SOLUTION: 12, quations of two adjacent sides of a rhombus are: 𝑦 = 2𝑥 + 4 and 𝑥 + 3𝑦 = 12, 1.2.1 If (12, 0) is one vertex and all vertices have positive coordinates, find the

Algebra ->  Length-and-distance -> SOLUTION: 12, quations of two adjacent sides of a rhombus are: 𝑦 = 2𝑥 + 4 and 𝑥 + 3𝑦 = 12, 1.2.1 If (12, 0) is one vertex and all vertices have positive coordinates, find the      Log On


   

Question 1201240: 12,
quations of two adjacent sides of a rhombus are:
𝑦 = 2𝑥 + 4 and 𝑥 + 3𝑦 = 12,
1.2.1 If (12, 0) is one vertex and all vertices have positive coordinates, find the
coordinates of the other three vertices. Leave your answers in surd form
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


(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.)

(2) Use the Pythagorean Theorem and the first two vertices to find the length of the side of the rhombus is 4%2Asqrt%2810%29.

(3) A third vertex is on the line y = 2x + 4, a distance 4%2Asqrt%2810%29 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%2Asqrt%2810%29. Use that to find that x is 4%2Asqrt%282%29.

Now you have enough to find the coordinates of the third and fourth vertices.

I leave the details of the calculations to you.

VERTICES:
(12,0)
(0,4)
(4*sqrt(2),4+8*sqrt(2))
(12+4*sqrt(2),8*sqrt(2))

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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.