The combined equation of the two lines $a x+b y+c=0$ and $a^{\prime} x+b^{\prime} y+c^{\prime}=0$ can be written as $(a x+b y+c)\left(a^{\prime} x+b^{\prime} y+c^{\prime}\right)=0$
The equation of the angle bisectors of the lines represented by the equation $2 x^2+x y-3 y^2=0$ is
The equation of the angle bisectors of the lines represented by the equation $2 x^2+x y-3 y^2=0$ is
- $3 x^2+5 x y+2 y^2=0$
- $x^2-y^2-10 x y=0$
- $3 x^2+x y-2 y^2=0$
- $x^2-y^2+10 x y=0$
The Correct Option is B
Solution and Explanation
To find the equation of the angle bisectors of the lines represented by the equation \(2x^2 + xy - 3y^2 = 0\), we first need to understand that this equation is a homogeneous equation of second degree representing two lines through the origin.
The given equation can be written as two separate lines:
\(2x^2 + xy - 3y^2 = 0\)
We assume the equation to be of the form \(ax^2 + 2hxy + by^2 = 0\). Here, \(a = 2\), \(h = \frac{1}{2}\), and \(b = -3\).
Using the formula for angle bisectors of lines represented as \(ax^2 + 2hxy + by^2 = 0\), which is:
\(\frac{x^2 - y^2}{a - b} = \frac{xy}{h}\)
Substitute the values \(a = 2\), \(b = -3\), and \(h = \frac{1}{2}\) into the formula:
\(\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}\)
Which simplifies to:
\(\frac{x^2 - y^2}{5} = 2xy\)
On clearing the denominators, this becomes:
\(x^2 - y^2 = 10xy\)
Re-writing gives:
\(x^2 - y^2 - 10xy = 0\)
Thus, the correct answer is:
$x^2-y^2-10 x y=0$
