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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| α‘αားαα
္αα္ α‘α်ား $ \displaystyle 2a$ αွိေαာ αံုးαားαီ ααိαံ၏ α‘ေျαα‘αား (α‘αား αα
္αα္) αα္ $ \displaystyle y$ αα္αိုးေαααြα္ αွိαΏαီး ၎α‘αား၏ α‘αα္αွα္αα္ origin $ \displaystyle (0, 0)$ ျαα
္αွ်α္ ααိαံ၏ ေαာα့္α
ြα္းαွα္α်ားαို αွာαါ။ ေαးαားα်α္α‘α ... $ \displaystyle (0,0)$ α α‘ေျαα‘αားαဲ့ α‘αα္αွα္ျαα ္αိုα ေαာα္α ြα္းαွα္ ႏွα ္αုα αα္ $ \displaystyle (0, a)$ αဲα αα္ $ \displaystyle (0, -a)$ ျαα ္αါαα္။ αံုးαားαီ ααိαံ ျαα ္αိုα α်α္αဲ့ ေαα့္α ြα္းαွα္α αα္ $ \displaystyle x$ αα္αိုးေαααွာ αွိαါαα္။ αါ့ေαΎαာα့္ α်α္ေαာα့္α ြα္းαွα္α αα္ $ \displaystyle (x,0)$ αို αိုα αားαိုα္αါαα္။ αါαိုαα္ αုα ာα¦αဲ့ ေαးα်α္α‘α αα္ $ \displaystyle x$ αို αွာေαးαα္ ααါαΏαီ။ αံုαိုαΎαα့္αါ။ $ \displaystyle (x, 0)$ αာ positive $ \displaystyle x$-axis αွာ αွိႏိုα္ααို negative $ \displaystyle x$-axis αα္αွာαဲ αွိႏိုα္αါαα္။ αါေααα့္ αုα ာα¦αြα္αာ αွ ႏွα ္αုαြဲ αα္αွα္αိုαααိုαဲ α‘ေျααွာ $ \displaystyle \pm x$ αဲα αြα္αာαွာ ျαα ္αါαα္။ $ \displaystyle ΞAOC$ αာ ေαာα့္αွα္ααိαံျαα ္αိုα $ \displaystyle OC$ αဲ့ α‘α်ားαို Pythagoras' Theorem αဲα αွာαα္αိုαα္ $ \displaystyle x$ αα္αိုးα်ား ααါαΏαီ။ αြα္αΎααေα‘ာα္... 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 ΞAOC$ 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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| $\displaystyle (7, 6)$ αဲα $\displaystyle (3, 4)$ αိုααွ αူαီα
ြာ αြာေαးေαာ $\displaystyle x$ αα္αိုးေαααွိ α‘αွα္αို αွာαါ။ $\displaystyle x$ αα္αိုးေαααွိ α‘αွα္ αိုα ေျαာαားαိုα $\displaystyle y$-coordinate α $\displaystyle 0$ αိုα αားαα္αားααါαα္္။ αိုα်α္αဲ့ α‘αွα္αို $\displaystyle (a, 0)$ αိုα αားαိုα္αα္။ αါαိုαα္ αုα ာα¦α‘α $\displaystyle (a, 0)$ αာ $\displaystyle (7, 6)$ αဲα $\displaystyle (3, 4)$ αေαα‘αြာα‘ေαးαူαါαα္။ αံုαို αΎαα့္αါ။ αါေαΎαာα့္ distance formula αဲα distance ႏွα ္αုαို αွာαΏαီး αီေαးαိုα္αα္ α‘ေျαααါαΏαီ။ αြα္αΎααေα‘ာα္... 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)$ α‘αွα္ေαးαွα္αို αα္αားေαာ α
αုαံαα
္αုαွိαα္။ ၎α
αုαံ၏ α‘αားα်ား၏ α‘αα္αွα္α်ားαα္ α‘αားαΏαိဳα္α
