One vertex of the equilateral triangle with c | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Triangle formed by the lines $3x + y + 4 = 0$ , $3x + 4y -15 = 0$ and $24x -7y = 3$ is a/an

The equations of two equal sides of an isosceles triangle are $7x - y + 3 = 0$ and $x + y - 3 = 0$ and the third side passes through the point $(1, -10)$. The equation of the third side is

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the equation to one diagonal is $11x + 7y = 9$, then the equation to the other diagonal is :-

Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are