Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(\left(a_{1}-a_{2}\right) x+\left(b_{1}-b_{2}\right) y\) (A) \(\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}\right)\) (B) \(a_{1}^{2}+a_{2}^{2}-b_{1}^{2}-b_{2}^{2}\) (C) \(\frac{1}{2}\left(a_{1}^{2}+a_{2}^{2}-b_{1}^{2}-b_{2}^{2}\right)\) (D) \(\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}\)

Step-by-Step Solution

Answer

Option (A) is correct: \(\frac{1}{2}(a_2^2+b_2^2-a_1^2-b_1^2)\).

Step 1: Understand the concept of locus

The locus of a point equidistant from two fixed points \((a_1, b_1)\) and \((a_2, b_2)\) is the perpendicular bisector of the line segment joining these two points. The equation of this locus should represent all the points \((x, y)\) that are equidistant from these points.

Step 2: Use the distance formula

The distance from a point \((x, y)\) to a point \((a_1, b_1)\) is given by \[ \sqrt{(x-a_1)^2 + (y-b_1)^2} \] and to a point \((a_2, b_2)\) is \[ \sqrt{(x-a_2)^2 + (y-b_2)^2} \].For the point \((x, y)\) to be equidistant from these points, these distances must be equal.

Step 3: Set up the equation

Equate the two distances: \[ \sqrt{(x-a_1)^2 + (y-b_1)^2} = \sqrt{(x-a_2)^2 + (y-b_2)^2} \].Square both sides to eliminate the square roots:\[ (x-a_1)^2 + (y-b_1)^2 = (x-a_2)^2 + (y-b_2)^2 \].

Step 4: Expand and simplify the equation

Expand both sides of the equation:\[ x^2 - 2a_1x + a_1^2 + y^2 - 2b_1y + b_1^2 \] \[ = x^2 - 2a_2x + a_2^2 + y^2 - 2b_2y + b_2^2 \].Now cancel out \(x^2\) and \(y^2\) from both sides and rearrange terms.

Step 5: Isolate the terms involving x and y

By canceling \(x^2\) and \(y^2\), we have:\[ -2a_1x - 2b_1y + a_1^2 + b_1^2 = -2a_2x - 2b_2y + a_2^2 + b_2^2 \].Rearranging the terms gives:\[ (a_2-a_1)x + (b_2-b_1)y = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) \].

Step 6: Compare with given options

The equation \[ (a_2-a_1)x + (b_2-b_1)y = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) \] matched with the option (A) \(\frac{1}{2}(a_2^2+b_2^2-a_1^2-b_1^2)\) is the correct option.

Key Concepts

Distance Formula

The distance formula is essential in geometry to calculate the distance between two points in a plane. It comes from the Pythagorean theorem and is expressed as: \[\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\] This formula is used to determine the distance between a point \((x_1, y_1)\) and another point \((x_2, y_2)\). The theory is straightforward: you find the horizontal and vertical distances between the points, square them, sum them, and then take the square root.

• Horizontal Distance: \(x_2 - x_1\)
• Vertical Distance: \(y_2 - y_1\)
• Sum of Squares: \( (x_2 - x_1)^2 + (y_2 - y_1)^2 \)
• Square Root: The final step to finding the actual distance, ensuring the measure isn't negative.

Using this formula ensures accuracy when measuring straight-line distances, useful in determining points equidistant to others, such as in our exercise.

Perpendicular Bisector

A perpendicular bisector is a line that divides another line segment into two equal parts at a right angle (90 degrees). When given two points \((a_1, b_1)\) and \((a_2, b_2)\), the perpendicular bisector of the line segment between these points has several critical properties.

• It is perpendicular to the original line segment.
• It passes through the midpoint of the segment.
• Any point on the perpendicular bisector is equidistant from the two endpoints of the segment.

To find the equation of a perpendicular bisector, first calculate the midpoint using: \[\left(\frac{a_1 + a_2}{2}, \frac{b_1 + b_2}{2}\right)\]Then, find the slope of the original line segment, which is \((b_2 - b_1) / (a_2 - a_1)\), and derive the negative reciprocal for the perpendicular slope, as this slope will be perpendicular to the original. These steps lead toward developing the line's equation, critical for understanding the locus of equidistant points.

Equidistant Points

Equidistant points are all the points that are equally distanced from a specific location or axis. In geometrical exercises, finding equidistant points often revolves around forming loci, which are sets of all points satisfying a given condition. For instance, if you need to find all points equidistant from two distinct points, you would look for points lying on the perpendicular bisector of the segment connecting these points.

• **Locus Concept**: The locus of points equidistant from two points \((a_1, b_1)\) and \((a_2, b_2)\) forms a line or plane, depending on the dimensions involved.
• **Use of Distance Formula**: This formula helps validate that all points on the locus assure equidistance.
• **Practical Applications**: Useful in various fields such as engineering, architecture, and more, wherever symmetry or balance is crucial.

When tackling problems involving equidistant points, breaking down the requirement as shown - like identifying the bisector or utilizing the distance formula - simplifies finding the necessary solutions.