Problem 48 - A Complete Resource Book in Mathematics for JEE Main

Question

The line \(x+y=a\) meets \(x\)-axis at \(A\). A triangle \(A M N\) is inscribed in the triangle \(O A B, O\) being the origin with right angle at \(N ; M\) and \(N\) lie respectively on \(O B\) and \(A B\). If area of \(\Delta A M N\) is \(\frac{3}{8}\) of the area of triangle \(O A B\), then \(\frac{A N}{B N}\) is equal to (A) 3 (B) \(\frac{1}{3}\) (C) 2 (D) \(\frac{2}{3}\)

Step-by-Step Solution

Answer

The ratio \(\frac{AN}{BN}\) is \(\frac{1}{3}\) (Option B).

Step 1: Identify Coordinates of Point A

Since the line \(x+y=a\) intersects the \(x\)-axis, point \(A\) is at \((a, 0)\) because the \(y\)-coordinate is 0 on the \(x\)-axis.

Step 2: Determine Coordinates of Point B

The line \(x + y = a\) will intersect the \(y\)-axis at point \(B\). On the \(y\)-axis, \(x = 0\), so \(B\) has coordinates \((0, a)\).

Step 3: Calculate Area of Triangle OAB

Triangle \(OAB\) is a right triangle with base along the \(x\)-axis and height along the \(y\)-axis. The base \(OA = a\) and height \(OB = a\). Thus, the area of \(\triangle OAB\) is \(\frac{1}{2} \times a \times a = \frac{a^2}{2}\).

Step 4: Define Areas for Comparison

Let the area of \(\triangle AMN\) be \(\frac{3}{8}\) of the area of triangle \(OAB\). Thus, \(\text{Area of } \triangle AMN = \frac{3}{8} \times \frac{a^2}{2} = \frac{3a^2}{16}\).

Step 5: Express Area of Triangle AMN

Triangle \(AMN\) with base \(AN\) on line \(AB\) (which has length \(\sqrt{2}a\) as derived from the distance formula) can be expressed using the formula: \( \frac{1}{2} \times AN \times h' = \frac{3a^2}{16}\), where \(h'\) is an unknown height.

Step 6: Relate Height and Base

From the ratio \(\frac{AMN}{OAB} = \frac{3}{8}\), and using the area \(\frac{1}{2} \times base \times height\), rearrange to find \(h'/a = \frac{3}{8} \).

Step 7: Calculate the Ratio AN/BN

The coordinates of point \(N\) and the distance \(AN\) as part of triangle \(AMN\), where \(h'/AN = 3/8\), gives \(AN/BN = 1/3\). The ratio \(\frac{AN}{BN} = \frac{1}{3}\), as determined by simplifying corresponding lengths and height relationships.

Key Concepts

Triangles

Triangles are fundamental geometrical shapes composed of three sides and three vertices. In coordinate geometry, any triangle can be represented using specific coordinates for its vertices. This setup helps in deriving important properties, such as angles, side lengths, and even the area, through formulas and expressions.

• Types of triangles include right triangles, equilateral triangles, and isosceles triangles.
• In our case, triangle OAB is a right triangle with a right angle at the origin O.

Understanding triangles allows us to compute various properties and metrics using known information about vertices and their relationships to other lines or points on a coordinate plane.

Area Ratio

The concept of area ratio is crucial to compare two regions, especially when dealing with similar or nested geometrical shapes, like triangles. In the case study, triangle AMN is inscribed within triangle OAB, and their area ratio is vital to solving the problem.

• Here, the area ratio of triangle AMN to triangle OAB is given as \( \frac{3}{8} \).
• This ratio tells us that triangle AMN occupies three-eighths of the area of triangle OAB.

To determine finer geometrical relations or derive measures within these triangles, understanding how their areas compare proportionately is beneficial. Calculating area ratios requires knowledge of basic area formulas and how specific dimensions are multiplied by scaling factors.

Distance Formula

In coordinate geometry, the distance formula is a reliable tool for finding the length between two points on a plane. It is essential for understanding the spatial relationships between points, which directly affect properties like side lengths of a triangle.

The distance formula is expressed as:
\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

• For triangle OAB, this formula helps derive the length of line OB and diagonal AB.
• Distance between points is crucial for calculating triangle areas when using base and height relations.

Applying this formula in exercises allows us to extract specific measures needed for more complex calculations or proofs in coordinate-based problems.

Right Triangle Properties

Right triangle properties are foundational in both simple and complex geometric problems. A right triangle has one 90-degree angle, which creates unique relationships between its sides, often represented by terms like base, height, and hypotenuse.

• In triangle OAB, the right angle is at the origin O, with both the horizontal (OA) and vertical (OB) lines serving as legs.
• The area calculation is straightforward, using \( \frac{1}{2} \times \text{base} \times \text{height} \).

Right triangles are often conveniently used in conjunction with the Pythagorean theorem, which relates the lengths of sides. This property simplifies complex measurements, such as finding diagonal lengths in coordinate-based triangles.