1.         The base of an equilateral triangle with side $ \displaystyle 2a$ lies along the $ \displaystyle y$-axis such that the mid-point of the base is at the origin. Find the vertices of triangle.

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အနားတစ္ဖက္ အလ်ား $ \displaystyle 2a$ ရွိေသာ သံုးနားညီ ႀတိဂံ၏ အေျခအနား (အနား တစ္ဖက္) သည္ $ \displaystyle y$ ၀င္႐ိုးေပၚတြင္ ရွိၿပီး ၎အနား၏ အလယ္မွတ္သည္ origin $ \displaystyle (0, 0)$ ျဖစ္လွ်င္ ႀတိဂံ၏ ေထာင့္စြန္းမွတ္မ်ားကို ရွာပါ။

ေပးထားခ်က္အရ ...

$ \displaystyle (0,0)$ က အေျခအနားရဲ့ အလယ္မွတ္ျဖစ္လို႔ ေထာင္စြန္းမွတ္ ႏွစ္ခုက သည္ $ \displaystyle (0, a)$ နဲ႔ သည္ $ \displaystyle (0, -a)$ ျဖစ္ပါတယ္။

သံုးနားညီ ႀတိဂံ ျဖစ္လို႔ က်န္တဲ့ ေထင့္စြန္းမွတ္က သည္ $ \displaystyle x$ ၀င္႐ိုးေပၚမွာ ရွိပါမယ္။

ဒါ့ေၾကာင့္ က်န္ေထာင့္စြန္းမွတ္က သည္ $ \displaystyle (x,0)$ ကို လို႔ ထားလိုက္ပါမယ္။

ဒါဆိုရင္ ပုစာၦရဲ့ ေမးခ်က္အရ သည္ $ \displaystyle x$ ကို ရွာေပးရင္ ရပါၿပီ။

ပံုကိုၾကည့္ပါ။

cg-01

$ \displaystyle (x, 0)$ ဟာ positive $ \displaystyle x$-axis မွာ ရွိႏိုင္သလို negative $ \displaystyle x$-axis ဘက္မွာလဲ ရွိႏိုင္ပါတယ္။ ဒါေပမယ့္ ပုစာၦတြက္ရာ မွ ႏွစ္ခုခြဲ သတ္မွတ္ဖို႔မလိုပဲ အေျဖမွာ $ \displaystyle \pm x$ နဲ႔ ထြက္လာမွာ ျဖစ္ပါတယ္။ $ \displaystyle ΔAOC$ ဟာ ေထာင့္မွတ္ႀတိဂံျဖစ္လို႔ $ \displaystyle OC$ ႐ဲ့ အလ်ားကို Pythagoras' Theorem နဲ႔ ရွာမယ္ဆိုရင္ $ \displaystyle x$ တန္ဖိုးမ်ား ရပါၿပီ။

တြက္ၾကရေအာင္...

Let the base of the triangle be $ \displaystyle AB$.

Since the mid-point of $ \displaystyle AB$ is origin, $ \displaystyle AO=OB=a$.

$ \displaystyle \therefore \ \ $ The points $ \displaystyle A$ and $ \displaystyle B$ are $ \displaystyle (0, a)$ and $ \displaystyle (0,-a).$

Since $ \displaystyle ΔAOC$ is a right triangle, by Pythagoras' theorem,

$\displaystyle \begin{array}{l}O{{C}^{2}}=A{{C}^{2}}-A{{O}^{2}}\\\\\ \ \ \ \ \ \ \ \ ={{\left( {2a} \right)}^{2}}-{{a}^{2}}\\\\\ \ \ \ \ \ \ \ \ =4{{a}^{2}}-{{a}^{2}}\\\\\ \ \ \ \ \ \ \ \ =3{{a}^{2}}\\\\\therefore OC=\pm \sqrt{{3a}}\\\\\therefore C=(\sqrt{{3a}},0)\ \operatorname{and}\ D=(-\sqrt{{3a}},0)\end{array}$

There fore The vertices of the triangle are $ \displaystyle (0,a)$, $ \displaystyle (0, -a)$, $ \displaystyle (\sqrt{3}a, 0)$ or $ \displaystyle (0,a)$, $ \displaystyle (0, -a)$, $ \displaystyle (-\sqrt{3}a, 0)$.

