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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Choose the correct answer from the given four options. One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is \[\] [Hint: Let ABC be the equilateral triangle with vertex A (h, k) and let D ($ \alpha , \beta $ ) be the point on BC. Then $ \large\frac{2\alpha + h}{3}$$=0 = \large\frac{2\beta + k}{3}$. Also $ \alpha + \beta -2 = 0$ and $ \bigg( \large\frac{k-0}{h-0} \bigg) $$ \times (-1) =-1]$

$\begin {array} {1 1} (A)\;(-1,-1) & \quad (B)\;(2,2) \\ (C)\;(-2, -2) & \quad (D)\;(2,-2) \end {array}$

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Toolbox:
  • If two lines are perpendicular then the product of their slopes is $m_1m_2=-1$
  • The centroid of a triangle ABC divides the line AD in the ratio 2 : 1.
Step 1 :
Let ABC be the triangle and let the coordinates of A be (h, k).
Let D be any point on BC and let the coordinates of D be ( $ \alpha, \beta $).
It is given that the coordinates of the centroid $O$ is (0,0)
$Ao : oD $ is 2 : 1
$ \therefore \large\frac{2\alpha+h}{2+1}$$=0 $ and $ \large\frac{2 \beta + k}{2+1}$$=0$
$ \Rightarrow 2 \alpha + h =0$ and $ 2\beta + k = 0$
$ \therefore 2 \alpha + h = 2 \beta + k$
$ \Rightarrow 2( \alpha - \beta ) = k - h$---------(1)
Also given BC = $x+y-2=0$ Since it passes through $( \alpha, \beta ); \alpha + \beta -2 = 0$-------(2)
Slope of the line AD is $ \large\frac{k-0}{h-0}$$ = \large\frac{k}{h}$
Slope of the line BC $x+y-2$ is -1
$ \therefore \large\frac{k}{h}$$ \times -1 = -1$
$ \Rightarrow k = h $-------(3)
$ \therefore \alpha - \beta = 0$
$ \alpha + \beta = 2$
Solving the above equations we get,
$ 2\alpha = 2$
$ \alpha = 1 $ and $ \beta = 1$
Substituting for $ \alpha $ and $ \beta $ we get,
$ 2 \alpha + h = 0$
$ 2(1) +h=0 \qquad \Rightarrow h = -2$
and $2\beta +k =0$
$ 2(1) + k = 0 \qquad \Rightarrow k = -2$
Hence the coordinates of the vertex A is (-2,-2).
Hence the correct option is 'C'.
answered Jul 6, 2014 by thanvigandhi_1
edited Jul 6, 2014 by thanvigandhi_1
 

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