Let us start with these two equations of two lines: x + y = 4 x − y = 2They intersect at ( x , y ) = ( 3 , 1 )Let us now translate (move) both lines so that they intersect at (0, 0). We need to move both lines by -3 along x-axis and by -1 along y-axis. So the equations of the lines become. ( x + 3 ) + ( y + 1 ) = 4 ( x + 3 ) − ( y + 1 ) = 2This is equivalent to x + y = 0 x − y = 0Why do the RHS become 0 for both equations? This happens no matter which two intersecting lines we begin with. What is the geometrical interpretation of this?

Question

Let us start with these two equations of two lines:  x  +  y  =  4  x  −  y  =  2They intersect at   (  x  ,  y  )  =  (  3  ,  1  )Let us now translate (move) both lines so that they intersect at (0, 0). We need to move both lines by -3 along x-axis and by -1 along y-axis. So the equations of the lines become.  (  x  +  3  )  +  (  y  +  1  )  =  4  (  x  +  3  )  −  (  y  +  1  )  =  2This is equivalent to  x  +  y  =  0  x  −  y  =  0Why do the RHS become 0 for both equations? This happens no matter which two intersecting lines we begin with. What is the geometrical interpretation of this?

Tatiana Gentry · Accepted Answer

Step 1The RHS is always zero for any line that goes through the origin.Think about it this way, if  a  x  +  b  y  =  cfor   c  ≠  0 then (0, 0) isn&#039;t a solution to this equation since  a  (  0  )  +  b  (  0  )  =  0  ≠  c  .Therefore the only way for the point (0, 0) to be on your line is for the RHS to be 0.

Let us start with these two equations of two lines: x + y = 4 x −<

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