Question

The centroid of an equilateral triangle is (0, 0). If two vertices of the triangle lie on $x + y = 2\sqrt 2 $, then one of them will have its coordinates
$\left( A \right)\left( {\sqrt 2 + \sqrt 6 ,\sqrt 2 - \sqrt 6 } \right)$
$\left( B \right)\left( {\sqrt 2 + \sqrt 3 ,\sqrt 2 - \sqrt 3 } \right)$
$\left( C \right)\left( {\sqrt 2 + \sqrt 5 ,\sqrt 2 - \sqrt 5 } \right)$
$\left( D \right)\left( {\sqrt 2 ,\sqrt 2 } \right)$

Answer 1

Hint: In this particular type of question use the concept that in an equilateral triangle if the centroid is at origin then circumcenter, incenter and orthocenter is also lie on the origin later on in the solution use the concept that the distance between two points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ so use these concepts to reach the solution of the question.

Complete step-by-step answer:

Let us suppose ABC is the equilateral triangle as shown in the figure.
It is given that the centroid of the triangle is (0, 0) as shown in the figure.
Now as we all know that if the centroid of equilateral triangle is origin than the circumcenter, incenter and orthocenter is also line on the origin.
So if the incenter is at origin then the line joining the incenter and the vertices of the triangle bisects.
Therefore, angle OBD = angle OBA
As we all know that in an equilateral triangle all of the angles are equal = 60 degrees.
Therefore, $\angle OBD = \angle OBA = {30^o}$
And we all know that the orthocenter is the intersection of all the medians of the triangle, as the orthocenter is also O, therefore, angle ABD = 90 degrees.
Now let us suppose that the given equation of line is line BC.
So line BC is $x + y = 2\sqrt 2 $
And line AD is perpendicular to it.
Now the slope of line BC by comparing to the standard equation of line, y = mx + c.
Therefore, m = -1
Now as we know that if two lines are perpendicular then the slope multiplication is -1.
Let the slope of line AD = ${m_1}$
Therefore, $m{m_1} = - 1$
Therefore, ${m_1} = 1$
Now the equation of line passing through (0, 0) and slope 1 is given as,
$\left( {y - {y_1}} \right) = {m_1}\left( {x - {x_1}} \right)$
Where (${x_1},{y_1}$) are the coordinates of the origin.
Therefore, $\left( {y - 0} \right) = 1\left( {x - 0} \right)$
Therefore, y = x is the equation of the line AD.
Now the intersection of line BC and AD is point D.
So substitute y = x in equation of line BC we have,
$ \Rightarrow x + x = 2\sqrt 2 $
$ \Rightarrow x = \sqrt 2 $
$ \Rightarrow x = \sqrt 2 = y$
So the coordinates of point D = $\left( {\sqrt 2 ,\sqrt 2 } \right)$
Now as we know that the distance between two pints $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as,
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
So the distance of OD, where O = $\left( {{x_1},{y_1}} \right)$ = (0, 0) and D = $\left( {{x_2},{y_2}} \right)$ = $\left( {\sqrt 2 ,\sqrt 2 } \right)$
So the distance OD = $\sqrt {{{\left( {\sqrt 2 - 0} \right)}^2} + {{\left( {\sqrt 2 - 0} \right)}^2}} = \sqrt {2 + 2} = \sqrt 4 = 2$
Now in triangle OBD we have,
$ \Rightarrow \tan {30^o} = \dfrac{{OD}}{{BD}}$
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{2}{{BD}}$
$ \Rightarrow BD = 2\sqrt 3 $
Now let the x coordinate of point B is p.
So this coordinate satisfies the equation of line BC.
So substitute we have,
 $ \Rightarrow p + y = 2\sqrt 2 $
So the y coordinate of the point B is, \[y = 2\sqrt 2 - p\] so the coordinates of point B is (p,\[\left( {2\sqrt 2 - p} \right)\])
Now again use distance formula between two points this time let
B = $\left( {{x_1},{y_1}} \right)$ = (p,\[\left( {2\sqrt 2 - p} \right)\]) and D = $\left( {{x_2},{y_2}} \right)$ = $\left( {\sqrt 2 ,\sqrt 2 } \right)$
So the distance BD is
$ \Rightarrow BD = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
$ \Rightarrow 2\sqrt 3 = \sqrt {{{\left( {\sqrt 2 - p} \right)}^2} + {{\left( {\sqrt 2 - \left( {2\sqrt 2 - p} \right)} \right)}^2}} $
$ \Rightarrow 2\sqrt 3 = \sqrt {{{\left( {\sqrt 2 - p} \right)}^2} + {{\left( {p - \sqrt 2 } \right)}^2}} $
$ \Rightarrow 2\sqrt 3 = \sqrt {2{{\left( {\sqrt 2 - p} \right)}^2}} $
Now squaring on both sides we have,
$ \Rightarrow {\left( {2\sqrt 3 } \right)^2} = 2\left( {\sqrt 2 - p} \right)$
$ \Rightarrow 12 = 2\left( {\sqrt 2 - p} \right)$
$ \Rightarrow 6 = {\left( {\sqrt 2 - p} \right)^2}$
Now take square root on both sides we have,
$ \Rightarrow \sqrt 2 - p = \pm \sqrt 6 $
Therefore, $p = \left( {\sqrt 2 - \sqrt 6 } \right),\left( {\sqrt 2 + \sqrt 6 } \right)$
So the coordinates of B is when, p = $\left( {\sqrt 2 - \sqrt 6 } \right)$
$\left( {\sqrt 2 - \sqrt 6 } \right),\left( {2\sqrt 2 - \left( {\sqrt 2 - \sqrt 6 } \right)} \right)$ = $\left( {\sqrt 2 - \sqrt 6 } \right),\left( {\sqrt 2 + \sqrt 6 } \right)$
So the coordinates of B is when, p = $\left( {\sqrt 2 + \sqrt 6 } \right)$
$\left( {\sqrt 2 + \sqrt 6 } \right),\left( {2\sqrt 2 - \left( {\sqrt 2 + \sqrt 6 } \right)} \right)$ = $\left( {\sqrt 2 + \sqrt 6 } \right),\left( {\sqrt 2 - \sqrt 6 } \right)$
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that if two lines are perpendicular to each other than the multiplication of the respective slopes is always (-1), so first find the equation of line OD then find the distance of OD as above, then in triangle OBD apply tan rule and find the distance BD as above, then assume any variable be the x coordinate of point B, then substitute this value in line BC and find the y coordinate of point B, then apply distance formula between point B and D as above and simplify we will get the required coordinate of B.