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decide the geometric meaning of 

x2 - y+ x + y = 0

 

Answer sheet

x = 0 

x + 4y + 3 = 0

 

 

How i do this? 

please help, it has to do with straight lines that intersect each other 90 degrees

 Apr 7, 2019
edited by Guest  Apr 7, 2019
 #1
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\(x^2 - y^2 + x + y = 0\\ (x^2 + x) - (y^2 - y) = 0\\ \left(x^2 + x +\dfrac 1 4 - \dfrac 1 4 \right)- \left(y^2 - y +\dfrac 1 4 - \dfrac 1 4\right) = \\ \left(x+\dfrac 1 2\right)^2 - \dfrac 1 4 -\left(y-\dfrac 1 2 \right)^2 +\dfrac 1 4 = 0\\ \left(x+\dfrac 1 2\right)^2 -\left(y-\dfrac 1 2 \right)^2= 0\\ \left(x+\dfrac 1 2\right)^2 =\left(y-\dfrac 1 2 \right)^2\\ \left(x+\dfrac 1 2\right)=\pm \left(y-\dfrac 1 2 \right)\)

 

\(y = x+1,~y= -x - 1\\ \text{you can see that these two lines are perpendicular as }1 = -\left(\dfrac{1}{-1}\right) \)

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 Apr 7, 2019

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