Problem 21 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(O\) be the origin and let \(A(2,0), B(0,2)\) be two points. If \(P(x, y)\) is a point such that \(x y>0\) and \(x+y<\) 2 , then (A) \(P\) lies either inside the triangle \(O A B\) or in the third quadrant (B) \(P\) cannot be inside the triangle \(O A B\) (C) \(P\) lies inside the triangle \(O A B\) (D) none of these

Step-by-Step Solution

Answer

Option (A) is correct; "P" lies either inside triangle OAB or in the third quadrant.

Step 1: Understanding the conditions

To solve the problem, we need to analyze the conditions given: \(xy > 0\) means that \(P\) must be in either the first quadrant (both \(x\) and \(y\) positive) or the third quadrant (both \(x\) and \(y\) negative). The inequality \(x+y < 2\) further restricts the location of \(P\) in these quadrants.

Step 2: Equations of lines: Triangle OAB

Firstly, identify the lines forming triangle \(OAB\). The line OA is along the x-axis where \(y = 0\) and extends from \(O(0,0)\) to \(A(2,0)\). The line OB is along the y-axis where \(x = 0\) and extends from \(O(0,0)\) to \(B(0,2)\). The line AB has the equation \(y = -x + 2\) obtained from its endpoints.

Step 3: Constraints analysis for first quadrant

In the first quadrant (where \(x > 0\) and \(y > 0\)), the inequality \(y = -x + 2\) from line AB, and the inequality \(x + y < 2\) are compared. Substituting a point (like \(P(1,0.5)\)) which both satisfies \(xy > 0\) and \(x+y < 2\) suggests \(P\) could lie inside \(OAB\) because it fulfills all conditions set by triangle lines and given inequalities.

Step 4: Constraints analysis for third quadrant

In the third quadrant (where \(x < 0\) and \(y < 0\)), \(xy > 0\) is satisfied, as is \(x+y < 2\) since both \(x\) and \(y\) are negative and their sum will also be negative and thus less than 2. This shows \(P\) can theoretically exist here but is outside the triangle \(OAB\) because \(OAB\) is not defined in third quadrant.

Step 5: Conclusion: Option elimination

Given that points satisfying the conditions lie inside the triangle in the first quadrant or theoretically in the third quadrant when \(xy > 0\), option (A) is correct because \(P\) lies either inside triangle \(OAB\) or in the third quadrant.

Key Concepts

Triangles

Triangles are basic shapes in geometry, consisting of three edges and three vertices. In coordinate geometry, we can easily describe triangles using points and lines. For a triangle like \(\triangle OAB\), we define the vertices at points \(O(0,0)\), \(A(2,0)\), and \(B(0,2)\). These points are critical, as they dictate the borders of the triangle.

To understand a triangle's position on a coordinate plane, we need to look at the lines connecting these points. For triangle \(OAB\), OB forms a vertical line at \(x = 0\) and OA forms a horizontal line at \(y = 0\). The other side, the diagonal AB, follows the equation \(y = -x + 2\), arising from the basic principle that a line's equation can be derived from two known points.

When dealing with inequalities involving a triangle, it's essential to use these lines as boundaries. They help determine whether a point is inside or outside the triangle. By substituting any coordinate point into the equation of these lines, we can check its position relative to the triangle.

Quadrants

The coordinate plane is divided into four sections called quadrants. These quadrants are helpful to locate points based on the signs of their coordinates (x, y).

• Quadrant I: Both \(x\) and \(y\) are positive.
• Quadrant II: \(x\) is negative, \(y\) is positive.
• Quadrant III: Both \(x\) and \(y\) are negative.
• Quadrant IV: \(x\) is positive, \(y\) is negative.

For the exercise given, any point \(P(x, y)\) must fulfill \(xy > 0\). This means \(P\) can only be in either the first or third quadrant.

Quadrants play a key role in problems involving location constraints. For example, when a condition suggests \(x + y < 2\), it restricts points' positions in these quadrants. The use of quadrants helps understand where points might be located, based on given conditions.

Inequalities

Inequalities are statements about the relative size or order of two objects. In coordinate geometry, they help define areas on the plane that satisfy certain conditions. In problems involving triangles, inequalities can constrain solutions within certain regions or lines.

The condition \(xy > 0\) means both coordinates share the same sign. It indicates points in the first and third quadrants, where \(x\) and \(y\) are either both positive or both negative.

Another condition from the exercise, \(x+y < 2\), helps further narrow down the region that \(P\) can occupy. This linear inequality implies that any point in these quadrants must lie below the line \(x + y = 2\). By intersecting these constraints with the triangle's boundaries, you can determine positions allowed by both the triangle and inequalities.

Point Location

Point location involves analyzing the position of a point concerning geometrical objects or restrictions on the coordinate plane. In the context of the exercise, determining the location of point \(P(x, y)\) requires checking if it satisfies multiple conditions.

First, confirm that \(P\) satisfies \(xy > 0\), meaning it resides in either Quadrant I or III. However, if \(x + y < 2\), check its presence relative to the triangle \(OAB\).

For a point \(P\) to be inside \(\triangle OAB\) in Quadrant I, substitute its coordinates into \(y = -x + 2\) and \(x + y < 2\). If both conditions hold, \(P\) lies inside this bounded area. However, in Quadrant III, despite satisfying \(xy > 0\), \(P\) cannot be inside the triangle, because triangle \(OAB\) does not cover this quadrant.