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

If the locus of the point, whose distances from the point $(2,1)$ and $(1,3)$ are in the ratio $5: 4$, is $a x^2+b y^2+c x y+d x+e y+170=0$, then the value of $\mathrm{a}^2+2 \mathrm{~b}+3 \mathrm{c}+4 \mathrm{~d}+\mathrm{e}$ is equal to:

Two sides of a rhombus are along the lines, $x -y+ 1 = 0$ and $7x-y-5 =0.$ If its diagonals intersect at $(-1,-2),$ then which one of the following is a vertex of this rhombus?

The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

The sides $AB,BC,CD$ and $DA$ of a quadrilateral are $x + 2y = 3,\,x = 1,$ $x - 3y = 4,\,$ $\,5x + y + 12 = 0$ respectively. The angle between diagonals $AC$ and $BD$ is ......$^o$

The area of the triangle bounded by the straight line $ax + by + c = 0,\,\,\,\,(a,b,c \ne 0)$ and the coordinate axes is