αုαံ αα
္αု၏ ေαာα့္α
ြα္းαွα္α်ား ျαα
္ေαΎαာα္း αα္ေαျααါ။ αα္αိုα္αာ α‘αားα်ား၏ α‘αα္αွα္αိုαွာαα္။ α‘αွα္ေαြαို αြဲαဲ့α‘αါ α‘α α₯္αိုα္αြဲαိုααိုαါαα္။ αα ္αုေα်ာ္ αြဲαိုα္αα္ α‘αားေαြ ααုα္αဲ ေαာα့္ျαα္α်α₯္းေαြ ျαα ္αြားαα္။ α‘αα္αွα္ေαြαို αα ္α ံုα ီαြဲαΏαီး αα္αိုα္αာ α်α₯္းျαα္α်ားαဲ့ slope αို αွာαα္။ slope ေαးαုαွာ ႏွα ္αုαα ္α ံုα ီ αီαဲ့αα္αိုαα္ α‘αားαΏαိဳα္α αုαံ ျαα ္αΏαီေαါ့။ αံုαိုαΎαα့္αါ။ αြα္αΎαα့္αေα‘ာα္။ 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.$
(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 ΞABC$ ၏ ေαာα့္α
ြα္းαွα္α်ား αα္ $ \displaystyle A(3,8)$, $ \displaystyle B(−1,2)$ and $ \displaystyle C(6,−6)$ αိုα ျαα
္αΎααွ်α္... (i) $ \displaystyle BC$ ၏ slope αို αွာαါ။ (ii) $ \displaystyle C$ αို ျαα္αြားေαာ α‘ျαα့္α်α₯္း ၏ slope αို αွာαါ။ (iii) $ \displaystyle A$ αို ျαα္αြားေαာ α‘αα္α်α₯္း ၏ slope αို αွာαါ။ (iv) $ \displaystyle CA$ ေαααွိ ေαာα့္αွα္α် αα္αα္αိုα္းα်α₯္း၏ slope αို αွာαါ။ slope αဲ့ αံုေαးαα္း $ \displaystyle m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$ αို αိααါαα္။ $\displaystyle {{l}_{1}}\bot {{l}_{2}}$ ျαα ္αဲ့α‘αါ $ \displaystyle {{m}_{1}}=-\frac{1}{{{{m}_{2}}}}$ αိုαာαို αိααါαα္။ 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)$ αိုαာαို αိααါαα္။ αံုαို αΎαα့္αါ။ (i) α αံုေααα္းαို αိုα္αိုα္α‘αံုးα်αΏαီး αွာαံုαဲ ျαα ္αါαα္။ (ii) α $ \displaystyle CF$ αဲ့ slope αို αွာαိုα္းαာ ျαα ္αါαα္။ $ \displaystyle CF$ α $ \displaystyle AB$ ေαααို ေαာα့္αα္α်αိုα $ \displaystyle {{m}_{{CF}}}=-\frac{1}{{{{m}_{{AB}}}}}$ αိုααိααါαα္။ (iii) α $ \displaystyle AD$ αဲ့ slope αို αွာαိုα္းαာ ျαα ္αါαα္။ $ \displaystyle AD$ α median ျαα ္αာေαΎαာα့္ $ \displaystyle D$ α $ \displaystyle BC$ αဲ့ α‘αα္αွα္ျαα ္αα္αိုα αိααါαα္။ $ \displaystyle D$ αို midpoint formula αို αံုးαΏαီး αွာααါαα္။ $ \displaystyle D$ αို ααα္ $ \displaystyle AD$ αဲ့ slope αို αွာႏိုα္αါαΏαီ။ (iii) α green dotted line $ \displaystyle (l)$ αဲ့ slope αို αွာαိုα္းαာ ျαα ္αါαα္။ $ \displaystyle l$ α $ \displaystyle AC$ ေαααာα့္αα္α်αိုα $ \displaystyle {{m}_{l}}=-\frac{1}{{{{m}_{{AC}}}}}$ ျαα ္αα္αိုα αိααα္။ slope αို αာ αွာαိုα္းျαα္း ျαα ္αိုα bisector αိုαဲ့ α αားαံုးαို αα့္αြα္းα α₯္းα ားαα္ ααိုေαာ့αါ။ αြα္αΎαα့္αΎαα ိုα။ (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)$ ႏွα့ $ \displaystyle B(0,−4)$ αိုα၏ α‘αα္αွα္ႏွα့္ Origin $ \displaystyle (0, 0)$ αိုααို ျαα္αြားေαာ α်α₯္းαα
