Quadratic Functions: Miscellaneous Exercise (Grade 10, New Syllabus)

ဖုန်းဖြင့်ကြည့်သည့်အခါ စာများကို အပြည့်မမြင်ရလျှင် screen ကို ဘယ်ညာ ဆွဲကြည့်နိုင်ပါသည်။

Important Notes

Quadratic Function (Standard Form)	$f(x)=a x^{2}+b x+c, a \neq 0$
Graph	Parabola $a>0$ (opens upward) $a<0$ (opens downward)
Axis of Symmetry	$x=-\displaystyle\frac{b}{2 a}$
Vertex	$\left(-\displaystyle\frac{b}{2 a}, f\left(-\displaystyle\frac{b}{2 a}\right)\right)=\left(-\displaystyle\frac{b}{2 a},-\displaystyle\frac{b^{2}-4 a c}{4 a}\right)$
y-intercept	(0, c)
Discriminant	$b^{2}-4 a c$ $b^{2}-4 a c>0 \Rightarrow$ two $x$ intercepts (cuts $x-$ axis at two points) $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (touch $x$ -axis at one point $)$ $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (does not intersect $x$ -axis)
Quadratic Equation	$a x^{2}+b x+c=0, a \neq 0$
Quadratic Formula	$x=\displaystyle\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Quadratic Function (Vertex Form)	$f(x)=a(x-h)^{2}+k, a \neq 0$
Vertex	$(h, k)$
Axis of Symmetry (Vertex Form)	$x=h$
Quadratic Function (Intercept Form when discriminant $>0$)	$f(x)=a(x-p)(x-q), a \neq 0$
$X$ -intercept points	$(p, 0)$ and $(q, 0)$
Axis of Symmetry	$x=\displaystyle\frac{p+q}{2}$
Quadratic Inequality	$a x^{2}+b x+c>0$ $a x^{2}+b x+c \geq 0$ $a x^{2}+b x+c<0$ $a x^{2}+b x+c \leq 0$
$a>0$ and $b^2-4ac<0$	The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\mathbb{R}$
$a<0$ and $b^2-4ac<0$	The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R}$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\varnothing$
$a>0$ and $b^2-4ac=0$	The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$
$a<0$ and $b^2-4ac=0$	The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\varnothing$
$a>0$ and $b^2-4ac>0$	The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$
$a<0$ and $b^2-4ac>0$	The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$

Exercises

ယခု post တွင်ပါဝင်သော မေးခွန်းများသည် grade 10 သင်ရိုးသစ်ပြဌာန်းချက်နှင့် သက်ဆိုင်သော သင်ရိုးလေ့ကျင့်ခန်း ပြင်ပမေးခွန်းများ ဖြစ်ပြီး Target Mathematics Grade 10 (Volume 1) တွင် Miscellaneous Exercise အဖြစ်ထည့်သွင်းပေးခဲ့ပါသည်။

1. The figure shows the graph of $y=x^2$ shifted to four new positions. Write an equation for each new graph.

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2. The figure shows the graph of $y=-x^2$ shifted to four new positions. Write an equation for each new graph.

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3. Find the standard form of the equation for the quadratic function for each diagram shown below.

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4. A quadratic function has an equation in the form $y=x^2+ax+a$ and passes through the point (1, 9). Calculate the value of $a$.

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5. The graph of quadratic function $y=ax^2+bx+c$ passes through the points (1,1), (0, 0) and (−1,1). Calculate the value of $a, b$ and $c$.

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6. A parabola has its vertex at the point $V (1,1)$ and passes through the point $(0,2)$. Find its equation in the form of $y=x^2+bx+c$.

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7. The graph of the function $f(x)=ax^2+bx+c$ has vertex at $(1, 4)$ and passes through the point $(−1,−8)$. Find $a, b$ and $c$.

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8. Given that $y = x^2 + 2x − 8$.
   

