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:
The question asks for the area of a triangle enclosed by three lines. One line is given explicitly as a horizontal line, $y = 9$. The other two lines are combined within a homogeneous second-degree equation representing a pair of straight lines passing through the origin.

Step 2: Key Formula or Approach:
1. Factorize the joint equation $y^2 - 9xy + 18x^2 = 0$ into two distinct linear equations of the form $y = m_1x$ and $y = m_2x$.
2. Determine the three vertices of the triangle by finding the points of intersection between these lines.
3. Use the coordinate geometry formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:
$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Step 3: Detailed Explanation:
Let's first split the pair of straight lines: $$y^2 - 9xy + 18x^2 = 0$$ Splitting the middle term: $$y^2 - 6xy - 3xy + 18x^2 = 0$$ $$y(y - 6x) - 3x(y - 6x) = 0$$ $$(y - 3x)(y - 6x) = 0$$ This gives two separate straight lines: Line 1: $y = 3x$
Line 2: $y = 6x$
The third line is Line 3: $y = 9$.
Now, let's find the intersection points (vertices): Intersection of Line 1 and Line 2: Both pass through the origin, so Vertex 1 is $(0, 0)$. Intersection of Line 1 ($y = 3x$) and Line 3 ($y = 9$): Substituting $y = 9$ into $y = 3x$ gives $9 = 3x \implies x = 3$. So, Vertex 2 is $(3, 9)$. Intersection of Line 2 ($y = 6x$) and Line 3 ($y = 9$): Substituting $y = 9$ into $y = 6x$ gives $9 = 6x \implies x = \frac{9}{6} = \frac{3}{2}$. So, Vertex 3 is $(\frac{3}{2}, 9)$. Now apply the area formula with coordinates $(0,0)$, $(3,9)$, and $(\frac{3}{2},9)$: $$\text{Area} = \frac{1}{2} \left| 0(9 - 9) + 3(9 - 0) + \frac{3}{2}(0 - 9) \right|$$ $$\text{Area} = \frac{1}{2} \left| 0 + 27 - \frac{27}{2} \right|$$ $$\text{Area} = \frac{1}{2} \left| \frac{27}{2} \right| = \frac{27}{4}$$

Step 4: Final Answer:
The area of the triangle is $\frac{27}{4}$ square units, which corresponds to option (C).