Problem 48 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(\triangle A M N\) 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 correct answer is (A) 3.

Step 1: Identify Key Points

We are given the line equation \(x+y=a\) that intersects the x-axis at point \(A\). Since on the \(x\)-axis, \(y=0\), substituting into the equation gives \(x+a=a\), so \(A = (a,0)\). The origin \(O\) is \((0,0)\) and point \(B\) can be determined as \(B = (0,a)\) because the line also meets the \(y\)-axis where \(x=0\).

Step 2: Triangle Reference Calculation

The triangle \(OAB\) is a right triangle with vertices at \(O(0,0)\), \(A(a,0)\), and \(B(0,a)\). Using the area formula for a right triangle \(\frac{1}{2} \times \text{base} \times \text{height}\), the area of \(\triangle OAB\) is \(\frac{1}{2} \times a \times a = \frac{a^2}{2}\).

Step 3: Link Area Condition

According to the problem, \(\triangle AMN\) is \(\frac{3}{8}\) the area of \(\triangle OAB\). Hence, the area of \(\triangle AMN\) is \(\frac{3}{8} \times \frac{a^2}{2} = \frac{3a^2}{16}\).

Step 4: Understand Line Segments

Since \(M\) is on \(OB\) and \(N\) is on \(AB\), we introduce \(M = (0,ma)\) and \(N = (ka, (1-k)a)\), where \(0 \leq k \leq 1\) and \(0 \leq m \leq 1\) because \(M\) and \(N\) lie along the segments from \(O\) to \(B\) and \(A\) to \(B\) respectively.

Step 5: Calculate Area of \(\triangle AMN\)

\(\triangle AMN\) can be computed using the determinant formula: \[\text{Area} = \frac{1}{2} \left| a(1-k)a + 0 - makk \right| = \frac{1}{2} a^2 (1-k) \].Using the given condition, \[ \frac{1}{2}a^2(1-k) = \frac{3a^2}{16} \].

Step 6: Solve for \(k\)

Equating and solving, \( \frac{1}{2}(1-k) = \frac{3}{16} \Rightarrow 1-k = \frac{3}{8} \Rightarrow k = \frac{5}{8} \).

Step 7: Compute Ratio \(\frac{AN}{BN}\)

Since \(N = (ka, (1-k)a)\), \(AN\) and \(BN\) are line segments such that \(AN = ka = \frac{5}{8}a\) and \(BN = (1-k)a = \frac{3}{8}a\).Hence, \(\frac{AN}{BN} = \frac{\frac{5}{8}a}{\frac{3}{8}a} = \frac{5}{3}.\)

Key Concepts

Right Triangle

A right triangle is one that includes a 90-degree angle. In our problem, triangle \(OAB\) is a right triangle where point \(O\) is the origin at (0, 0), point \(A\) is where the line \(x + y = a\) meets the x-axis at \(A = (a, 0)\), and point \(B\) is where the line meets the y-axis at \(B = (0, a)\). These points form the perfect setup for a right triangle because:

• \(OA\) and \(OB\) are at right angles to each other.
• The hypotenuse is \(AB\), spanning from the x-axis to the y-axis.

Right triangles have special properties that make calculations easier, such as using one leg as the base and the other as the height, simplifying area calculations.

Area Calculation

To find the area of a triangle, we use the formula \( \frac{1}{2} \times \text{base} \times \text{height} \). In a right triangle, this formula becomes very straightforward because the base and height are simply the legs of the triangle, which stand at right angles.
For triangle \(OAB\), both the base \(OA\) and the height \(OB\) are of length \(a\). So, the area is:\[\frac{1}{2} \times a \times a = \frac{a^2}{2}\]Next, triangle \(AMN\) is given as \(\frac{3}{8}\) of the area of triangle \(OAB\). This yields:\[\frac{3}{8} \times \frac{a^2}{2} = \frac{3a^2}{16}\]Thus, knowing a part of a triangle's area helps us find various points and relations in geometric problems.

Line Intersection

Line intersections occur when two lines meet at a single point. In our scenario, we're dealing with the intersection of the line \(x+y=a\) with both x and y-axes. This intersection gives us points \(A\) and \(B\) which, respectively, are:

• \(A = (a, 0)\)
• \(B = (0, a)\)

The line intersects the x-axis where \(y = 0\), substituting into the line equation gives \(x = a\), resulting in point \(A\). When intersecting the y-axis, \(x = 0\), hence \(y = a\), which gives us point \(B\). Understanding these intersections helps us to form the vertices of the triangle \(OAB\) and solve the problem.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, entwines algebra and geometry by using coordinate points to express geometric shapes. In triangle \(AMN\), using coordinates is vital for determining the vertices based on intersections and proportional distances. Points \(M\) and \(N\) lie on segments \(OB\) and \(AB\) respectively. We describe \(M\) as \((0, ma)\) and \(N\) as \((ka, (1-k)a)\), where \(0 \leq k, m \leq 1\). This lets us express possible positions using variables along the known segments.
Coordinate geometry helps solve problems by transforming them into algebraic terms. Using proportions or divisions along lines from two points can determine locations within a plane, aiding in calculating distances or ratios like \(\frac{AN}{BN}\). It turns visual understanding into quantifiable mathematics.