   (a)	What is the vertex of $y$?
   (b)	What are the $x$-intercepts of the graph of $y$ ?
   (c)	Find the $x$-coordinates of the ponts on the graph $y = x^2 + 2x − 8$ when $y = −8$.


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9. If a piece of string of fixed length is made to enclose a rectangle, show that the enclosed area is the greatest when the rectangle is a square.

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10. A farmer has $2000$ yards of fence to enclose a rectangular field. What are the dimensions of the rectangle that encloses the largest area?

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11. The profit (in millions of kyats) of a company is given by $P(x)= 5000+1000x−5x^2$ where $x$ is the amount ( in millions of kyats) the company spends on advertising.
    

    (a)	Find the amount, $x$, that the company has to spend to maximize its profit.
    (b)	Find the maximum profit.


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12. Find the integral values of $x$ that satisfy the inequality $x^2 + 48 < 16x$.

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13. Find the suitable domain of the function $f(x) = 1 + 3x − 2x^2$ for which the curve of $f(x)$ lies completely above the line $y = −1$.
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Exercise (9.1) - Sloutions (Grade 10, New Syllabus)

Exercise (9.1) ၏ မေးခွန်း ပုစ္ဆာများကို ဖြေရှင်းရန် ပြဌာန်းစာအုပ်ပါ Theorem (1) မှ Theorem (4) အထိ သီအိုရမ် မှန်ကန်ချက်များကို ဦးစွာလေ့လာထားရမည် ဖြစ်ပါသည်။ Theorem (1) မှ Theorem (4) အထိ ရှင်းလင်းချက်များကို ဒီ post မှာ လေ့လာနိုင်ပါသည်။

Ex 9.1-No (1)

The point $O$ is the centre of the given circle. Find the values of $x$, $y$ and $z$.

(a)

$\begin{array}{l} x=\angle A D C(\angle \mathrm{s} \text { on arc } A C) \\\\ \therefore x=70^{\circ}\\\\ A B C D \text { is cyclic quadrilateral. }\\\\ 70^{\circ}+\beta=180^{\circ}\\\\ \therefore \beta=180^{\circ}-70^{\circ}=110^{\circ} \\\\ \text { Since } A D / / \mathrm{BC}, \\\\ \theta=\angle C A D \\\\ \therefore \quad \theta=30^{\circ} \\\\ \therefore \quad y\ =180^{\circ}-(\beta+\theta) \\\\ \quad \quad\quad =180^{\circ}-\left(110^{\circ}+30^{\circ}\right) \\\\ \quad\quad\quad =40^{\circ} \end{array}$

(b)

$\begin{array}{l} \text { Since } A C=A D, \angle A C D=x \\\\ \text{In}\ \Delta A C D, x+x=150^{\circ} \\\\ \therefore\ 2 x=150^{\circ} \Rightarrow x=75^{\circ}\\\\ A B C D \text { is a cyclic quadrilateral. }\\\\ \therefore x+\phi=180^{\circ}\\\\ \therefore\ \ \phi\ =180^{\circ}-x \\\\ \quad\quad =180^{\circ}-75^{\circ} \\\\ \quad\quad =105^{\circ} \\\\ \text { Since } A B=B C, \angle B A C=y . \\\\ \text { In } \Delta A B C, 2 y+\phi=180^{\circ} \\\\ \therefore\ 2 y=180^{\circ}-\phi \\\\ \quad\quad\ =180^{\circ}-105^{\circ} \\\\ \quad\ y\ \ =37.5^{\circ} \end{array}$

(c)

$\begin{aligned} \angle B O D &=360^{\circ}-240^{\circ} \\\\ & =120^{\circ} \\\\ x &=\displaystyle\frac{1}{2} \angle B O D \\\\ &=\displaystyle\frac{1}{2}\left(120^{\circ}\right) \\\\ &=60^{\circ} \\\\ \theta &=\displaystyle\frac{1}{2}\left(240^{\circ}\right) \\\\ \therefore\quad\theta &=120^{\circ} \\\\ \therefore\quad y &=360^{\circ}-\left(120^{\circ}+120^{\circ}+55^{\circ}\right) \\\\ &=65^{\circ} \end{aligned}$