2.         Find a point on the x-axis which is equidistant from the points $ \displaystyle (7, 6)$ and $ \displaystyle (3, 4).$

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$\displaystyle (7, 6)$ နဲ႔ $\displaystyle (3, 4)$ တို႔မွ တူညီစြာ ကြာေ၀းေသာ $\displaystyle x$ ၀င္႐ိုးေပၚရွိ အမွတ္ကို ရွာပါ။

$\displaystyle x$ ၀င္႐ိုးေပၚရွိ အမွတ္ လို႔ ေျပာထားလို႔ $\displaystyle y$-coordinate က $\displaystyle 0$ လို႔ နားလည္ထားရပါမယ္္။

လိုခ်င္တဲ့ အမွတ္ကို $\displaystyle (a, 0)$ လို႔ ထားလိုက္မယ္။

ဒါဆိုရင္ ပုစာၦအရ $\displaystyle (a, 0)$ ဟာ $\displaystyle (7, 6)$ နဲ႔ $\displaystyle (3, 4)$ ကေနအကြာအေ၀းတူပါတယ္။

ပံုကို ၾကည့္ပါ။

cg-02

ဒါေၾကာင့္ distance formula နဲ႔ distance ႏွစ္ခုကို ရွာၿပီး ညီေပးလိုက္ရင္ အေျဖရပါၿပီ။

တြက္ၾကရေအာင္...

Let the required point be $\displaystyle (a, 0)$.

$ \displaystyle \ \ \ \ $ Distance between $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$

$ \displaystyle \ =\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$

$ \displaystyle \therefore \ \ $ Distance between $\displaystyle (7, 6)$ and $\displaystyle (a, 0)$

$ \displaystyle \begin{array}{l}\ =\sqrt{{{{{\left( {a-7} \right)}}^{2}}+{{{\left( {6-0} \right)}}^{2}}}}\\\\\ =\sqrt{{{{a}^{2}}-14a+49+36}}\\\\\ =\sqrt{{{{a}^{2}}-14a+85}}\end{array}$

$\displaystyle \therefore \ \ $ Distance between $\displaystyle (3, 4)$ and $\displaystyle (a, 0)$

$ \displaystyle \begin{array}{l}\ =\sqrt{{{{{\left( {a-3} \right)}}^{2}}+{{{\left( {4-0} \right)}}^{2}}}}\\\\\ =\sqrt{{{{a}^{2}}-6a+9+16}}\\\\\ =\sqrt{{{{a}^{2}}-6a+25}}\end{array}$

By the problem,

$ \displaystyle \begin{array}{l}\ \ \ \sqrt{{{{a}^{2}}-14a+85}}=\sqrt{{{{a}^{2}}-6a+25}}\\\\\ \ \ {{a}^{2}}-14a+85={{a}^{2}}-6a+25\\\\\therefore \ 8a=60\\\\\therefore \ a=\displaystyle \frac{{15}}{2}\end{array}$

Thefore the required point on $ \displaystyle x$-axis is $ \displaystyle \left( {\frac{{15}}{2},0} \right).$

3.         A quadrilateral has the vertices at the points$ \displaystyle (-4, 2)$, $ \displaystyle (2, 6)$, $ \displaystyle (8, 5)$ and $ \displaystyle (9, -7).$ Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram.

