Question

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

• A. \(\frac{12}{5}\)
• B. \(\frac{5}{12}\)
• C. \(\frac{1}{12}\)
• D. \(\frac{1}{6}\)

Hint:

For pair of lines \(ax^2+2hxy+by^2=0\), factor it into two lines, then find triangle vertices and apply coordinate geometry area formula.

Answer 1

The Correct Option is C

Step 1: Factor the pair of straight lines.
\[ x^2-3xy+2y^2=0 \Rightarrow (x-y)(x-2y)=0 \]
So lines are:
\[ x-y=0,\quad x-2y=0 \]
Step 2: Find intersection points with \(x+y+1=0\).
With \(x-y=0\Rightarrow x=y\):
\[ x+x+1=0\Rightarrow 2x=-1\Rightarrow x=-\frac{1}{2} \]
So:
\[ P\left(-\frac{1}{2},-\frac{1}{2}\right) \]
With \(x-2y=0\Rightarrow x=2y\):
\[ 2y+y+1=0\Rightarrow 3y=-1\Rightarrow y=-\frac{1}{3} \]
\[ x=2y=-\frac{2}{3} \]
So:
\[ Q\left(-\frac{2}{3},-\frac{1}{3}\right) \]
Intersection of \(x-y=0\) and \(x-2y=0\):
From \(x=y\) and \(x=2y\Rightarrow y=0,x=0\).
So:
\[ R(0,0) \]
Step 3: Area of triangle \(PQR\).
Using determinant formula:
\[ \Delta=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \]
Here:
\[ P\left(-\frac{1}{2},-\frac{1}{2}\right),\; Q\left(-\frac{2}{3},-\frac{1}{3}\right),\; R(0,0) \]
\[ \Delta=\frac{1}{2}\left|-\frac{1}{2}\left(-\frac{1}{3}-0\right)+\left(-\frac{2}{3}\right)(0+\frac{1}{2})+0\right| \]
\[ =\frac{1}{2}\left|\frac{1}{6}-\frac{1}{3}\right| =\frac{1}{2}\cdot\frac{1}{6} =\frac{1}{12} \]
Final Answer:
\[ \boxed{\frac{1}{12}} \]