(d)

In $\Delta A D E$, $\theta=180^{\circ}-\left(30^{\circ}+40^{\circ}\right)=110^{\circ}$ $A C D E$ is a cyclic quadrilateral. $\begin{aligned} \therefore y+ \theta &=180^{\circ} \\\\ \therefore y \quad &=180^{\circ}-\theta \\\\ &=180^{\circ}-110^{\circ} \\\\ &=70^{\circ} \end{aligned}$ In $\Delta A C D, A C =A D$. $\therefore\quad \angle A D C=y$ $A C D E$ is a cyclic quadrilateral. $\begin{aligned} \therefore\quad x+y & =180^{\circ} \\\\ \therefore\quad x \quad &=180^{\circ}-y \\\\ &=180^{\circ}-70^{\circ} \\\\ &=110^{\circ} \end{aligned}$

(e)

Since $O A=O B\ \text{( radii )}$ $\therefore\quad \theta=50^{\circ}$ $A B D E$ is a cyclic quadrilateral. $\therefore\quad x+60^{\circ}+\theta=180^{\circ}$ $\begin{aligned} \therefore \quad x & =180^{\circ}-\left(60^{\circ}+\theta\right) \\\\ & =180^{\circ}-\left(60^{\circ}+50^{\circ}\right) \\\\ & =70^{\circ} \end{aligned}$ $C E$ is a diameter. $\therefore \angle C D E=90^{\circ}$ $\begin{aligned} \therefore \quad x+y & =90^{\circ}\\\\ \therefore \quad y & =90^{\circ}-x\\\\ & =90^{\circ}-70^{\circ}\\\\ & =20^{\circ} \end{aligned}$

(f)

$A B D E$ is a cyclic quadrilateral. $\therefore y=110^{\circ}$ Since $\text{arc}\ A D=\text{arc}\ B C,$ $A D=B C$ $\therefore\quad A B D E$ is isosceles trapezium. $\therefore\quad x=y=110^{\circ}$ $x+z=180^{\circ}$ $\therefore\quad z=180^{\circ}-x$ $\quad\quad =180^{\circ}-110^{\circ}$ $\quad\quad =70^{\circ}$

Ex 9.1-No (2)

$ABC$ is an acute triangle inscribed in $\odot O$, and $OD$ is the perpendicular drawn $O$ to $BC$. Prove that $\angle BOD = \angle BAC$.

$\begin{array}{l} \text { Join }\ O C.\\\\ \text { In }\ \triangle BOD\ \text { and }\ \triangle COD,\\\\ \angle BDO = \angle CDO \ \text{(right angles)}\\\\ OB=OC \ \text{(radii)}\\\\ OD=OD \ \text{(common side)}\\\\ \therefore\quad \triangle B O D \cong \triangle C O D\\\\ \therefore\quad\angle B O D=\angle C O D=\displaystyle\frac{1}{2} \angle B O C\\\\ \text { But }\ \angle B A C=\displaystyle\frac{1}{2} \angle B O C\\\\ \therefore\quad\angle B O D=\angle BAC \end{array}$

Ex 9.1-No (3)

In $\odot O$, two chords $A B$ and $C D$ intersect in the circle at $P$. Show that $\angle A P D=\displaystyle\frac{1}{2}(\angle A O D+\angle B O C)$.

Join $A C$ In $\triangle A P C$, $\begin{aligned} \angle A P D & =\angle P C A+\angle P A C \\\\ & =\angle D C A+\angle B A C \\\\ & =\displaystyle\frac{1}{2} \angle A O D+\displaystyle\frac{1}{2} \angle B O C \\\\ & =\displaystyle\frac{1}{2}(\angle A O D+\angle B O C) \end{aligned}$

Ex 9.1-No (4)

Two circles intersect at $M, N$. From $M$, diameters $M A, M B$ are drawn in each circle. If $A, B$ are joined to $N$, prove that $A N B$ is a straight line.