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$ \displaystyle (-4,2)$, $ \displaystyle (2,6)$, $ \displaystyle (8,5)$ and $ \displaystyle (9,-7)$ အမွတ္ေလးမွတ္ကို ဆက္ထားေသာ စတုဂံတစ္ခုရွိသည္။ ၎စတုဂံ၏ အနားမ်ား၏ အလယ္မွတ္မ်ားသည္ အနားၿပိဳင္စတုဂံ တစ္ခု၏ ေထာင့္စြန္းမွတ္မ်ား ျဖစ္ေၾကာင္း သက္ေသျပပါ။


သက္ဆိုင္ရာ အနားမ်ား၏ အလယ္မွတ္ကိုရွာမယ္။

အမွတ္ေတြကို တြဲတဲ့အခါ အစဥ္လိုက္တြဲဖို႔လိုပါတယ္။ တစ္ခုေက်ာ္ တြဲလိုက္ရင္ အနားေတြ မဟုတ္ပဲ ေထာင့္ျဖတ္မ်ဥ္းေတြ ျဖစ္သြားမယ္။

အလယ္မွတ္ေတြကို တစ္စံုစီတြဲၿပီး သက္ဆိုင္ရာ မ်ဥ္းျပတ္မ်ားရဲ့ slope ကို ရွာမယ္။

slope ေလးခုမွာ ႏွစ္ခုတစ္စံုစီ ညီခဲ့မယ္ဆိုရင္ အနားၿပိဳင္စတုဂံ ျဖစ္ၿပီေပါ့။

ပံုကိုၾကည့္ပါ။

cg-03

တြက္ၾကည့္ရေအာင္။

Let $ \displaystyle A=(-4,2)$, $ \displaystyle (2,6)$, $ \displaystyle (8,5)$ and $ \displaystyle (9,-7)$.

Let the mid-points of $ \displaystyle AB$, $ \displaystyle BC$, $ \displaystyle CD$, and $ \displaystyle DA$ be $ \displaystyle P, Q, R $, and $ \displaystyle S$ respectively.

$ \displaystyle \therefore P=\left( {\frac{{-4+2}}{2},\frac{{2+6}}{2}} \right)$

$ \displaystyle \ \ \ \ \ \ =\left( {-1,4} \right)$

$ \displaystyle \ \ \ \ Q=\left( {\frac{{2+8}}{2},\frac{{6+5}}{2}} \right)$

$ \displaystyle \ \ \ \ \ \ =\left( {5,\frac{{11}}{2}} \right)$

$ \displaystyle \ \ \ \ R=\left( {\frac{{8+9}}{2},\frac{{5-7}}{2}} \right)$

$ \displaystyle \ \ \ \ \ \ =\left( {\frac{{17}}{2},-1} \right)$

$ \displaystyle \ \ \ \ S=\left( {\frac{{9-4}}{2},\frac{{-7+2}}{2}} \right)$

$ \displaystyle \ \ \ \ \ \ =\left( {\frac{5}{2},-\frac{5}{2}} \right)$

$ \displaystyle \begin{array}{l}\therefore \ \ {{m}_{{PQ}}}=\displaystyle \frac{{\displaystyle \frac{{11}}{2}-4}}{{5-(-1)}}\\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{4}\\\\\ \ \ {{m}_{{QR}}}=\displaystyle \frac{{-1-\displaystyle \frac{{11}}{2}}}{{\displaystyle \frac{{17}}{2}-5}}\\\\\ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{13}}{7}\\\\\ \ \ {{m}_{{RS}}}=\displaystyle \frac{{-\displaystyle \frac{5}{2}-(-1)}}{{\displaystyle \frac{5}{2}-\displaystyle \frac{{17}}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{4}\\\\\ \ \ {{m}_{{PS}}}=\displaystyle \frac{{-\displaystyle \frac{5}{2}-4}}{{\displaystyle \frac{5}{2}-(-1)}}\\\\\ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{13}}{7}\\\\\therefore \ \ {{m}_{{PQ}}}={{m}_{{RS}}}\\\\\therefore \ \ PQ\parallel RS\\\\\therefore \ \ {{m}_{{QR}}}={{m}_{{PS}}}\\\\\therefore \ \ QR\parallel PS \end{array}$

$ \displaystyle \therefore P, Q, R$ and $ \displaystyle S$ are vertices of a parallelogram.

