Problem 25 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation to the sides \(A B, B C, C A\) of a \(\triangle A B C\) are \(x+y=1,4 x-y+4=0\) and \(2 x+3 y=6 .\) Circles are drawn on \(A B, B C, C A\) as diameters. The point of concurrence of the common chords is (A) centroid of the triangle (B) orthocentre (C) circumcentre (D) incentre

Step-by-Step Solution

Answer

The point of concurrence is the circumcentre (Option C).

Step 1: Find the vertices of the triangle

First, we need to find the points where each pair of lines intersects, as these are the vertices of the triangle. 1. To find point A: solve the equations of the lines \(x+y=1\) and \(4x-y+4=0\). Substitute \(y=1-x\) into the second equation:\[4x - (1-x) + 4 = 0\]Solve for \(x\):\[4x + x - 1 + 4 = 0\]\[5x + 3 = 0\]\[x = -\frac{3}{5}\]Substitute \(x = -\frac{3}{5}\) back into \(y = 1-x\):\[y = 1 - \left(-\frac{3}{5}\right) = \frac{8}{5}\]So, point A is \((-\frac{3}{5}, \frac{8}{5})\).2. To find point B: solve \(4x-y+4=0\) and \(2x+3y=6\). Substitute \(y=4x+4\) into \(2x+3y=6\):\[2x + 3(4x + 4) = 6\]\[2x + 12x + 12 = 6\]\[14x = -6\]\[x = -\frac{3}{7}\]Substitute \(x = -\frac{3}{7}\) back into \(y = 4x + 4\):\[y = 4(-\frac{3}{7}) + 4 = \frac{4}{7}\]So, point B is \((-\frac{3}{7}, \frac{4}{7})\).3. To find point C: solve \(x+y=1\) and \(2x+3y=6\). Substitute \(y=1-x\) into \(2x+3y=6\):\[2x + 3(1-x) = 6\]\[2x + 3 - 3x = 6\]\[-x = 3\]\[x = -3\]Substitute \(x = -3\) back into \(y = 1-x\):\[y = 4\]So, point C is \((-3, 4)\).

Step 2: Determine the circumcentre

The point of concurrence of all the perpendicular bisectors of a triangle is the circumcentre. In a triangle, the perpendicular bisectors are the lines that bisect the sides and are perpendicular to the sides. To find the circumcentre algebraically when the triangle vertices are known, find the perpendicular bisectors, and solve for their intersection point. The formulas result in the concurrency of these bisectors at the circumcentre based on the calculation results.

Step 3: Identify the point of concurrence

For circles drawn with diameters as triangle sides, the common chords intersect at the circumcentre. This is because the circumcentre is equidistant from all the vertices, which is the center of the circle with any side of the triangle as a diameter. Thus, the point of concurrence of the common chords of circles drawn on the triangle's sides as diameters corresponds to the circumcentre.

Key Concepts

Triangle

A triangle is one of the basic shapes in geometry. It consists of three sides and three angles. In the given exercise, we have a triangle with vertices at specific coordinates determined by solving the intersections of three lines. This process involves solving pairs of linear equations to find each vertex. By substituting and solving these equations, we discover the exact positions of these vertices. Understanding these principles is crucial because the placement of these vertices determines the various centers and properties of the triangle. In particular, the circumcenter's position is directly linked to these vertices.

Perpendicular Bisectors

Perpendicular bisectors are lines that divide each side of a triangle into two equal parts at a 90-degree angle. In any triangle, there are three perpendicular bisectors, one for each side. These bisectors are significant because they always meet at a single point called the circumcentre. For the given triangle, you would calculate the midpoint of each side and then find the line that is both perpendicular to the side and passes through this midpoint. This can be achieved by using the slope of the line segment, finding the negative reciprocal of this slope, and then using the midpoint to form the equation of the bisector. This meeting point of the bisectors is crucial for various geometric constructions and proofs.

Concurrency

Concurrency refers to the situation where three or more lines intersect at a single point. In the context of triangles, the most common points of concurrency are the centroid, incenter, orthocenter, and circumcenter. In this exercise, we specifically focus on the circumcenter, which is the point of concurrency for the perpendicular bisectors of a triangle's sides. This intersection point holds unique properties; it is equidistant from all the vertices of the triangle. Therefore, this point also serves as the center for the circle that can be circumscribed around the triangle, with the triangle's sides acting as diameters for circles drawn around it. Thus, concurrency in this context plays a critical role in defining the geometric properties and symmetries of the triangle.