$\begin{array}{lll} \text{Given} &: & \odot O\ \text{and}\ \odot P\ \text{intersect at}\ M \text{and}\ N.\\\\ & & M A\ \text{and}\ M B\ \text{are diameters of}\ \odot O\ \text{and}\\\\ & & \odot P\ \text{and respectively.}\\\\ \text{To Prove} &: & A N B\ \text{is a straight line.}\\\\ \text{Proof} &: & \text{Join}\ M N.\\\\ & & \text{In}\ \odot O, \angle A N M=90^{\circ} \quad(AM\ \text{is a diameter} )\\\\ & & \text{In}\ \odot P, \angle B N M=90^{\circ} \quad(BM\ \text{is a diameter} )\\\\ & & \therefore\quad \angle A N M+\angle B N M=180^{\circ}\\\\ & & \therefore\quad A N B\ \text{ is a straight line.} \end{array}$

Ex 9.1-No (5)

$OA$ and $O B$ are two radii of a circle meeting at right angles. From $A, B$ two parallel chords $A X, B Y$ are drawn. Prove that $A Y \perp B X$.

$\begin{array}{lll} \text{Given} &: & O A \perp O B, A X \parallel B Y\\\\ \text{To Prove} &: & A Y \perp B X .\\\\ \text{Proof} &: & \text{Let}\ A Y\ \text{cut}\ B X\ \text{at}\ D \\\\ & & \alpha=\gamma \quad(\angle\mathrm{s}\ \text{stand on arc}\ AB )\\\\ & & \text{But}\ \alpha=\gamma=\displaystyle\frac{1}{2} \angle A O B \\\\ & & \alpha=\gamma=\displaystyle\frac{1}{2}\left(90^{\circ}\right)=45^{\circ}\\\\ & & \text{Since} A X \parallel B Y,\\\\ & & \alpha=\beta \quad (\text{alternating} \angle \mathrm{s})\\\\ & & \therefore\quad \beta=45^{\circ}\\\\ & & \text{Since} \angle A D B=\beta+\gamma,\\\\ & & \angle A D B=90^{\circ}\\\\ & & \therefore\quad A Y \perp B X.\\\\ \end{array}$

Ex 9.1-No (6)

Two circles intersect at $R$ and $S$. Two straight lines $A R B$ and $C S D$ are drawn meeting one circle at $A, C$ and the other at $B, D$. Prove that $A C \parallel B D .$ If $A B \parallel C D$, show that $A B=C D$.

$ACSR$ is a cyclic quadrilateral. $\therefore\quad \alpha=\theta$ $BDSR$ is also a cyclic quadrilateral. $\therefore\quad \beta+\theta = 180^{\circ}$ $\therefore\quad \alpha+\theta = 180^{\circ}$ $\therefore\quad A C \parallel B D$ If $A B \parallel C D$, $ABDC$ is a parallelogram. then we can say $A B=C D$

Circles - Definitions and Theorems (Part 1)

Circle and Centre of a Circle

A circle is the set of all points that are at a fixed distance from a fixed point. The fixed point producing a circle is called the centre of the circle.

A circle with centre $O$ is called circle $O$ and denoted by $\odot O$.

ရွေ့လျားမှုမရှိသော အမှတ်တစ်ခုမှ အကွာအဝေးတူ အမှတ်များဖြင့် ဖွဲ့စည်းထားသော အမှတ်အစုအဝေးကို စက်ဝိုင်းဟုခေါ်သည်။ စက်ဝိုင်းတစ်ခုကို ဖြစ်စေသည့် အဆိုပါ ရွေ့လျားမှုမရှိသော အမှတ်ကို ဗဟိုဟုခေါ်သည်။

$O$ ဗဟိုရှိသော စက်ဝိုင်းကို $\odot O$ ဟုခေါ်သည်။

Concentric Circles

Circles having the same centre are called concentric circles.

