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(A) To find the equation of the pair of lines joining the origin to the intersection points,we homogenize the equation of the curve $3x^2 + 4xy - 4x +

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$\pi/2$

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(A) To find the equation of the pair of lines joining the origin to the intersection points,we homogenize the equation of the curve $3x^2 + 4xy - 4x + 1 = 0$ using the line $2x + y = 1$ (i.e.,$2x + y = 1$).Substituting $1 = (2x + y)$ into the curve equation:$3x^2 + 4xy - 4x(2x + y) + (2x + y)^2 = 0$Expanding the terms:$3x^2 + 4xy - 8x^2 - 4xy + 4x^2 + 4xy + y^2 = 0$Simplifying the expression:$-x^2 + 4xy + y^2 = 0$ or $x^2 - 4xy - y^2 = 0$Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = -4$ (so $h = -2$),and $b = -1$.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 denominator is $0$,which implies $	an 	heta = \infty$.Therefore,$	heta = \pi/2$,meaning the lines are perpendicular.

The straight lines joining the origin to the points of intersection of the line $2x + y = 1$ and the curve $3x^2 + 4xy - 4x + 1 = 0$ include an angle of:

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