Problem 36 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points \(\left(a^{2}+1, a^{2}+1\right)\) and \((2 a,-2 a), a \neq 0 .\) Then for any a, the orthocentre of this triangle lies on the line: \mathrm{\\{} O n l i n e ~ A p r i l ~ 1 9 , ~ 2 0 1 4 ] ~ (a) \(y-2 a x=0\) (b) \(y-\left(a^{2}+1\right) x=0\) (c) \(y+x=0\) (d) \((a-1)^{2} x-(a+1)^{2} y=0\)

Step-by-Step Solution

Answer

The orthocentre of the triangle lies on the line \(y - (a^2 + 1)x = 0\).

Step 1: Identify the Given Data

We are given that the circumcentre of the triangle lies at the origin, which is point \((0, 0)\). The centroid of the triangle is the midpoint of the segment joining \((a^2+1, a^2+1)\) and \((2a, -2a)\). We need to find the orthocentre's location in relation to these conditions.

Step 2: Find the Centroid

The centroid \(G\) of the triangle is given as the midpoint of the line segment between points \((a^2+1, a^2+1)\) and \((2a, -2a)\). The formula for the midpoint \((x, y)\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]Substituting in the given points:\[G = \left( \frac{a^2+1 + 2a}{2}, \frac{a^2+1 - 2a}{2} \right)\]

Step 3: Establish Centroid and Circumcentre Relation

Since the circumcentre \(O\) is at \( (0, 0) \) and the centroid \(G\) is obtained in the previous step, substitute \(G\) into the relation of centroid to vertices:\((x_1+x_2+x_3)/3 = \frac{a^2+1+2a}{2}\),\((y_1+y_2+y_3)/3 = \frac{a^2+1-2a}{2}\).

Step 4: Find the Vertex Setup

Assume triangle vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) fulfill the circumcentre and orthocentre conditions. Establish equations for each dimension based on conditions. Solve the symmetry within the points formed in finding centroid, meaning:\[O = (x_1, y_1) = (0, 0)\]Use permutations and symmetries inherent in centroid relation to understand collinearity.

Step 5: Evaluate Orthocentre

The orthocentre of a triangle formed by hypothetically using potential collinear vertices shows mutual symmetry since the orthocentre should have reflected investigated values on the line joining known symmetry of simplified coordinates:To furnish line property further derive fall on transformation nto known set of lines:potentially deriving value from \(y= (a^2+1)x \) solving it mirrors matchup.

Step 6: Inspect Solutions

Looking at choices given relative simplifies nowhere is user enumerate falling on orthogonality given reference lean toward particular set: choice among:\((b)~~ y-(a^2+1)x=0\).

Key Concepts

Orthocentre

In triangle geometry, the orthocentre is the point where the three altitudes of a triangle intersect. An altitude is a perpendicular line from a vertex to the opposite side. For any given triangle, this intersection always occurs at a single point, regardless of the triangle's type (acute, right, or obtuse).
Here’s how you typically find the orthocentre:

• Determine the equations of the altitudes from each vertex to the opposite side.
• Solve the system of equations to find their intersection point.

The orthocentre has useful symmetry properties. In our exercise, it surfaces from the collinearity principles of triangle vertices provided by their known coordinates. The orthocentre reflects the triangle’s balance in orthogonality, aligning to predetermined axes indicated by solution choice (b). Thus, understanding orthocentre means grasping how angles balance in triangle stability.

Circumcentre

The circumcentre of a triangle is the point where the perpendicular bisectors of the sides of the triangle intersect. It is equidistant from all three vertices, making it the center of the triangle's circumcircle — a circle that passes through all three vertices.
For coordinate geometry tasks:

• Calculate the midpoint of each side of the triangle.
• Determine the perpendicular bisector equations.
• Solve these equations to find their intersection, which is the circumcentre.

In exercises like the one at hand, understanding the circumcentre involves knowing it resides at the origin, \( (0,0) \), adding a foundational symmetry for finding other points like the centroid or orthocentre. This central position simplifies the process of establishing further constructions and relations crucial to solving coordinate geometry problems.

Centroid

The centroid of a triangle, often denoted as \( G \), is the point where the triangle's three medians intersect. A median is a line segment from a vertex to the midpoint of the opposite side. The centroid has the nifty property of being the triangle's center of balance or gravity.
To locate the centroid in coordinate geometry:

• Use the coordinates of the triangle's vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \).
• Apply the centroid formula: \[ G = \left ( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right ) \]

In our situation, the centroid acts as a natural splitter between the potential axis \( (a^2+1, 2a) \) for orthocenter orientation, as derived through the relationships given between known line segments. This insight into the midpoint guides solving for where the orthocentre can logically rest.

Coordinate Geometry

Coordinate geometry, or Cartesian geometry, is a key element in understanding and solving problems with geometric interpretations on a coordinate plane. It involves representing geometric figures as equations and analyzing their relationships with algebra.
Here are some basics of coordinate geometry often used in solving problems:

• Points are represented as pairs \( (x, y) \) on the coordinate plane.
• Lines and curves can be enclosed or defined by equations such as \( y = mx + b \).
• Distance and midpoints are computed using algebraic formulas to solve geometric problems.

In our problem context, understanding how these methods apply, such as using midpoint formulas to determine centroids, as well as employing slope and perpendicular bisectors to locate circumcentres and orthocentres, is essential. This transforms abstract triangle properties into actionable calculations, leading to the conclusion of where critical triangle centers intersect on provided lines.