Question

Area of the triangle formed by the lines $y^2 - 9xy + 18x^2 = 0$ and $y = 9$ is

• A. $\frac{27}{3}$ sq. units
• B. $\frac{27}{2}$ sq. units
• C. $\frac{27}{4}$ sq. units
• D. $27$ sq. units

Hint:

For a triangle formed by lines $y = m_1x$, $y = m_2x$, and a horizontal line $y = c$, the area can be calculated directly using the short formula:
$$\text{Area} = \frac{c^2}{2} \left| \frac{1}{m_1} - \frac{1}{m_2} \right|$$ Substituting $c = 9$, $m_1 = 3$, and $m_2 = 6$:
$$\text{Area} = \frac{9^2}{2} \left| \frac{1}{3} - \frac{1}{6} \right| = \frac{81}{2} \times \frac{1}{6} = \frac{27}{4}\ \text{sq. units.}$$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the area of the triangle formed by the pair of lines y² - 9xy + 18x² = 0 and the horizontal line y = 9.

Step 2: Key Formula or Approach:
Factor the joint equation into two lines through the origin. Find intersection points with y = 9 to get three vertices, then apply the coordinate area formula: Area = ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|.

Step 3: Detailed Explanation:
Factorizing: y² - 6xy - 3xy + 18x² = (y - 3x)(y - 6x) = 0. Lines: y = 3x and y = 6x. Intersections: Line 1 ∩ Line 2 → (0,0). Line 1 ∩ y = 9 → 9 = 3x → (3,9). Line 2 ∩ y = 9 → 9 = 6x → (3/2, 9). Area = ½|0(9-9) + 3(9-0) + (3/2)(0-9)| = ½|27 - 27/2| = ½|27/2| = 27/4.

Step 4: Final Answer:
The area is 27/4 sq. units, option (C).