Question

One vertex of the equilateral triangle with a centroid at the origin and one side as $x+y-2$ is
A. $(-1,-1)$
B. $(2,2)$
C. $(-2,-2)$
D. None of these

Answer 1

Hint: For this, we need to go through the concept of median in a triangle and their ratio in which they divide each other. Also, we know the concept that the perpendicular lines' slopes are their opposite reciprocals.

Formula Used: The line segment connecting points $A$ and $B$ is internally divided by a point $P$ whose coordinates are $(x,y)$ in the ratio $m_1\colon\:m_2$ is given by the Section formula:
$(x,y)=(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2})$

Complete step by step solution: Let $\triangle PQR$ be the equilateral triangle with centroid $O$ at the origin.
$PS \bot RQ$

Given the equation of side, $RQ$ is
$x+y-2=0 …(i)$
Slope of $x+y-2=0$ is $-1$.
As you can see the centroid $O$ of an equilateral triangle divides $PS$ in the ratio of $2\colon1$.
So, using the Section Formula we get
$0=\dfrac{1\times h+2\times a}{1+2}\\
h+2a=0 ...(ii)$
Also,
$0=\dfrac{1\times k+2\times b}{1+2}\\
k+2b=0 ...(iii)$

Point $S$ passes through $(a,b)$ and lies on the line $x+y-2=0$
Therefore,
$a+b-2=0$
The slope of $PO$
$=\dfrac{k-0}{h-0}\\
=\dfrac{k}{h}$
Since $PS \bot RQ$ So,
$-1\times \dfrac{k}{h}=-1\\
h=k$

From eq (ii) and (iii) we get
$h+2a=0\\
\Rightarrow2a=-h$
And
$k+2b=0\\
\Rightarrow2b=-k\\
\therefore a=b$
Substitute $a=b$ in $a+b-2=0$
$\Rightarrow a+a-2=0\\
\Rightarrow 2a-2=0\\
a=1;b=1$
Substitute the values of $a$ and $b$ to obtain the vertex
$h=-2\times1=-2$
$k=-2\times1=-2$
Hence, the required vertex is $(-2,-2)$.

So, option C is correct.

Note: The coordinate order is extremely important in the ordered pair $(x,y)$. Keep in mind that x stands for the x coordinate and y for the y-coordinate. When substituting in the formula, keep in mind that the order in which the points are taken into account and the ratio is important. Frequent mistakes include miscalculating the slope of a parallel and perpendicular line.