Question

The combined equation of lines parallel to the coordinate axes and passing through the point of intersection of lines represented by \(x^2-6xy+5y^2+10x-14y+9 = 0\) is \(\ldots\)

• A. \(xy+2x+y+2 = 0\)
• B. \(xy+2x-y-2 = 0\)
• C. \(xy-2x+y-2 = 0\)
• D. \(xy-2x-y+2 = 0\)

Hint:

Factor the pair of lines, find their meeting point, then write (x-h)(y-k)=0.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A second degree equation of the form \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) can represent a pair of straight lines. Their point of intersection is the point we need. The lines through it parallel to the axes are \(x = h\) and \(y = k\).

Step 2: Key Formula or Approach:
Factor \(x^2 - 6xy + 5y^2 + 10x - 14y + 9\) as \((x - y + a)(x - 5y + b)\), because \(x^2 - 6xy + 5y^2 = (x-y)(x-5y)\).

Step 3: Detailed Explanation:
Expand and compare the coefficients of x, y and the constant term.
Coefficient of x: \(a + b = 10\).
Coefficient of y: \(-(b + 5a) = -14\), so \(5a + b = 14\).
Subtract: \(4a = 4\), so \(a = 1\) and \(b = 9\). The constant term is \(ab = 9\), which matches.
The two lines are \(x - y + 1 = 0\) and \(x - 5y + 9 = 0\).
Subtract them: \(4y - 8 = 0\), so \(y = 2\) and \(x = 1\). The point of intersection is \((1, 2)\).
Lines through \((1,2)\) parallel to the axes are \(x = 1\) and \(y = 2\). Their combined equation is
\[ (x - 1)(y - 2) = 0 \Rightarrow xy - 2x - y + 2 = 0 \]

Final Answer:
The combined equation is \(xy - 2x - y + 2 = 0\), option (D). \[ \boxed{xy - 2x - y + 2 = 0 \text{ (D)}} \]