One vertex of equilateral triangle with centroid at origin and one side as x + y

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9.Straight Line

hard

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

$( - 1, - 1)$

$(2,2)$

$( - 2, - 2)$

None of these

Solution

$⇒$ $\frac{{k – 0}}{{h – 0}} \times \frac{{ – 1}}{1} = – 1 \Rightarrow k = h$…..$(i)$

Let the coordinates of $D$ be$(\alpha ,\beta )$. Then the co-ordinates of $O$ are $\left( {\frac{{2\alpha + h}}{{2 + 1}},\frac{{2\beta + k}}{{2 + 1}}} \right)$. 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 $x + y – 2 = 0$

$⇒$ $\alpha + \beta – 2 = 0$

$⇒$ $ – h/2 – k/2 – 2 = 0$==>$h + k + 4 = 0$

$⇒$ $2h + 4 = 0 \Rightarrow h = k = – 2$,[from (i)]

Hence the coordinates of vertex $A$ are $( – 2, – 2)$.

Standard 11

Mathematics

Let the equation of two sides of a triangle be $3x\,-\,2y\,+\,6\,=\,0$ and $4x\,+\,5y\,-\,20\,=\,0.$ If the orthocentre of this triangle is at $(1, 1),$ then the equation of its third side is

hard

(JEE MAIN-2019)

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equation to the side $AB \,\,ane\,\, AC$ are $px + qy = 1 \,\,ane\,\, qx + py = 1$ . The equation of the median through $A$ is :

normal

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

hard

(JEE MAIN-2023)

The medians $AD$ and $BE$ of a triangle with vertices $A\;(0,\;b),\;B\;(0,\;0)$ and $C\;(a,\;0)$ are perpendicular to each other, if

medium

The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is

medium

(IIT-1995)

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

Solution

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