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Differential Equation Giving a Pair of Lines | JEE

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Question
The solution of dydx=x−2y+32x−y+5\dfrac{dy}{dx} = \dfrac{x - 2y + 3}{2x - y + 5} is f(x,y,c)=0f(x, y, c) = 0. For suitable values of cc, f(x,y,c)=0f(x, y, c) = 0 represents a pair of lines whose point of intersection is:
A(−73,23)\left(-\dfrac{7}{3}, \dfrac{2}{3}\right)
B(−73,43)\left(-\dfrac{7}{3}, \dfrac{4}{3}\right)
C(−73,13)\left(-\dfrac{7}{3}, \dfrac{1}{3}\right)correct
D(73,−13)\left(\dfrac{7}{3}, -\dfrac{1}{3}\right)
Solution
Step 1: Cross-multiply and group:
(x−2y+3) dx−(2x−y+5) dy=0(x - 2y + 3)\,dx - (2x - y + 5)\,dy = 0
⇒x dx+y dy−2(y dx+x dy)+3 dx−5 dy=0.\Rightarrow x\,dx + y\,dy - 2(y\,dx + x\,dy) + 3\,dx - 5\,dy = 0.
 Step 2: Integrate term by term:
x22+y22−2xy+3x−5y=c′⇒x2+y2−4xy+6x−10y+c=0.\frac{x^2}{2} + \frac{y^2}{2} - 2xy + 3x - 5y = c' \Rightarrow x^2 + y^2 - 4xy + 6x - 10y + c = 0.
 Step 3: For a pair of lines, the point of intersection is the centre, found from ∂f∂x=∂f∂y=0\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0:
2x−4y+6=0⇒x−2y+3=0,2x - 4y + 6 = 0 \Rightarrow x - 2y + 3 = 0,
−4x+2y−10=0⇒2x−y+5=0.-4x + 2y - 10 = 0 \Rightarrow 2x - y + 5 = 0.
 Step 4: Solve: from the first, x=2y−3x = 2y - 3; substituting, 2(2y−3)−y+5=0⇒3y=1⇒y=132(2y - 3) - y + 5 = 0 \Rightarrow 3y = 1 \Rightarrow y = \dfrac{1}{3}, x=−73x = -\dfrac{7}{3}. Correct answer: (3)
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