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Find the equations of the two lines represented by the equation $$2x^2+3xy+y^2+5x+2y-3=0$$.

My Attempt: $$2x^2+3xy+y^2+5x+2y-3=0$$ $$2x^2+2xy+xy+y^2+5x+2y-3=0$$ $$2x(x+y)+y(x+y)+5x+2y-3=0$$ $$(x+y)(2x+y)+5x+2y-3=0$$

how do I complete it without using the quadratic formula?

pi-π
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5 Answers5

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You can obtain that $0=(x+y+3)(2x+y-1)$. Therefore the equations are $x+y+3=0$ and $(2x+y-1)=0$.

Student
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  • How did you factorize that? I want to know the procedure of factorization. – pi-π Nov 14 '17 at 13:44
  • @blue_eyed_... look for a pair of straight lines on google. – prog_SAHIL Nov 14 '17 at 13:45
  • Just some tries. Due to the coefficients you should have something like (x+y+number) times (2x+y+number), and the numbers should be factors of 3. – Student Nov 14 '17 at 13:46
  • @blue_eyed_... just to give you a go. Assume this to be a quadratic equation in either x or y and solve for 2 roots. 2 different roots will provide you two straight lines. – prog_SAHIL Nov 14 '17 at 13:46
  • It is also not difficult to use quadratic forms, you can also have a try :) – Student Nov 14 '17 at 13:48
  • @Cuteboy Yes usually the discriminant comes out to be a perfect square when dealing with easily factorizable pair of straight lines. – prog_SAHIL Nov 14 '17 at 13:51
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Note that you are almost close. A simple adding of $x + y $ on both sides will do. $$(x+y)(2x+y) + 5x + 2y-3 = 0$$ $$\implies (x+y)(2x+y) + 6x + 3y - 3 = (x+y) $$ $$\implies (x+y)(2x+y-1) + 6x + 3y - 3 = 0$$ $$\implies (x+y)(2x + y-1) + 3 (2x + y-1) = 0 $$ $$\implies ?? $$

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You have dealt with the part of order $2$ and are now looking for an equation of the form $$(x+y+a)(2x+y+b)=2x^2+3xy+y^2+5x+2y-3$$

Because this is how the equation will break down into two lines.

Multiplying out, this gives $2a+b=5, a+b=2, ab=-3$ and immediately $a=3, b=-1$

Mark Bennet
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Hint:

Write $$5x+2y=3(2x+y)-(x+y)$$

Now Google for middle term factor in case you are not aware of it.

lab bhattacharjee
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This conic has the rank-2 matrix $$A=\begin{bmatrix}2&\frac32&\frac52\\\frac32&1&1\\\frac52&1&-3\end{bmatrix}.$$ If the point of intersection is $\mathbf p$, then there is a scalar $\alpha$ such that $A+\alpha[\mathbf p]_\times$ has rank one and represents the same conic. (Here, $[\mathbf p]_\times$ denotes the “cross-product matrix” of $\mathbf p$.) This rank-one matrix is some scalar multiple of the outer product of the two intersecting lines, so we can read them directly from the matrix.

The intersection point is $(4,-7)$, which can be computed in various ways, such as differentiation. We then have $$A+\alpha[\mathbf p]_\times = \begin{bmatrix}2&\frac32&\frac52\\\frac32&1&1\\\frac52&1&-3\end{bmatrix}+\alpha\begin{bmatrix}0&-1&-7\\1&0&-4\\7&4&0\end{bmatrix} = \begin{bmatrix}2&-\alpha+\frac32&-7\alpha+\frac52 \\ \alpha+\frac32 & 1 & -4\alpha+1 \\ 7\alpha+\frac52 & 4\alpha+1 & -3\end{bmatrix}.$$ This will have rank 1 when all its $2\times2$ determinants vanish. Taking the lower-right submatrix gives the equation $16\alpha^2-4=0$, so $\alpha=\pm\frac12$. Either root will do; the negative root produces the matrix $$\begin{bmatrix}2&2&6\\1&1&3\\-1&-1&-3\end{bmatrix}.$$ The two lines are given by any row and column of this matrix, so finally we have $x+y+3 = 0$ and $2x+y-1=0$.

This procedure is overkill for this particular equation, which can be factored directly without too much trouble, but can be handy to know when factorization isn’t as straightforward. There’s also a variation of this method that computes a suitable rank-1 matrix directly without having to solve any equations at all.

amd
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