4.        The vertices of a triangle are $ \displaystyle A(3, 8)$, $ \displaystyle B(-1, 2)$ and $ \displaystyle C(6, -6).$ Find the slope of

(i) side $ \displaystyle BC;$

(ii) altitude through $ \displaystyle C;$

(iii) median through $ \displaystyle A;$

(iv) perpendicular bisector of side $ \displaystyle CA.$

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            $ \displaystyle ΔABC$ ၏ ေထာင့္စြန္းမွတ္မ်ား သည္ $ \displaystyle A(3,8)$, $ \displaystyle B(−1,2)$ and $ \displaystyle C(6,−6)$ တို႔ ျဖစ္ၾကလွ်င္...

(i) $ \displaystyle BC$ ၏ slope ကို ရွာပါ။

(ii) $ \displaystyle C$ ကို ျဖတ္သြားေသာ အျမင့္မ်ဥ္း ၏ slope ကို ရွာပါ။

(iii) $ \displaystyle A$ ကို ျဖတ္သြားေသာ အလယ္မ်ဥ္း ၏ slope ကို ရွာပါ။

(iv) $ \displaystyle CA$ ေပၚရွိ ေထာင့္မွတ္က် ထက္၀က္ပိုင္းမ်ဥ္း၏ slope ကို ရွာပါ။


slope ရဲ့ ပံုေသးနည္း $ \displaystyle m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$ ကို သိရပါမယ္။

$\displaystyle {{l}_{1}}\bot {{l}_{2}}$ ျဖစ္တဲ့အခါ $ \displaystyle {{m}_{1}}=-\frac{1}{{{{m}_{2}}}}$ ဆိုတာကို သိရပါမယ္။

Midpoint between $ \displaystyle (x_1,y_1)$ and $ \displaystyle (x_2,y_2)$ = $ \displaystyle \left( {\frac{{{{x}_{1}}+{{x}_{2}}}}{2},\frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$ ဆိုတာကို သိရပါမယ္။

ပံုကို ၾကည့္ပါ။


cg-04abc

(i) က ပံုေသနည္းကို တိုက္႐ိုက္အသံုးခ်ၿပီး ရွာ႐ံုပဲ ျဖစ္ပါတယ္။

(ii) က $ \displaystyle CF$ ရဲ့ slope ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။

$ \displaystyle CF$ က $ \displaystyle AB$ ေပၚကို ေထာင့္မတ္က်လို႔ $ \displaystyle {{m}_{{CF}}}=-\frac{1}{{{{m}_{{AB}}}}}$ လို႔သိရပါမယ္။ (iii) က $ \displaystyle AD$ ရဲ့ slope ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။

$ \displaystyle AD$ က median ျဖစ္တာေၾကာင့္ $ \displaystyle D$ က $ \displaystyle BC$ ရဲ့ အလယ္မွတ္ျဖစ္တယ္လို႔ သိရပါမယ္။

$ \displaystyle D$ ကို midpoint formula ကို သံုးၿပီး ရွာရပါမယ္။ $ \displaystyle D$ ကို ရရင္ $ \displaystyle AD$ ရဲ့ slope ကို ရွာႏိုင္ပါၿပီ။

(iii) က green dotted line $ \displaystyle (l)$ ရဲ့ slope ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။

$ \displaystyle l$ က $ \displaystyle AC$ ေပၚထာင့္မတ္က်လို႔ $ \displaystyle {{m}_{l}}=-\frac{1}{{{{m}_{{AC}}}}}$ ျဖစ္တယ္လို႔ သိရမယ္။

slope ကို သာ ရွာခိုင္းျခင္း ျဖစ္လို႔ bisector ဆိုတဲ့ စကားလံုးကို ထည့္သြင္းစဥ္းစားရန္ မလိုေတာ့ပါ။

တြက္ၾကည့္ၾကစို႔။



(i) $ \displaystyle \ \ \ \ \ \ B=(−1,2)$ and $ \displaystyle C=(6,−6)$

$ \displaystyle \begin{array}{l}\therefore \ \ {{m}_{{BC}}}=\displaystyle \frac{{-6-2}}{{6-(-1)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-8}}{7}\end{array}$

(ii) Let the altitude through $ \displaystyle C$ be $ \displaystyle CF$.

