Question

If the equation \(16x^2-24xy+9y^2-8x+6y-35 = 0\) represents a pair of straight lines, then the equation of the locus of points equidistant from these two lines is..........

• A. \(4x-3y-1 = 0\)
• B. \(4x-3y+1 = 0\)
• C. \(8x-6y-1 = 0\)
• D. \(8x-6y+1 = 0\)

Hint:

Factor the second-degree expression into two parallel lines and take their midline.

Answer 1

The Correct Option is A

Step 1: Use a substitution:
Since $-8x + 6y = -2(4x - 3y)$, the whole equation depends only on $u = 4x - 3y$: $u^2 - 2u - 35 = 0$.

Step 2: Roots:
Discriminant $4 + 140 = 144$, so $u = \frac{2 \pm 12}{2} = 7$ or $-5$.

Step 3: Midline:
The two lines are $u = 7$ and $u = -5$, so equidistant points satisfy $u = 1$. The average of the constants gives 1, so $4x - 3y = 1$.

Final Answer:
The required line is 4x - 3y - 1 = 0, option (A). \[ \boxed{4x-3y-1=0} \]