Question

The point of intersection of the lines represented by the equation $2x^{2} + 3y^{2} + 7xy + 8x + 14y + 8 = 0$is

• A. $(0,2)$
• B. $(1,2)$
• C. $( - 2,0)$
• D. $( - 2,1)$

Answer 1

Answer: (C)

$( - 2,0)$

Answer 2

Let$\varphi \equiv 2x^{2} + 3y^{2} + 7xy + 8x + 14y + 8 = 0$

$\frac{\partial\varphi}{\partial x} = 4x + 7y + 8 = 0$ and $\frac{\partial\varphi}{\partial y} = 6y + 7x + 14 = 0$

On solving these equations, we get $x = - 2,y = 0$

Trick : If the equation is $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$The points of intersection are given by $\left\{ \frac{hf - bg}{ab - h^{2}},\frac{hg - af}{ab - h^{2}} \right\}$. Hence point is (– 2, 0)