Since $ \displaystyle CF\bot AB$, $ \displaystyle {{m}_{{CF}}}=-\frac{1}{{{{m}_{{AB}}}}}.$

$ \displaystyle \begin{array}{l}\ \ \ \ \ {{m}_{{AB}}}=\displaystyle \frac{{8-2}}{{3-(-1)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{6}{4}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{3}{2}\\\\\therefore \ \ \ {{m}_{{CF}}}=\displaystyle -\frac{2}{3}\end{array}$

(iii) Let the median through $ \displaystyle A$ be $ \displaystyle AD$ where $ \displaystyle D$ is the midpoint of $ \displaystyle BC.$

$ \displaystyle \begin{array}{l}\therefore \ \ D=\left( { \displaystyle \frac{{-1+6}}{2}, \displaystyle \frac{{2-6}}{2}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( { \displaystyle \frac{5}{2},-2} \right)\\\\\therefore \ \ {{m}_{{AD}}}= \displaystyle \frac{{-2-8}}{{\frac{5}{2}-3}}\\\\\ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{-10}}{{ \displaystyle -\frac{1}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ =20\end{array}$

(iv) Let the perpendicular bisector of $ \displaystyle CA$ be $ \displaystyle l.$

Since $ \displaystyle l\bot CA$, $ \displaystyle {{m}_{{l}}}=-\frac{1}{{{{m}_{{CA}}}}}.$

$ \displaystyle \begin{array}{l}\therefore \ \ {{m}_{{CA}}}=\displaystyle \frac{{8-(-6)}}{{3-6}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle -\frac{{14}}{3}\\\\\therefore \ \ {{m}_{l}}=\displaystyle \frac{3}{{14}}\end{array}$

5.         Find the equation of a line which passes through the origin and mid-point of the line segment joining the points $ \displaystyle A(8, 0)$ and$ \displaystyle B(0, -4).$

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            $ \displaystyle A(8,0)$ ႏွင့ $ \displaystyle B(0,−4)$ တို႔၏ အလယ္မွတ္ႏွင့္ Origin $ \displaystyle (0, 0)$ တို႔ကို ျဖတ္သြားေသာ မ်ဥ္းတစ္ေၾကာင္း ၏ ညီမွ်ျခင္း (line equation) ကို ရွာေပးပါ။


line equation ကို ရွာဖို႔ slope-intercept form, point-slope form, two-point form တစ္ခုခုကို သံုးရပါတယ္။

ရွာလိုေသာ line ဟာ Origin $ \displaystyle (0, 0)$ နဲ႔ $ \displaystyle AB$ ရဲ့ အလယ္မွတ္ ကို ျဖတ္သြားတယ္လို႔ ေျပာထားလို႔ two-point form ကို သံုးရင္ ရပါၿပီ။

two-point form ဆိုတာက အမွတ္ႏွစ္ခုကိုသံုးၿပီး slope ကို ရွာမယ္။ ၿပီးေတာ့ point-slope form နဲ႔ line equation ကို ဆက္ၿပီး ရွာမယ္။

ပံုကို ၾကည့္ပါ။

cg-05 $ \displaystyle AB$ ရဲ့ အလယ္မွတ္ ကို ရွာမယ္။ Origin နဲ႔ အလယ္မွတ္ကို ဆက္သြယ္ထားတဲ့ $ \displaystyle l$ ရဲ့ slope ကို ရွာမယ္။ ျမင္တတ္ရင္ $ \displaystyle (x_1, y_1)=(0,0)$ ျဖစ္လို႔ $ \displaystyle m= \frac{y_2}{x_2}$ လို႔ တန္းသိႏိုင္ပါတယ္။

တြက္ၾကည့္ရေအာင္။


Let the mid-point of $ \displaystyle AB$ be $ \displaystyle M.$

$ \displaystyle \begin{array}{l}\therefore \ \ \ M=\left( {\frac{{8+0}}{2},\displaystyle \frac{{0-4}}{2}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ =\left( {4,-2} \right)\end{array}$

Let the line joining the origin and $ \displaystyle M$ be $ \displaystyle l.$

$ \displaystyle \begin{array}{l}\therefore \ \ \ {{m}_{{_{l}}}}=\displaystyle \frac{{-2-0}}{{4-0}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle -\frac{1}{2}\end{array}$

