Question 1201240
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(1) The second vertex is the intersection of the two lines forming a side of the rhombus.  Solve the pair of equations to find that second vertex is (0,4).  (I leave the details to you.)<br>
(2) Use the Pythagorean Theorem and the first two vertices to find the length of the side of the rhombus is {{{4*sqrt(10)}}}.<br>
(3) A third vertex is on the line y = 2x + 4, a distance {{{4*sqrt(10)}}} to the right and above (0,4).  Since the slope of that line is 2, this third vertex is x units to the right and 2x units above (0,4).  This gives us a right triangle with legs of length x and 2x and a hypotenuse of length {{{4*sqrt(10)}}}.  Use that to find that x is {{{4*sqrt(2)}}}.<br>
Now you have enough to find the coordinates of the third and fourth vertices.<br>
I leave the details of the calculations to you.<br>
VERTICES:
(12,0)
(0,4)
(4*sqrt(2),4+8*sqrt(2))
(12+4*sqrt(2),8*sqrt(2))<br>
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Note that in fact the problem as stated is impossible: two of the vertices have a coordinate that is NOT positive. To solve the problem we have to ignore the instruction that says all coordinates of the vertices are positive.<br>