Question

If the equation \(mx^{2}-y^{2}+2y+n=3\) represents a pair of perpendicular lines, then the values of \(m\) and \(n\) respectively are...

• A. \(1,3\)
• B. \(1,2\)
• C. \(1,-3\)
• D. \(1,-2\)

Hint:

For perpendicular lines the coefficients of x squared and y squared add to zero, and the determinant must vanish.

Answer 1

The Correct Option is B

Step 1: Rewrite
\(mx^2-y^2+2y+(n-3)=0\). Compare with \(ax^2+2hxy+by^2+2gx+2fy+c=0\): \(a=m,\ b=-1,\ h=0,\ g=0,\ f=1,\ c=n-3\).

Step 2: Perpendicular condition
\(a+b=0\Rightarrow m-1=0\Rightarrow m=1\).

Step 3: Pair of lines condition
\(abc+2fgh-af^2-bg^2-ch^2=0\) gives \(m(-1)(n-3)-m(1)=0\).

Step 4: Solve
With \(m=1\): \(-(n-3)-1=0\Rightarrow n=2\).

Final Answer:
The values are m = 1 and n = 2. \[ \boxed{\text{(B) }1,2} \]