Therefore, the equation of $ \displaystyle l$ is

$ \displaystyle \begin{array}{l}\ \ \ \ y-0= \displaystyle -\frac{1}{2}(x-0)\\\\\therefore \ \ y= \displaystyle -\frac{1}{2}x\end{array}$

6.         If $ \displaystyle A(5, -3)$, $ \displaystyle B(8, 2)$, $ \displaystyle C(0, 0)$ are the vertices of a triangle, show that median from $ \displaystyle A$ is perpendicular to $ \displaystyle BC.$

Show/Hide Solution
$ \displaystyle ΔABC$ တြင္ အမွတ္ $ \displaystyle A$ ကို ျဖတ္သြားေသာ အလယ္မ်ဥ္းသည္ $ \displaystyle BC$ ေပၚတြင္ ေထာင့္မတ္က်ေၾကာင္း သက္ေသျပပါ။

$ \displaystyle BC$ ရဲ့ အလယ္မွတ္က $ \displaystyle M$ လို႔ဆိုပါစို႔။ ဒါဆိုရင္ $ \displaystyle AM$ က သတ္မွတ္ထားတဲ့ အလယ္မ်ဥ္း ျဖစ္မယ္။

ပံုကိုၾကည့္ပါ။

cg-06

အလယ္မွတ္ $ \displaystyle M$ ကို ရွာမယ္။

$ \displaystyle AM$ ရဲ့ slope ($ \displaystyle {{m}_{{_{{AM}}}}}$) ကို ရွာမယ္။

$ \displaystyle BC$ ရဲ့ slope ($ \displaystyle {{m}_{{_{{BC}}}}}$) ကို ရွာမယ္။

$ \displaystyle {{m}_{{_{{AM}}}}}=\displaystyle -\frac{1}{{\text{ }{{m}_{{_{{BC}}}}}}}$ ျဖစ္ရင္ $ \displaystyle AM\bot BC$ ျဖစ္ၿပီေပါ့။

တြက္ၾကည့္ရေအာင္...


Let $ \displaystyle M$ be the mid-point of $ \displaystyle BC$.

$ \displaystyle \begin{array}{l}\therefore \ \ M=\left( {\displaystyle \frac{{8+0}}{2},\displaystyle \frac{{2+0}}{2}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {4,1} \right)\end{array}$

Hence, $ \displaystyle AM$ is the median from $ \displaystyle A.$

$ \displaystyle \begin{array}{l}\therefore \ \ {{m}_{{_{{AM}}}}}=\displaystyle \frac{{1-(-3)}}{{4-5}}=-4\\\\\ \ \ \ {{m}_{{_{{BC}}}}}=\displaystyle \frac{{0-2}}{{0-8}}=\displaystyle \frac{1}{4}\\\\\therefore \ \ {{m}_{{_{{AM}}}}}=\displaystyle -\frac{1}{{{{m}_{{_{{BC}}}}}}}\\\\\therefore \ \ AM\bot BC\end{array}$

7.         The points $ \displaystyle (1, 3)$ and $ \displaystyle (5, 1)$ are the opposite vertices of a rectangle. The other two vertices lie on the line $ \displaystyle y = 2x + c.$ Find $ \displaystyle c$ and the remaining vertices.

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cg-07

Let the given rectangle be $ \displaystyle ABCD$ where $ \displaystyle A$ and $ \displaystyle C$ be $ \displaystyle (1, 3)$ and $ \displaystyle (5, 1)$ respectively.

By the problem, $ \displaystyle B$ and $ \displaystyle D$ lie on the line $ \displaystyle y = 2x + c.$

Let $ \displaystyle M$ be the midpoint of $ \displaystyle AC$.

$ \displaystyle \therefore \ \ M=\left( {\frac{{1+5}}{2},\frac{{3+1}}{2}} \right)=\left( {3,2} \right)$

Since $ \displaystyle ABCD$ is a rectangle, $ \displaystyle AC$ and $ \displaystyle BD$ bisects each other.

