Question

# The equation of the base of an equilateral triangle is X +Y =2 and vertex is (2, -1) length of its side is :

A
(12)
B
(32)
C
(23)
D
2

Solution

## The correct option is C $$\displaystyle \sqrt{\left( \frac{2}{3} \right)}$$$$Given-\\ P({ x }_{ 1 },{ y }_{ 1 })=(2,-1)\quad is\quad the\quad vetex\quad of\quad an\quad equilateral\quad triangle.\\ The\quad equation\quad of\quad its\quad base\quad is\quad x+y=2\Longrightarrow x+y-2=0.\\ To\quad find\quad out-\quad \\ the\quad side\quad of\quad the\quad given\quad equilateral\quad triangle=?\\ Solution-\\ The\quad pependicular\quad distace\quad of\quad the\quad vertex\quad from\quad the\quad base\\ of\quad a\quad triangle\quad is\quad its\quad height.\\ For\quad an\quad equilateral\quad triangle\quad of\quad side\quad a,\\ the\quad height\quad h=\dfrac { \sqrt { 3 } }{ 2 } a\Longrightarrow a=\dfrac { 2h }{ \sqrt { 3 } } .\\ Now\quad we\quad apply\quad the\quad formula\quad p=\dfrac { a{ x }_{ 1 }+b{ y }_{ 1 }+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \quad when\quad \\ P({ x }_{ 1 },{ y }_{ 1 })\quad is\quad the\quad point\quad and\quad ax+by+c=0\quad is\quad the\quad equation\quad \\ of\quad the\quad line.\\ Here\quad P({ x }_{ 1 },{ y }_{ 1 })=(2,-1),\quad a=1,\quad b=1\quad \& \quad c=-2.\\ \therefore \quad p=h=\left| \dfrac { 1\times 2+1\times (-1)+(-2) }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \right| units=\dfrac { 1 }{ \sqrt { 2 } } units.\\ \therefore \quad The\quad side\quad a=\dfrac { 2h }{ \sqrt { 3 } } =\dfrac { 2\times \dfrac { 1 }{ \sqrt { 2 } } }{ \sqrt { 3 } } units=\sqrt { \dfrac { 2 }{ 3 } } units.\\$$Mathematics

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