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(C) The equation of the line is $x + 2y + 1 = 0$,which can be written as $-(x + 2y) = 1$. Substituting this into the homogeneous equation of the curve

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(C) The equation of the line is $x + 2y + 1 = 0$,which can be written as $-(x + 2y) = 1$. Substituting this into the homogeneous equation of the curve $2x^2 - 2xy + 3y^2 + (2x - y)(1) - (1)^2 = 0$,we get: $2x^2 - 2xy + 3y^2 + (2x - y)(-(x + 2y)) - (-(x + 2y))^2 = 0$. Expanding this: $2x^2 - 2xy + 3y^2 - (2x^2 + 4xy - xy - 2y^2) - (x^2 + 4xy + 4y^2) = 0$. $2x^2 - 2xy + 3y^2 - 2x^2 - 3xy + 2y^2 - x^2 - 4xy - 4y^2 = 0$. $-x^2 - 9xy + y^2 = 0$,or $x^2 + 9xy - y^2 = 0$. This is of the form $ax^2 + 2hxy + by^2 = 0$,where $a = 1$,$2h = 9$,$b = -1$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. Since $a + b = 1 - 1 = 0$,the lines are perpendicular. Therefore,$	heta = \frac{\pi}{2}$.

The angle between the lines joining the origin to the points of intersection of the line $x + 2y + 1 = 0$ and the curve $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ is

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