္ေαΎαာα္း ၏ αီαွ်ျαα္း (line equation) αို αွာေαးαါ။ line equation αို αွာαိုα slope-intercept form, point-slope form, two-point form αα ္αုαုαို αံုးααါαα္။ αွာαိုေαာ line αာ Origin $ \displaystyle (0, 0)$ αဲα $ \displaystyle AB$ αဲ့ α‘αα္αွα္ αို ျαα္αြားαα္αိုα ေျαာαားαိုα two-point form αို αံုးαα္ ααါαΏαီ။ two-point form αိုαာα α‘αွα္ႏွα ္αုαိုαံုးαΏαီး slope αို αွာαα္။ αΏαီးေαာ့ point-slope form αဲα line equation αို αα္αΏαီး αွာαα္။ αံုαို αΎαα့္αါ။ $ \displaystyle AB$ αဲ့ α‘αα္αွα္ αို αွာαα္။ Origin αဲα α‘αα္αွα္αို αα္αြα္αားαဲ့ $ \displaystyle l$ αဲ့ slope αို αွာαα္။ ျαα္αα္αα္ $ \displaystyle (x_1, y_1)=(0,0)$ ျαα ္αိုα $ \displaystyle m= \frac{y_2}{x_2}$ αိုα αα္းαိႏိုα္αါαα္။ αြα္αΎαα့္αေα‘ာα္။ 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.$
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| αံုαိုαΎαα့္αါ။ α‘αα္αွα္ $ \displaystyle M$ αို αွာαα္။ $ \displaystyle AM$ αဲ့ slope ($ \displaystyle {{m}_{{_{{AM}}}}}$) αို αွာαα္။ $ \displaystyle BC$ αဲ့ slope ($ \displaystyle {{m}_{{_{{BC}}}}}$) αို αွာαα္။ $ \displaystyle {{m}_{{_{{AM}}}}}=\displaystyle -\frac{1}{{\text{ }{{m}_{{_{{BC}}}}}}}$ ျαα ္αα္ $ \displaystyle AM\bot BC$ ျαα ္αΏαီေαါ့။ αြα္αΎαα့္αေα‘ာα္... 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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| 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)$ αα္ ေα်αုα်αံုα ံျαα ္αα့္α‘αြα္ $ \displaystyle B$ ႏွα့္ $ \displaystyle D$ αα္αα့္α‘αွα္αို ααို αိုα္α ားျαဳႏိုα္αါαα္။ αိုαေαာ္ αုα ာα¦αြα္αာαြα္ αိα်ေαာ αေαာေαာα္ေα αα္ αာαα္αို P αု αα္αွα္ျαα္း ျαα ္αါαα္။] 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 ΞABC.$ 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 ΞABC$ 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 ΞOPQ?$
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$.