ဗဟိုတစ်ခုတည်းမှ ဖြစ်ပေါ်လာသော စက်ဝိုင်းများကို ဗဟိုတူစက်ဝိုင်းများ ဟုခေါ်သည်။

Radius

A radius is a segment joining the centre and a point on the circle.

ဗဟိုနှင့် စက်ဝန်းပေါ်ရှိ အမှတ်တစ်ခုကု ဆက်သွယ်ထားသော မျဉ်းပြတ်ကို အချင်းဝက် ဟုခေါ်သည်။

Chord

A segment joining two points on a circle is called a chord of the circle.

စက်ဝန်းပေါ်ရှိ အမှတ်နှစ်ခုကို ဆက်သွယ်ထားသော မျဉ်းပြတ်ကို လေးကြိုးဟု ခေါ်သည်။

Diameter

A diameter is a chord passing through the centre of a circle. In a circle, a diameter is a longest chord.

ဗဟိုကိုဖြတ်၍ဆွဲသော လေးကြိုးကို အချင်းမျဉ်းဟု ခေါ်သည်။ အချင်းမျဉ်းသည် စက်ဝိုင်းတစ်ခု၏ အရှည်းဆုံး လေးကြိုးလည်း ဖြစ်သည်။

Congruent Circles

Circles having the same radius are called congruent circles.

အချင်းဝက် တူညီသော စက်ဝိုင်းများကို ထပ်တူညီစက်ဝိုင်းများ ဟုခေါ်သည်။

Secant

A secant of a circle is a line that intersects the circle at two points.

စက်ဝန်းပေါ်ရှိ အမှတ်နှစ်ခုကို ဖြတ်သွားသော မျဉ်းတစ်ကြောင်းကိုဝန်းဖြတ်မျဉ်း ဟုခေါ်သည်။

Tangent

A line touching the circle at one point only is called a tangent to the circle. This touching point is called the point of contact.

စက်ဝိုင်းကို အမှတ်တစ်ခု၌ ထိသွားသော မျဉ်းတစ်ကြောင်းကို ဝန်းထိမျဉ်း ဟုခေါ်သည်။

Arc, Semicircle, Minor Arc, Major Arc

An arc is a part of a circle.

စက်ဝိုင်း၏ အစိပ်အပိုင်းကို အဝန်းပိုင်း ဟုခေါ်သည်။

A semicircle is a half part of a circle.

စက်ဝိုင်း၏ တစ်ဝက်တိတိရှိသော အစိပ်အပိုင်းကို စက်ဝိုင်းခြမ်း ဟုခေါ်သည်။

Arc $ACB$ is a semicircle.

A minor arc is an arc shorter than a semicircle.

စက်ဝိုင်းခြမ်းအောက် အလျားတိုသော အဝန်းပိုင်းကို minor arc ဟုခေါ်သည်။

A major arc is an arc longer than a semicircle.

စက်ဝိုင်းခြမ်းထက် အလျားရှည်သော အဝန်းပိုင်းကို major arc ဟုခေါ်သည်။

Arc $AB$ is a minor arc and arc $ACB$ is a major arc.

Central Angles and Inscribed Angles

A central angle is an angle subtended by an arc (or chord) of a circle at the centre.

ထောင့်လက်တံနှစ်ခု၏ အစွန်းမှတ်များသည် အဝန်းပိုင်း၏ အစွန်းနှစ်ဘက် (သို့မဟုတ်) လေးကြိုး၏ အစွန်းနှစ်ဘက်၌ ရှိပြီး ထောင်းစွန်းမှတ်သည် စက်ဝိုင်းတစ်ခု၏ ဗဟို၌ရှိသော ထောင့်ကို ဗဟိုခံထောင့်ဟု ခေါ်သည်။

An inscribed angle is an angle subtended by an arc (or chord) of a circle at a point on the other arc.