So, $ \displaystyle M$ also lies on $ \displaystyle BD$ and hence $ \displaystyle M$ lies on the line $ \displaystyle y = 2x + c.$

And, the point $ \displaystyle \left( {3,2} \right)$ satisfies the equation $ \displaystyle y = 2x + c.$

$ \displaystyle \therefore \ \ 2=2(3)+c\Rightarrow c=-4$

$ \displaystyle \therefore \ \ $ The given line is $ \displaystyle y = 2x - 4.$

Let $ \displaystyle P\left( {x,2x-4} \right)$ represents any point $ \displaystyle B$ or $ \displaystyle D.$

[$ \displaystyle \left( {x,2x-4} \right)$ သည္ ေယ်ဘုယ်ပံုစံျဖစ္သည့္အတြက္ $ \displaystyle B$ ႏွင့္ $ \displaystyle D$ မည္သည့္အမွတ္ကို မဆို ကိုယ္စားျပဳႏိုင္ပါသည္။ သို႔ေသာ္ ပုစာၦတြက္ရာတြင္ တိက်ေသာ သေဘာေဆာင္ေစရန္ နာမည္ကို P ဟု သတ္မွတ္ျခင္း ျဖစ္ပါသည္။]

Since $ \displaystyle ABCD$ is a rectangle,

$ \displaystyle \begin{array}{l}\ \ \ \ AP\bot CP\\\\\therefore \ \ {{m}_{{_{{AP}}}}}\cdot {{m}_{{_{{CP}}}}}=-1\\\\\therefore \ \ \displaystyle \frac{{2x-4-3}}{{x-1}}\ \cdot \displaystyle \frac{{2x-4-1}}{{x-5}}=-1\\\\\therefore \ \ \displaystyle \frac{{2x-7}}{{x-1}}\ \cdot \displaystyle \frac{{2x-5}}{{x-5}}=-1\\\\\therefore \ \ \displaystyle \frac{{4{{x}^{2}}-24x+35}}{{{{x}^{2}}-6x+5}}\ =-1\\\\\therefore \ \ 5{{x}^{2}}-30x+40\ =0\\\\\therefore \ \ {{x}^{2}}-6x+8\ =0\\\\\therefore \ \ (x-2)(x-4)=0\\\\\therefore \ \ x=2\ (\text{or})\ x=4\\\\\therefore \ \ y=0\ (\text{or})\ y=4\end{array}$

Therefore, the other vertices are $ \displaystyle (2, 0)$ and $ \displaystyle (4, 4).$

8.         Show that the points $ \displaystyle (5, 1),$ $ \displaystyle (1, -1)$ and $ \displaystyle (11, 4)$ lie on a straight line. Also find the equation of that line.


9.         The vertices of a triangle are $ \displaystyle A(10, 4)$, $ \displaystyle B(-4, 9)$ and $ \displaystyle C(-2, -1).$ Find the equation of the altitude through $ \displaystyle A.$


10.      $ \displaystyle A(10, -4)$, $ \displaystyle B(-4, 9)$, $ \displaystyle C(-2, -1)$ are the vertices of $ \displaystyle ΔABC.$ Find the equation of the median through $ \displaystyle A.$


11.      The perpendicular from the origin to the line $ \displaystyle l$ meets at the point $ \displaystyle (-2, 9),$ find the $ \displaystyle x$- and $ \displaystyle y$-intercept of $ \displaystyle l$.


12.      Two consecutive sides of a parallelogram $ \displaystyle 4x + 5y = 0$ and $ \displaystyle 7x + 2y = 0.$ If the equation of one diagonal is $ \displaystyle 11x + 7y = 9,$ find the equation of the other diagonal.


13.      The points $ \displaystyle A(3, 7), B(6, 2)$ and $ \displaystyle C (2, k)$ are the vertices of a triangle. Find the possible values of $ \displaystyle k$ for which $ \displaystyle ΔABC$ is a right triangle.


