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Question: Three vertices of parallelogram WXYZ are \[{\text{W}}\left( {{\text{3,1}}} \right){\text{, X}}\left(...

Three vertices of parallelogram WXYZ are W(3,1), X(2,7),and Z(4,0).{\text{W}}\left( {{\text{3,1}}} \right){\text{, X}}\left( {{\text{2,7}}} \right){\text{,and Z}}\left( {{\text{4,0}}} \right). How do I find the coordinates of vertex Y?

Explanation

Solution

You can find the fourth vertex of the parallelogram with help of middle point formula. First assume the coordinates of the fourth vertex to be variables, then find midpoint of the diagonal of the parallelogram with help of two opposite vertices and then write the midpoint formula for diagonal of parallelogram with two vertices in which one is the unknown one, solve the formula for the unknown coordinates.

Complete step by step solution:
Let us consider the required coordinates to be (x,  y)(x,\;y)
We know that both diagonals of a parallelogram have same midpoint,
Taking the diagonals of parallelogram WXYZ to be WX and YZ, then from above information and midpoint formula, we can write
x1=3+22=4+x25=4+xx=1 y1=1+72=0+y28=0+yy=8  {x_1} = \dfrac{{3 + 2}}{2} = \dfrac{{4 + x}}{2} \Rightarrow 5 = 4 + x \Rightarrow x = 1 \\\ {y_1} = \dfrac{{1 + 7}}{2} = \dfrac{{0 + y}}{2} \Rightarrow 8 = 0 + y \Rightarrow y = 8 \\\
Where x1  and  y1{x_1}\;{\text{and}}\;{y_1} are coordinates of the midpoint of the diagonal.
We got value for coordinate of y(1,  8)y \equiv (1,\;8)
There is one more possibility, that if we take the diagonals to be WY and XZ, again finding coordinates of yy similarly as above process
x2=3+x2=2+423+x=2+4x=63=3 y2=1+y2=7+021+y=7+0y=71=6  {x_2} = \dfrac{{3 + x}}{2} = \dfrac{{2 + 4}}{2} \Rightarrow 3 + x = 2 + 4 \Rightarrow x = 6 - 3 = 3 \\\ {y_2} = \dfrac{{1 + y}}{2} = \dfrac{{7 + 0}}{2} \Rightarrow 1 + y = 7 + 0 \Rightarrow y = 7 - 1 = 6 \\\
Where x2  and  y2{x_2}\;{\text{and}}\;{y_2} are coordinates of the midpoint of diagonal.
So we got the value of the coordinate of y(3,  6)y \equiv (3,\;6) in the second possibility.

Formula used: Midpoint formula: x=a+c2  and y=b+d2,  where  (x,  y)x = \dfrac{{a + c}}{2}\;{\text{and }}y = \dfrac{{b + d}}{2},\;{\text{where}}\;(x,\;y) is vertex of midpoint and (a,  b)  and  (c,  d)(a,\;b)\;{\text{and}}\;(c,\;d) are vertices of diagonal of a parallelogram.

Note: x1  and  y1{x_1}\;{\text{and}}\;{y_1} and x2  and  y2{x_2}\;{\text{and}}\;{y_2} are two different coordinates of different middle points of two different diagonals of two different parallelograms. Basically they are coordinates in two different parallelograms so don’t get confused that if both are the coordinates of the middle point of the diagonal then why are they different? There is one more possibility for coordinates of yy when XY and WZ are taken as diagonals of the parallelogram.