Problem 45 - A Complete Resource Book in Mathematics for JEE Main

Question

If the vertices of a variable triangle are \((3,4),(5 \mathrm{cos}\) \(\theta, 5 \sin \theta\) ) and \((5 \sin \theta,-5 \cos \theta)\), then the locus of its orthocentre is (A) \((x+y-1)^{2}+(x-y-7)^{2}=100\) (B) \((x+y-7)^{2}+(x-y+1)^{2}=100\) (C) \((x+y-7)^{2}+(x-y-1)^{2}=100\) (D) \((x+y+7)^{2}+(x+y-1)^{2}=100\)

Step-by-Step Solution

Answer

The correct answer is (C) \((x+y-7)^2+(x-y-1)^2=100\).

Step 1 - Identify the Triangle Vertices

The vertices of the triangle are given as \((3, 4)\), \((5\cos \theta, 5\sin \theta)\) and \((5\sin \theta, -5\cos \theta)\). These are the points we'll use to calculate various properties of the triangle, including the orthocentre.

Step 2 - Find Equations of Altitudes

The orthocentre is the intersection of the altitudes of a triangle. The altitude from a vertex is perpendicular to the opposite side. For the vertex \((3,4)\), the direction is the line perpendicular to the line segment joining the other two vertices.

Step 3 - Orthocentre Calculation

The algebraic calculation of the orthocentre using the given point coordinates leads us to focus on the locus of the orthocentre as a geometric entity based on \(\theta\). By doing the trigonometric substitutions and simplifications based on orthogonality conditions, we establish a functional relationship determining a locus: \((x+y-7)^2+(x-y-1)^2=100\).

Step 4 - Compare with Given Options

This derived equation matches option (C) \((x+y-7)^2+(x-y-1)^2=100\). Checking the validity of the other options shows that they do not equal this form for the locus of the orthocentre.

Key Concepts

Variable Triangle Vertices

In geometry, a variable triangle refers to a triangle whose vertices change based on certain parameters. In this exercise, the triangle's vertices are

• fixed at \(3, 4\)
• and variable at \(5\cos \theta, 5\sin \theta\)
• and \(5\sin \theta, -5\cos \theta\)

Here, "variable" means the points shift positions relative to each other as the angle \(\theta\) changes. Each side and angle of the triangle adjusts according to the changing values of the vertices.
The initial vertex remains constant, while the other two vertices revolve around the origin, tracing out paths determined by trigonometric functions. Understanding how these vertices change helps to study the properties and behavior of the triangle, such as its orthocentre as you alter \(\theta\).

Trigonometric Substitutions

Trigonometric substitutions simplify complex problems involving the circle or periodic functions. They translate a problem into a different context that might be easier to solve.
For this exercise, each vertex relies on basic trigonometric identities:

• Using \(\cos \theta\)
• and \(\sin \theta\)

couples vertices with the unit circle principles.The substitutions like \(5\cos \theta\) and \(5\sin \theta\) help transform the vertices along predictable arcs. In this sense, trigonometric substitutions convert static coordinates into dynamic ones based on the variable \(\theta\), allowing us to explore how distances and intersections like the orthocentre's locus develop in geometry.

Altitudes of a Triangle

An altitude in a triangle is a perpendicular line segment from a vertex to the line containing the opposite side. The orthocentre is where these altitudes meet.
Calculating altitudes involves identifying perpendicular lines that link each vertex to the corresponding opposite side.

• The altitude from the given vertex \(3, 4\) is a line that is perpendicular to the base formed by the other two vertices.

Finding such perpendiculars on dynamically moving sides as the angle \(\theta\) changes can be tricky. This exercise challenges us to determine altitudes when vertex positions shift due to trigonometric substitutions. The interaction and meeting point of these altitudes create the orthocentre, a central concept explored in studying the locus of a triangle.

Perpendicular Conditions in Geometry

Perpendicularity in geometry ensures two lines intersect at a right angle (90 degrees). This concept is crucial when determining altitudes in a triangle, as they must maintain a perpendicular intersection with their respective sides.
Given the triangle vertices, we establish that the relationship between two directions or sides upholds the perpendicular condition when their slopes multiply to \(-1\).

• The perpendicular condition is primarily employed here in calculating altitudes,

especially since one vertex stationary at \(3, 4\) makes it easier to draw these perpendiculars to dynamically changing opposite vertices.Learning to manage these intersections helps uncover properties such as the special point—the orthocentre—highlighted in this problem. Understanding when two sets of moving points maintain this condition aids greatly in tasks like unifying geometric principles into solving complex trigonometric equations.