14.      If $ \displaystyle P$ lies on the $ \displaystyle y$-axis, $ \displaystyle Q$ has coordinates $ \displaystyle (4, 0),$ and $ \displaystyle PQ$ passes through the point $ \displaystyle R(2, 4),$ what is the area of $ \displaystyle ΔOPQ?$


15.      The diagram shows points $ \displaystyle A(1, 0)$, $ \displaystyle B(2, 4)$ and $ \displaystyle C(6, 1).$ The point $ \displaystyle D$ lies on $ \displaystyle BC$ such that $ \displaystyle AD\bot BC$.


cg-diagram1


(i) Show that the equation of $ \displaystyle BC$ is $ \displaystyle 3x + 4y - 22 = 0.$

(ii) Find the length of $ \displaystyle AD.$

(iii) Hence, or otherwise, find the area of $ \displaystyle ΔABC.$


(၉)တန္းဟာ အေျခခံကို စသင္ရသလို၊ ျမန္မာလို သင္ရာမွ အဂၤလိပ္လို ေျပာင္းသင္ရပါတယ္။ ဒီအခါမွာ အေျခခံ အားနည္းခဲ့တဲ့ ကေလးတခ်ိဳ႕ အတြက္ေတာ့ အခက္အခဲရွိပါတယ္။ အားနည္းတယ္ဆိုၿပီး အားနည္းသလို ထားလိုက္ဖို႔ မျဖစ္ႏိုင္ပါဘူး (မျဖစ္သင့္ပါဘူး)။ စကားလံုးေတြကို မွန္ကန္စြာသိဖို႔ ဖတ္တတ္ဖို႔ လိုပါတယ္။ သခၤ်ာသင္တာ အဂၤလိပ္စာက အေရးမႀကီးပါဘူးဆိုတဲ့ ယူဆခ်က္ဟာ မမွန္ႏိုင္ပါဘူး။ သူ႔ဘာသာစကားကို သင္ရင္ သူ႔စည္းမ်ဥ္းေတြကို လိုက္နာဖို႔ တာ၀န္ရွိပါတယ္။ စကားလံုးအသံုးအႏႈန္း (usage) နဲ႔ အသံထြက္ (pronunciation) က အေရးႀကီးပါတယ္။ ၿပီးၿပီးေရာ အသံထြက္လိုက္လို႔ တသက္လံုး အမွတ္မွားခဲ့ရင္ သင္ေပးလိုက္သူ ဆရာမွာ တာ၀န္မကင္းပါဘူး။

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အခုေပးလိုက္တဲ့ ပုစာၦေတြမွာ ပါတဲ့ စကားလံုး တခ်ိဳ႕ရဲ့ အသံထြက္ေတြကို voice clip နဲ႕ ပူးတြဲတင္ေပး လိုက္ပါတယ္။ မွန္ကန္ေသာ အသံထြက္မ်ားကို မွတ္သားႏိုင္ၾကပါေစေၾကာင္း ...


vertices                      ➨
[ေထာင့္စြန္းမွတ္မ်ား]

equilateral                  ➨
[သံုးနားညီ]


equidistant                 ➨
[တူညီစြာ ကြာေ၀းေသာ]


altitude                     ➨
[အျမင့္မ်ဥ္း]


median                     ➨
[အလယ္မ်ဥ္း]


bisector                    ➨
[ထက္၀က္ပိုင္းမ်ဥ္း]


perpendicular            ➨
[ေထာင့္မတ္က်ေသာ/ေထာင့္မတ္က်မ်ဥ္း]


parallel                     ➨
[ၿပိဳင္ေသာ]


segment                   ➨
[မ်ဥ္းျပတ္]


quadrilateral             ➨
[စတုဂံ]


rectangle                 ➨
[ေထာင့္မွန္ စတုဂံ]


parallelogram           ➨
[အနားၿပိဳင္ စတုဂံ]


trapezium                ➨
[ၾတာပီဇီယံ]


diagonal                  ➨
[ေထာင့္ျဖတ္မ်ဥ္း]


coordinate               ➨
[ကိုၾသဒိနိတ္]


perimeter                ➨
[ပတ္လည္အနား]


အေျဖကို ေနာက္ရက္မ်ားတြင္ တင္ေပးပါမည္။