Let A=(1 2) B=(3 4) and let C=(x y) be a point such that (x-1)(x-3) + (y-2)(y-4)=0.if area of triangle ABC equal to 1.then maximum number of position of C in the x-y plane

To solve the problem, we need to analyze the given equation and the conditions for the area of triangle ABC. The points A and B are fixed, while point C can vary in the x-y plane. The equation provided, \((x-1)(x-3) + (y-2)(y-4) = 0\), represents a specific geometric condition that point C must satisfy. Let's break this down step by step.

Understanding the Equation

The equation \((x-1)(x-3) + (y-2)(y-4) = 0\) can be rearranged to understand its geometric significance. This is a quadratic equation in both x and y, which describes a conic section. Specifically, it can be rewritten as:

• Expanding the terms: \(x^2 - 4x + 3 + y^2 - 6y + 8 = 0\)
• Combining like terms gives: \(x^2 + y^2 - 4x - 6y + 11 = 0\)

This represents a circle or an ellipse depending on the coefficients, but in this case, it simplifies to a circle centered at the point \((2, 3)\) with a certain radius.

Finding the Area of Triangle ABC

The area of triangle ABC can be calculated using the formula:

Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2|

Substituting the coordinates of points A, B, and C:

• A = (1, 2)
• B = (3, 4)
• C = (x, y)

The area becomes:

Area = 0.5 * |1(4 - y) + 3(y - 2) + x(2 - 4)|

Setting this equal to 1 (as given in the problem), we have:

|1(4 - y) + 3(y - 2) - 2x| = 2

Solving for Point C

Now, we need to solve this equation along with the conic section equation to find the maximum number of positions for point C. The area condition gives us two cases:

• Case 1: \(1(4 - y) + 3(y - 2) - 2x = 2\)
• Case 2: \(1(4 - y) + 3(y - 2) - 2x = -2\)

Both cases will yield linear equations in terms of x and y. We can solve these equations simultaneously with the conic section equation to find the intersection points, which represent the possible positions of point C.

Finding the Maximum Positions

Since the equation of the conic section is quadratic, and each linear equation can intersect it at most at two points, we can analyze the intersections:

• Each linear equation can intersect the circle at two points.
• Thus, for two linear equations, we can have a maximum of \(2 \times 2 = 4\) intersection points.

Therefore, the maximum number of positions for point C in the x-y plane, given the area of triangle ABC is equal to 1, is 4.

Final Thoughts

In summary, by analyzing the geometric implications of the equations and the area condition, we determined that point C can occupy up to four distinct positions in the x-y plane while satisfying all given conditions. This approach illustrates the interplay between algebra and geometry in solving such problems.