Question

The distance between the lines represented by \(16 x^2+9 y^2+48 x-24 x y-36 y+35=0\) is ......... units

• A. \(\frac{2}{5}\)
• B. \(\frac{35}{2}\)
• C. \(\frac{5}{2}\)
• D. \(\frac{7}{5}\)

Hint:

The quadratic is a perfect square in 4x - 3y, so factorise it into two parallel lines.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Here \(a=16,\ b=9,\ h=-12\), so \(h^{2}=ab\) and the equation represents two parallel lines.

Step 2: Factorise
\(16x^{2}-24xy+9y^{2}=(4x-3y)^{2}\) and \(48x-36y=12(4x-3y)\). With \(u=4x-3y\): \(u^{2}+12u+35=0\), so \((u+5)(u+7)=0\).

Step 3: Write the lines
The lines are \(4x-3y+5=0\) and \(4x-3y+7=0\).

Step 4: Distance
Distance \(=\dfrac{|7-5|}{\sqrt{4^{2}+3^{2}}}=\dfrac{2}{5}\) units.

Final Answer:
The lines are 2/5 units apart. \[ \boxed{\text{(A) }\dfrac{2}{5}} \]