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: Understanding the Concept:
The given equation represents a pair of lines. Look for a perfect square in the second-degree terms: \(16x^2 - 24xy + 9y^2 = (4x - 3y)^2\). This tells us that the lines are parallel.

Step 2: Factorise:
Let \(u = 4x - 3y\). The equation is \(u^2 - 2u - 35 = 0\), because \(-8x + 6y = -2(4x - 3y)\).
\[ (u - 7)(u + 5) = 0 \]
So the lines are \(4x - 3y = 7\) and \(4x - 3y = -5\).

Step 3: Locus:
Points equidistant from two parallel lines lie on the parallel line midway between them: \(4x - 3y = \frac{7 + (-5)}{2} = 1\).
So the locus is \(4x - 3y - 1 = 0\).

Step 4: Why the other options are wrong.
\(4x - 3y + 1 = 0\) corresponds to \(u = -1\), not the midpoint. Options (C) and (D) have coefficients \(8x - 6y\), which gives \(u = \frac12\) or \(-\frac12\), also not 1.

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