How does $x^2+4xy-6x+4y^2-12y+9=0$ represent a straight line.

Question

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?

Answer 1

$$x^2+4xy+4y^2-6x-12y+9=0$$ $$(x+2y)^2-6(x+2y)+9=0$$ $$(x+2y-3)^2=0$$ $$x+2y-3=0$$ is a stright line.

Answer 2

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.

Answer 3

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.

Answer 4

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.