Question

If the equation of the locus of a point equidistant from the points \(\left(a_{1}, b_{1}\right)\) and \(\left(a_{2}, b_{2}\right)\) is \(\left(a_{1}-a_{2}\right) x+\left(b_{1}, b_{2}\right) y+c=0\), then the value of \(\mathrm{C}\) will be (a) \((1 / 2)\left(\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}-\mathrm{a}_{1}^{2}-\mathrm{b}_{1}^{2}\right)\) (b) \(a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}\) (c) \((1 / 2)\left(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{b}_{1}^{2}+\mathrm{b}_{2}^{2}\right)\) (d) \(\sqrt{\left(a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}\right)}\)

Short answer

The value of C is (a) \(\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}\right)\).

Step-by-step solution

Understanding the Locus Equation

If a point is equidistant from two given points, then it lies on the perpendicular bisector of the line segment joining those two points. Thus, the equation of the locus will be a linear equation that represents the perpendicular bisector of the line segment.

Write down the locus equation given in the question

The given locus equation is \[\left(a_{1}-a_{2}\right) x + \left(b_{1} - b_{2}\right) y + c = 0.\]

Calculate the distance between two given points

Let P \((x,y)\) be the equidistant point from \((a_1, b_1)\) and \((a_2, b_2)\). Since the distance between P and both given points is equal, we can write the distance formula: \[d_1 = \sqrt{(x-a_1)^2+(y-b_1)^2}\] \[d_2 = \sqrt{(x-a_2)^2+(y-b_2)^2}\] Now, we have \(d_1 = d_2\).

Square both sides of the distance equation

Square both sides of the equation \(d_1 = d_2\) to get rid of the square root: \[(x-a_1)^2+(y-b_1)^2 = (x-a_2)^2+(y-b_2)^2\]

Simplify the equation to find the value of c

Expand and simplify the equation: \[x^2 -2a_1x +a_1^2 +y^2 -2b_1y + b_1^2 = x^2 -2a_2x +a_2^2 +y^2 -2b_2y + b_2^2\] Cancel out the similar terms: \[2a_2x - 2a_1x + 2b_2y - 2b_1y = a_2^2 - a_1^2 + b_2^2 - b_1^2\] Compare this equation with the given locus equation: \[\left(a_{1}-a_{2}\right) x + \left(b_{1} - b_{2}\right) y + c = 0\] We get the value of c: \[c = \frac{1}{2}(a_2^2 - a_1^2 + b_2^2 - b_1^2)\] From the given options, the correct answer is (a) \[\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}\right)\].

Key concepts

Perpendicular Bisector

The concept of a perpendicular bisector is fundamental in coordinate geometry, particularly when defining the locus of points equidistant from two separate points. Imagine you have a line segment with endpoints; the perpendicular bisector would be a line that divides this segment into two equal parts at a 90-degree angle.

When you have two fixed points, such as \(a_1, b_1\) and \(a_2, b_2\), the set of points that are equidistant from them will always lie on this perpendicular bisector. This makes it a vital tool for solving various geometric problems, including ones where we need to express a locus equation.

Coordinate Geometry

Within the realm of JEE Maths problems, coordinate geometry plays an integral role. It provides a framework where geometric problems are translated into algebraic form. Using coordinates, one can represent geometric figures, like points, lines, and circles, with equations.

Understanding how to derive and manipulate these equations allows us to solve complex problems related to loci, tangents, distances, and more. In the context of our problem, coordinate geometry was used to define the locus equation as a line representing the perpendicular bisector.

Distance Formula

The distance formula, \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\), derived from the Pythagorean theorem, is a powerful tool. It is used to calculate the distance between two points in a coordinate plane. This concept is frequently tested in JEE Maths problems and is crucial for problems involving loci.

As shown in the solution steps, squaring and rearranging the formula sets the foundation for calculating the value of 'c' in the equation of the locus. Avoiding common mistakes such as incorrect squaring or expansion of terms is essential when applying the distance formula.

JEE Maths Problems

Problems in the JEE Mathematics section often involve complex applications of coordinate geometry, including locus equations and the distance formula. To excel in these, a deep understanding and accurate application of the underlying principles are vital.

Students should focus on the process: breaking problems into smaller steps, like finding the equation of a perpendicular bisector or applying the distance formula correctly. Moreover, practicing a variety of problems can help in familiarizing with different problem patterns and enhance problem-solving speed and accuracy.