Question

The area (in square unit) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^2-3xy+2y^2=0$ is

• A. $\frac{7}{12}$
• B. $\frac{5}{12}$
• C. $\frac{1}{12}$
• D. $\frac{1}{6}$

Hint:

Area of triangle formed by $y = m_1x, y = m_2x$ and $lx + my + n = 0$ is $\frac{n^2 \sqrt{(m_1-m_2)^2}}{2|(l+m_1m)(l+m_2m)|}$.

Answer 1

The Correct Option is C

Step 1: Separate the Pair of Lines
$x^2 - 3xy + 2y^2 = (x - 2y)(x - y) = 0$.
Lines: $L_1: x - 2y = 0$ and $L_2: x - y = 0$. Both pass through the origin $(0, 0)$.
Step 2: Find Vertices
Vertex 1: $(0, 0)$.
Vertex 2: Intersection of $x - 2y = 0$ and $x + y + 1 = 0$.
$2y + y + 1 = 0 \Rightarrow y = -1/3, x = -2/3$. Point $A(-2/3, -1/3)$.
Vertex 3: Intersection of $x - y = 0$ and $x + y + 1 = 0$.
$y + y + 1 = 0 \Rightarrow y = -1/2, x = -1/2$. Point $B(-1/2, -1/2)$.
Step 3: Area Calculation
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |0(-1/2 + 1/3) - \frac{2}{3}(-1/2 - 0) - \frac{1}{2}(0 + 1/3)|$.
Area $= \frac{1}{2} |0 + 1/3 - 1/6| = \frac{1}{2} (\frac{1}{6}) = \frac{1}{12}$.
Final Answer: (c)