ထောင့်လက်တံနှစ်ခု၏ အစွန်းမှတ်များသည် အဝန်းပိုင်း၏ အစွန်းနှစ်ဘက် (သို့မဟုတ်) လေးကြိုး၏ အစွန်းနှစ်ဘက်၌ ရှိပြီး ထောင်းစွန်းမှတ်သည် စက်ဝိုင်းတစ်ခု၏ အဝန်း၌ရှိသော ထောင့်ကို အဝန်းခံထောင့်ဟု ခေါ်သည်။

Theorem 1

The central angle is twice the inscribed angle subtended by the same arc.

အဝန်းပိုင်းတစ်ခုမှ စက်ဝိုင်း၏ ဗဟိုတွင်ခံဆောင်ထားသော ထောင့်သည် အဆိုပါအဝန်းပိုင်းကိုပင် တစ်ဖက်အဝန်းပိုင်းတွင် ခံဆောင်ထားသော ထောင့်၏ နှစ်ဆရှိသည်။

Visual Proof

Corollary 1.1

Inscribed angles subtended by the same arc are equal.

တူညီသော အဝန်းပိုင်းတစ်ခုမှ အခြားအဝန်းပိုင်းတစ်ခုတွင် ခံဆောင်ထားသော အဝန်းခံထောင့်များ တူညီကြသည်။

Visual Proof

Corollary 1.2

An inscribed angle subtended by a diameter is a right angle.

အချင်းမျဉ်းကို အဝန်းပိုင်း၌ ခံဆောင်ထားသော အဝန်းခံထောင့်သည် ထောင့်မှန်တစ်ခု ဖြစ်သည်။

Cyclic Quadrilateral

A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral.

စက်ဝိုင်းပေါ်တွင် ထောင့်စွန်းမှတ်များရှိသော စတုဂံတစ်ခုကို စက်ဝိုင်းတွင်းကျစတုဂံ ဟုခေါ်သည်။

$ABCD$ is a cyclic quadrilateral.

Theorem 2

Opposite angles of a cyclic quadrilateral are supplementary.

စက်ဝိုင်းတွင်းကျစတုဂံ စတုဂံတစ်ခု၏ အတွင်းမျက်နှာချင်းဆိုင် ထောင့်တစ်စုံသည် ထောင့်ဖြောင့်ဖြည့်ဘက်များ ဖြစ်ကြသည်။

Corollary 2.1

The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle of the quadrilateral

စက်ဝိုင်းတွင်းကျစတုဂံ စတုဂံတစ်ခု၏ အနားတစ်ဖက်ကို ဆက်ဆွဲ၍ ဖြစ်ပေါ်လာသော အပြင်ထောင့်သည် အတွင်းမျက်နှာချင်းဆိုင်ထောင့်နှင့် အမြဲ‌ညီသည်။

Theorem 3

In the same circle or in congruent circles,

1. equal arcs subtend equal central angles,
2. arcs subtending equal central angles are equal.

စက်ဝိုင်းတစ်ခုတည်းတွင် ဖြစ်စေ၊ ထပ်တူညီသော စက်ဝိုင်းများတွင် ဖြစ်စေ၊ အဝန်းပိုင်းများ ထပ်တူညီလျှင် ဗဟိုခံထောင့်များ တူညီပြီး၊ အပြန်အလှန်အားဖြင့် ဗဟိုခံထောင့်များ တူညီလျှင် အဝန်းပိုင်းများ ထပ်တူညီသည်။

Theorem 4

In the same circle or in congruent circles, two inscribed angles are equal if and only if the corresponding arcs are equal.

စက်ဝိုင်းတစ်ခုတည်းတွင် ဖြစ်စေ၊ ထပ်တူညီသော စက်ဝိုင်းများတွင် ဖြစ်စေ၊ အဝန်းပိုင်းများ ထပ်တူညီလျှင် အဝန်းခံထောင့်များ တူညီပြီး၊ အပြန်အလှန်အားဖြင့် အဝန်းခံထောင့်များ တူညီလျှင် အဝန်းပိုင်းများ ထပ်တူညီသည်။

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

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

$\therefore\ y= -5$.