Question

If one of the lines \(my^2 + (1-m^2)xy - mx^2 = 0\) is a bisector of the angle between the lines \(xy = 0\), find \(m\).

• A. -2
• B. -1
• C. 1/2
• D. -1/2

Hint:

When dealing with a pair of lines \(ax^2 + 2hxy + by^2 = 0\), the angle bisector equation is \(\frac{x^2 - y^2}{a - b} = \frac{xy}{h}\). Using this on \(xy=0\) (where \(a=0, b=0, 2h=1\)) instantly gives \(x^2 - y^2 = 0\).

Answer 1

The Correct Option is B

Step 1: Identify the bisectors of $xy=0$.
The pair $xy=0$ is the two coordinate axes. Their angle bisectors are $y=x$ and $y=-x$, i.e. lines of slope $+1$ and $-1$.
Step 2: Factorise the given pair.
Rewrite $my^2+(1-m^2)xy-mx^2=0$ as $-mx^2+xy-m^2xy+my^2=0$. Grouping, $-x(mx-y)-my(mx-y)=0$, so $-(mx-y)(x+my)=0$.
Step 3: Read off the two lines.
The factors give $y=mx$ (slope $m$) and $y=-\dfrac{x}{m}$ (slope $-\tfrac1m$).
Step 4: Impose the bisector condition.
One of these must be a bisector, so its slope is $\pm1$. Testing the slope $m=\pm1$ is the natural route.
Step 5: Check $m=-1$.
With $m=-1$ the lines become $y=-x$ and $y=-\dfrac{x}{-1}=x$, exactly the two bisectors of the axes.
Step 6: Conclude.
So $m=-1$ satisfies the requirement, which is option 2.
\[ \boxed{m=-1} \]