(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 ΞABC.$
(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 ΞABC.$
(α)αα္းαာ α‘ေျααံαို α
αα္αααို၊ ျαα္αာαို αα္αာαွ α‘αၤαိα္αို ေျαာα္းαα္ααါαα္။ αီα‘αါαွာ α‘ေျααံ α‘ားαα္းαဲ့αဲ့ αေαးαα်ိဳα α‘αြα္ေαာ့ α‘αα္α‘αဲαွိαါαα္။ α‘ားαα္းαα္αိုαΏαီး α‘ားαα္းααို αားαိုα္αိုα αျαα
္ႏိုα္αါαူး (αျαα
္αα့္αါαူး)။ α
αားαံုးေαြαို αွα္αα္α
ြာαိαိုα αα္αα္αိုα αိုαါαα္။ ααၤ်ာαα္αာ α‘αၤαိα္α
ာα α‘ေαးαααီးαါαူးαိုαဲ့ αူαα်α္αာ ααွα္ႏိုα္αါαူး။ αူααာαာα
αားαို αα္αα္ αူαα
α္းα်α₯္းေαြαို αိုα္αာαိုα αာαα္αွိαါαα္။ α
αားαံုးα‘αံုးα‘ႏႈα္း (usage) αဲα α‘αံαြα္ (pronunciation) α α‘ေαးααီးαါαα္။ αΏαီးαΏαီးေαာ α‘αံαြα္αိုα္αိုα ααα္αံုး α‘αွα္αွားαဲ့αα္ αα္ေαးαိုα္αူ ααာαွာ αာαα္ααα္းαါαူး။
αြα္αဲ့αဲ့ αα္α
ုႏွα
္ ေαာα္α α‘αံαြα္αို αိαိုα dictionary αိုαာ α‘ားαိုးαဲ့αေαးαα့္ ααုα‘αါαွာ ေαာ့ α‘αြα္αြα္αူαြားαါαΏαီ။
α‘αုေαးαိုα္αဲ့ αုα
ာα¦ေαြαွာ αါαဲ့ α
αားαံုး αα်ိဳααဲ့ α‘αံαြα္ေαြαို voice clip αဲα αူးαြဲαα္ေαး αိုα္αါαα္။ αွα္αα္ေαာ α‘αံαြα္α်ားαို αွα္αားႏိုα္αΎααါေα
ေαΎαာα္း ...
vertices ➨
[ေαာα့္α ြα္းαွα္α်ား]
equilateral ➨
[αံုးαားαီ]
[αံုးαားαီ]
equidistant ➨
[αူαီα ြာ αြာေαးေαာ]
[αူαီα ြာ αြာေαးေαာ]
altitude ➨
[α‘ျαα့္α်α₯္း]
[α‘ျαα့္α်α₯္း]
median ➨
[α‘αα္α်α₯္း]
[α‘αα္α်α₯္း]
bisector ➨
[αα္αα္αိုα္းα်α₯္း]
[αα္αα္αိုα္းα်α₯္း]
perpendicular ➨
[ေαာα့္αα္α်ေαာ/ေαာα့္αα္α်α်α₯္း]
[ေαာα့္αα္α်ေαာ/ေαာα့္αα္α်α်α₯္း]
parallel ➨
[αΏαိဳα္ေαာ]
[αΏαိဳα္ေαာ]
segment ➨
[α်α₯္းျαα္]
[α်α₯္းျαα္]
quadrilateral ➨
[α αုαံ]
[α αုαံ]
rectangle ➨
[ေαာα့္αွα္ α αုαံ]
[ေαာα့္αွα္ α αုαံ]
parallelogram ➨
[α‘αားαΏαိဳα္ α αုαံ]
[α‘αားαΏαိဳα္ α αုαံ]
trapezium ➨
[αΎαာαီαီαံ]
[αΎαာαီαီαံ]
diagonal ➨
[ေαာα့္ျαα္α်α₯္း]
[ေαာα့္ျαα္α်α₯္း]
coordinate ➨
[αိုαΎααိαိα္]
[αိုαΎααိαိα္]
perimeter ➨
[αα္αα္α‘αား]
[αα္αα္α‘αား]
α
ာαα်αူ၏ α‘αြα်αို αေးα
ားα
ွာα
ောα့်αျှော်αျα်!
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