One vertex of the equilateral triangle with c | vedclass

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Question

(c) Let the co-ordinate of vertex $A$ be $(h,k)$. Then $AD$ is perpendicular to $BC$, therefore $OA\, \bot \,BC$

$&rArr;$  $\frac{{k - 0}}{{h

Vedclass · Accepted Answer

$( - 2, - 2)$

Vedclass · Answer

(c) Let the co-ordinate of vertex $A$ be $(h,k)$. Then $AD$ is perpendicular to $BC$, therefore $OA\, \bot \,BC$

$&rArr;$  $\frac{{k - 0}}{{h - 0}} 	imes \frac{{ - 1}}{1} = - 1 \Rightarrow k = h$.....$(i)$

Let the coordinates of $D$ be$(\alpha ,\beta )$. Then the co-ordinates of $O$ are $\left( {\frac{{2\alpha + h}}{{2 + 1}},\frac{{2\beta + k}}{{2 + 1}}} ight)$. Therefore $\frac{{2\alpha + h}}{3} = 0$ and $\frac{{2\beta + k}}{3} = 0$ $ \Rightarrow \alpha = - \frac{h}{2},\beta = \frac{{ - k}}{2}$.

Since $(\alpha ,\beta )$lies on $x + y - 2 = 0$

$&rArr;$ $\alpha + \beta - 2 = 0$

$&rArr;$  $ - h/2 - k/2 - 2 = 0$==>$h + k + 4 = 0$

$&rArr;$  $2h + 4 = 0 \Rightarrow h = k = - 2$,[from (i)]

Hence the coordinates of vertex $A$ are $( - 2, - 2)$.

One vertex of the equilateral triangle with centroid at the origin and one side as $x + y - 2 = 0$ is

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