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I need to show $x^2+4xy-6x+4y^2-12y+9=0$ is a straight line. But I only know of a straight line in the form $y=mx+c$. Any help?

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    $\begingroup$ Try calculating $(y-mx-c)^2$ $\endgroup$ – anderstood May 8 '15 at 14:33
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$$x^2+4xy+4y^2-6x-12y+9=0$$ $$(x+2y)^2-6(x+2y)+9=0$$ $$(x+2y-3)^2=0$$ $$x+2y-3=0$$ is a stright line.

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The equation is equivalent with $$(x+2y)^2-6(x+2y)+9 = (x+2y-3)^2 = 0$$ This is equivalent with $$x+2y-3=0$$ Which is a straight line.

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  • $\begingroup$ Why is this a straight line? Also how do you recognize how to factor this? $\endgroup$ – user238962 May 8 '15 at 14:36
  • $\begingroup$ @user238962 solve it with respect to $y$. $\endgroup$ – Bman72 May 8 '15 at 14:37
  • $\begingroup$ Ok got it. one last question how would you recognize that this is $(x+2y-3)^2$ $\endgroup$ – user238962 May 8 '15 at 14:41
  • $\begingroup$ Actually, because you said so. I just tried the most obvious factorization. There is a general method though. You can find your answer here: en.m.wikipedia.org/wiki/Degenerate_conic $\endgroup$ – Jef May 8 '15 at 15:18
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Note that \begin{align} 0 & = x^2+4y^2+9 + 4xy-6x-12y\\ & = x^2 + (2y)^2 + (-3)^2 + 2\cdot x \cdot (2y) + 2 \cdot x \cdot (-3) + 2 \cdot (2y) \cdot (-3)\\ & = (x+2y-3)^2 \end{align} This gives us $$x+2y-3 = 0$$ which indeed is a straight line.

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Since everyone has answered to your question. I will suggest you to study the concept pair of straight lines. After that you will learn that xy=0 represents the two coordinate axes! And much more. I suggest you read books on this or see tutorials.

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