Key Concepts and Formulas

• The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
• The equation of the angle bisectors of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.

Step-by-Step Solution

Step 1: Identify the coefficients in the given equation

The given equation is $2x^2 + xy - 3y^2 = 0$. We need to compare this with the general form $ax^2 + 2hxy + by^2 = 0$ to find the values of $a$, $b$, and $h$. Comparing the coefficients, we have: $a = 2$ $2h = 1$, which implies $h = \frac{1}{2}$ $b = -3$

Step 2: Apply the angle bisector formula

The formula for the angle bisectors is $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$. Substituting the values $a = 2$, $b = -3$, and $h = \frac{1}{2}$ into the formula, we get: $\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}$

Step 3: Simplify the equation

Simplify the equation obtained in the previous step: $\frac{x^2 - y^2}{2 + 3} = 2xy$ $\frac{x^2 - y^2}{5} = 2xy$ Multiply both sides by 5: $x^2 - y^2 = 10xy$ Rearrange the terms to get the equation in the standard form: $x^2 - y^2 - 10xy = 0$

Step 4: Rearrange the equation to match the target format

The target answer is of the form $3x^2 + xy - 2y^2 = 0$. However, our calculated result is $x^2 - y^2 - 10xy = 0$.

The given "Correct Answer" is (A) $3{x^2} + xy - 2{y^2} = 0$. Let us factor the given equation $2x^2+xy-3y^2=0$ to find the lines. $2x^2+xy-3y^2 = 2x^2 + 3xy - 2xy - 3y^2 = x(2x+3y) - y(2x+3y) = (x-y)(2x+3y)=0$. So the two lines are $x-y=0$ and $2x+3y=0$. The equation of the bisectors are given by: $\frac{x-y}{\sqrt{1^2+(-1)^2}} = \pm \frac{2x+3y}{\sqrt{2^2+3^2}}$ $\frac{x-y}{\sqrt{2}} = \pm \frac{2x+3y}{\sqrt{13}}$ $\sqrt{13}(x-y) = \pm \sqrt{2}(2x+3y)$ Squaring both sides, we have: $13(x-y)^2 = 2(2x+3y)^2$ $13(x^2-2xy+y^2) = 2(4x^2+12xy+9y^2)$ $13x^2-26xy+13y^2 = 8x^2+24xy+18y^2$ $5x^2 - 50xy - 5y^2 = 0$ $x^2-10xy-y^2=0$ However, this is not the angle bisector equation we want.

Let us try to factorize the equation in Option A. $3x^2+xy-2y^2 = 3x^2+3xy-2xy-2y^2=3x(x+y)-2y(x+y)=(3x-2y)(x+y)=0$. This implies the bisectors are $3x-2y=0$ and $x+y=0$.

The correct angle bisector equation should be: $\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm \frac{a'x+b'y+c'}{\sqrt{a'^2+b'^2}}$ In our case, $2x^2+xy-3y^2 = (2x+3y)(x-y) = 0$. Hence, we have the lines $2x+3y=0$ and $x-y=0$. $\frac{2x+3y}{\sqrt{2^2+3^2}} = \pm \frac{x-y}{\sqrt{1^2+(-1)^2}}$ $\frac{2x+3y}{\sqrt{13}} = \pm \frac{x-y}{\sqrt{2}}$ $\sqrt{2}(2x+3y) = \pm \sqrt{13}(x-y)$ Squaring both sides, $2(4x^2+12xy+9y^2) = 13(x^2-2xy+y^2)$ $8x^2+24xy+18y^2 = 13x^2-26xy+13y^2$ $5x^2-50xy-5y^2 = 0$ $x^2-10xy-y^2 = 0$.

This still doesn't match option A. There is an error in the provided options.

The correct equation for the angle bisectors is $x^2 - y^2 - 10xy = 0$, which is the same as ${x^2} - {y^2} - 10xy = 0$.

Common Mistakes & Tips

• Remember that the formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ applies only when the equation is in the form $ax^2 + 2hxy + by^2 = 0$.
• Be careful when identifying the value of $h$. It is half of the coefficient of the $xy$ term.
• Always simplify the equation after substituting the values to get the final answer.

Summary

We are given the equation $2x^2 + xy - 3y^2 = 0$. We identified $a = 2$, $h = \frac{1}{2}$, and $b = -3$. Using the formula for the angle bisectors, $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$, we obtained $\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}$. Simplifying this equation, we arrived at $x^2 - y^2 - 10xy = 0$. This matches option (B).

Final Answer The final answer is \boxed{{x^2} - {y^2} - 10xy = 0}, which corresponds to option (B).