Question

If the vertex of an equilateral triangle is the origin and the side opposite to it has the equation $ x + y = 1 $ , then the orthocenter of the triangle is:
(A) $ \left( {\dfrac{1}{3},\dfrac{1}{3}} \right) $
(B) $ \left( {\dfrac{{\sqrt 2 }}{3},\dfrac{{\sqrt 2 }}{3}} \right) $
(C) $ \left( {\dfrac{2}{3},\dfrac{2}{3}} \right) $
(D) None of these

Answer 1

Hint: In the given question, we are given an equilateral triangle whose one vertex is given to us as origin. Also, we are given that the side opposite to the origin has the equation $ x + y = 1 $ and we are required to find the orthocenter of the equilateral triangle. The question involves the concepts of analytical geometry and properties.

Complete step by step solution:
In the given question, we have an equilateral triangle with origin as one of the vertices and the equation of the side opposite to origin is $ x + y = 1 $ . So, we get the figure of the scenario as below.

Now, we have to find the orthocenter of the equilateral triangle. Orthocenter of a triangle is the point of intersection of the perpendiculars or altitudes from all the vertices of the triangle.
So, we first find the equation of perpendicular from the origin to the side opposite to the origin $ x + y = 1 $ .
The slope of the straight line $ x + y = 1 $ is $ \left( { - 1} \right) $ . Now, we know that the product of the slopes of the lines that are perpendicular to each other is $ \left( { - 1} \right) $ . Hence, the slope of the line perpendicular to the line with equation $ x + y = 1 $ must be $ 1 $ .
Also, the normal from origin to the side with equation $ x + y = 1 $ must pass through the origin.
So, we get the equation of the line as,
 $ \Rightarrow \dfrac{{y - 0}}{{x - 0}} = 1 $
 $ \Rightarrow y = x $
Now, we find the intersection point of the perpendicular lines $ x + y = 1 $ and $ y = x $ so as to find the foot of perpendicular.
So, substituting the value of y from equation $ y = x $ into the equation $ x + y = 1 $ , we get,
 $ \Rightarrow x + x = 1 $
 $ \Rightarrow x = \left( {\dfrac{1}{2}} \right) $
So, we get the value of x as $ \left( {\dfrac{1}{2}} \right) $ .
Putting the value of x in $ y = x $ , we get the value of y as $ \left( {\dfrac{1}{2}} \right) $ .
So, the foot of perpendicular from origin to the side opposite to it is $ \left( {\dfrac{1}{2},\dfrac{1}{2}} \right) $ .
Now, we know that in an equilateral triangle, the orthocenter, the centroid and the circumcenter all lie at the same point and the foot of the perpendicular also bisects the side.
Also, the centroid of a triangle divided the bisector in the ratio $ 2:1 $ .
Thus, we have, the coordinates of centroid of the triangle as $ \left( {\dfrac{{1\left( 0 \right) + 2\left( {\dfrac{1}{2}} \right)}}{3},\dfrac{{1\left( 0 \right) + 2\left( {\dfrac{1}{2}} \right)}}{3}} \right) $ . So, we get centroid as $ \left( {\dfrac{1}{3},\dfrac{1}{3}} \right) $ using the section formula.
Now, we know that in an equilateral triangle, the orthocenter, the centroid and the circumcenter all lie at the same point. So, we get the coordinates of the orthocenter as $ \left( {\dfrac{1}{3},\dfrac{1}{3}} \right) $ .
Thus, option (A) is correct.
So, the correct answer is “Option A”.

Note: Such problems where we have to find the orthocenter are generally solved by finding another perpendicular from a vertex and then finding out the intersection of those two altitudes. One must have a basic understanding of analytical geometry and properties of an equilateral triangle to solve the given question.