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Suvadip
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If each of the equations ax^2+2hxy+by^2+2gx+2fy+c=0 and ax^2+2hxy+by^2-2gx-2fy+c=0 represents a pair of straight lines , find the area of the parallelogram enclosed by them .
Please help
Please help
If the conic $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines then it must be of the form $(lx+my+n)(px+qy+r)=0$, where (comparing coefficients) $a = lp$, $h = \frac12(lq+mp)$, $b = mq$, ..., $c = nr.$ The conic $ax^2+2hxy+by^2-2gx-2fy+c=0$ is then given by $(lx+my-n)(px+qy-r)=0.$suvadip said:If each of the equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ and $ax^2+2hxy+by^2-2gx-2fy+c=0$ represents a pair of straight lines, find the area of the parallelogram enclosed by them.
I think you are misunderstanding the question. Each equation represents a pair of lines. For example [tex]a^2x^2- b^2y^2= (ax- by)(ax+by)= 0[/tex] gives the lines ax- by= 0 and ax+ by= 0.Fantini said:You should note that if these are lines, none of the terms $x^2, xy$ and $y^2$ should appear. This should tell you something about their respective coefficients. :)
The general equation of a pair of straight lines can be written as ax² + 2hxy + by² + 2gx + 2fy + c = 0, where a, b, and h are not all zero.
A pair of straight lines can have a maximum of two solutions.
The condition for a pair of straight lines to be perpendicular is that the product of their slopes must be -1.
No, a pair of straight lines can only intersect at one point. If they intersect at more than one point, then they are the same line.
The type of intersection between two pairs of straight lines can be determined by calculating the discriminant of their general equation. If the discriminant is positive, the lines intersect at two distinct points. If it is zero, they intersect at one point. And if it is negative, they do not intersect.