-1 \\\\ \text { Slope of } B C=m_{B C} &=\dfrac{2-(-2)}{4-2} \\\\ &=\dfrac{4}{2} \\\\ &=2 \\\\ \text { Slope of } C D=m_{C D} &=\dfrac{4-2}{2-4} \\\\ &=\dfrac{2}{-2} \\\\ &=-1 \\\\ \text { slope of } A D=m_{A D} &=\dfrac{4-2}{2-(-2)} \\\\ &=\dfrac{2}{4} \\\\ &=\dfrac{1}{2}\\\\ \text{ Slope of } A C=m &=\dfrac{2-2}{4-(-2)} \\\\ &=\dfrac{0}{6} \\\\ &=0\\\\ \text{ Slope of } B D=m_{B D} &=\dfrac{4-(-2)}{2-2} \\\\ &=\dfrac{6}{0} \\\\ &=\text { undefined } \\\\ m_{A B} &=m_{C D} \\\\ \therefore A B & \| C D \\\\ m_{A C} &=0 \end{aligned}$ $\therefore$ AC is horizontal line. $m_{B D}=$ undefined $\therefore B D$ is vertical line. $\therefore A C \perp B D$ $\therefore$ Quadrilateral $A B C D$ is trapezoid with perpendicular diagonals.



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



.$A(-5,4), B(3,5), C(7,-2)$ and $D(-1,-3)$ $\begin{aligned} \text{ Slope } =m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \text{ Slope of } A B=m_{A B}&=\dfrac{5-4}{3-(-5)}\\\\ &=\dfrac{1}{8}\\\\ \text{ Slope of } B C=m_{B C}&=\dfrac{-2-5}{7-3}\\\\ &=-\dfrac{7}{4}\\\\ \text{ Slope of }C D=m_{C D}&=\dfrac{-3-(-2)}{-1-7}\\\\ &=\dfrac{1}{8}\\\\ \text{ Slope of } D A=m_{D A}&=\dfrac{4-(-3)}{-5-(-1)}\\\\ &=-\dfrac{7}{4}\\\\ \text { slope of } A C=m_{A C}&=\dfrac{-2-4}{7-(-5)}\\\\ &=-\dfrac{6}{12}\\\\ &=-\dfrac{1}{2}\\\\ \text { Slope of } B D=m_{B D}&=\dfrac{-3-5}{-1-3}\\\\ &=\dfrac{-8}{-4}\\\\ &=2\\\\ m_{A B}&=m_{ CD }\\\\ \therefore A B &\| C D\\\\ m_{B C}&=m_{A D}\\\\ \therefore B C &\| A D\\\\ m_{A C} \times m_{B D}&=-\dfrac{1}{2} \times 2\\\\ &=-1\\\\ \therefore A C &\perp B D \end{aligned}$ $\therefore A B C D $is a rhombus.