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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, – 1), then find the length of the side of the triangle \[\] [Hint: Find length of perpendicular (p) from (2, – 1) to the line and use $p = l \: \sin 60^{\circ}$, where $l$ is the length of side of the triangle]

$\begin {array} {1 1} (A)\;\large\frac{2}{\sqrt 3} & \quad (B)\;\large\frac{2}{ 3} \\ (C)\;\large\frac{\sqrt 2}{ 3} & \quad (D)\;\sqrt{\large\frac{2}{3}} \end {array}$

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  • Length of the perpendicular from $(x_1, y_1)$ to the line $ax+by+c=0$ is $ p = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1 :
Let $a$ be the length of each side of the equilateral triangle.
The length of the perpendicular p from (2,-1) on $x+y-2=0$
$ \Rightarrow p = \bigg| \large\frac{2-1-2}{\sqrt{1^2+1^2}}\bigg|$
$ = \large\frac{1}{\sqrt 2}$
Step 2 :
Consider the $ \Delta $ ABD
$ \sin 60^{\circ} = \large\frac{p}{a}$
$ \Rightarrow \large\frac{\sqrt 3}{2} $$ = \large\frac{\Large\frac{1}{\sqrt 2}}{a}$
$ \Rightarrow a = \large\frac{2}{\sqrt 3}$$ \times \large\frac{1}{\sqrt 2}$
$ a = \large\frac{\sqrt 2}{3}$
hence the length of the side of the equilateral triangle is $ \large\frac{\sqrt 2}{3}$
answered Jul 1, 2014 by thanvigandhi_1
edited Jul 1, 2014 by thanvigandhi_1
 

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