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Only one plane passes through $y=x$ and $y=x+1$. If its wrong then why so? Is there any other plane that passes through these two lines?

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closed as off-topic by Xander Henderson, Jendrik Stelzner, TheSimpliFire, Namaste, Aretino Aug 25 '18 at 16:44

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    $\begingroup$ Any three noncollinear points define a plane. Can you find three noncollinear points on a pair of parallel lines? $\endgroup$ – Xander Henderson Aug 25 '18 at 13:12
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    $\begingroup$ What you write are planes in $R^3$. So the intersection, if any, is a line and from a line pass uncountable many planes. $\endgroup$ – dmtri Aug 25 '18 at 13:17
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    $\begingroup$ Give him some + to be abale to make comments. $\endgroup$ – mrs Aug 25 '18 at 13:22
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    $\begingroup$ @dmtri Oi... I didn't even think about that–I read the title of the question and kind of ignored the rest of it. Prashanth CN: This does need clarification. You ask about two parallel lines, but the equations that you have given do you not define lines in higher dimensional spaces. In three-dimensional space, each of those equations defines a plane. More generally, those equations define hyperplanes of codimension 1. $\endgroup$ – Xander Henderson Aug 25 '18 at 13:23
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    $\begingroup$ @ResidentDementor New users cannot generally comment, but they can comment on their own questions and answers. This is explained in this question on the main meta. $\endgroup$ – Xander Henderson Aug 25 '18 at 13:24
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Yes, two parallel lines is one way to determine a plane. But, as stated in the comments, the lines must be given in their 3D version.

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If the problem is 2D, there is only one plane and any line lies in it.

If the problem is 3D, you didn't give the equations of two lines !

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