# The points W(-2,-2), X(-6,2) and Y(2,5) are three verticies of paralellogram WXYZ. Find the coordinates of "Z".

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### 2 Answers

Recall that a parallelogram is a quadrilateral that has two pairs of parallel sides and the two opposite sides are equal in length. So, if you'll plot the given points:see that line XY and line WZ are the parallel lines and line WX and line YZ are parallel too. From that, you can solve for the slope by:

`m = (y_2-y_1)/(x_2-x_1)`

where m = slope

`x_1 and x_2` = x-coordinates of the points

`y_1 and y_2` = y-coordinates of the points

Let's start with line WX. W(-2,-2) and X(-6,2).

`m_(WX)= (2-(-2))/(-6-(-2))`

`m_(WX) = 4/(-4)`

`m_(WX) = -1`

So the slope of line WX = -1. Let's leave this for awhile. We'll be using this later. Next, solve for the slope of line XY. Same formula will apply. X(-6,2) and Y(2,5).

`m_(XY) = (5-2)/(2-(-6))`

`m_(XY) = 3/8`

Now that we have the slopes of the two lines, we can establish our working equation. We said earlier that line WX is parallel with line YZ and line XY is parallel with line WZ. Recall also that parallel lines have the same or equal slope. So,

`m_(YZ) = m_(WX) = -1`

`m_(WZ) = m_(XY) = 3/8`

To get the working equations or the equation of the line YZ and WZ, use the point-slope formula:

`y-y_2 = m(x-x_2)`

Let y be the y-coordinate of point Z

x be the x-coordinate of point Z

Start with line YZ. Since it is passing through Y, y2 = 5 and x2 = 2.

`y - y_2 = m_(YZ) (x-x_2)`

`y-5 = -1(x-2)`

`y-5 = -x +2`

`y = -x+7`

From the second line, -1 was distributed to the terms inside (), that's why we have the third line. Then, combine similar terms. The similar terms are 2 and 5 (from the 3rd line) since they are constants. To combine, they must be on same side, so move 5 to the right and moving will make the sign opposite and it became +5. So 2 + 5 = 7.

Then, get the equation of line WX. Same process. For this, y2 = -2 and x2 =-2, since the line WZ is passing through point W. Note that y and x are still the same since it is still pertaining to point Z.

`y - y_2 = m_(WZ) (x-x_2)`

`y - (-2) = 3/8(x-(-2))`

`y+2 =(3x)/8 + 6/8`

`8y +16 = 3x + 6`

`8y = 3x -10`

That's the second equation. For the second line, the -2 is preceeded by -, so it became +. Distribute 3/8 into (). In multiplying fractions, just multiply the numerators and denominators of each fraction. Since 2 is a whole number, 1 is considered as its denominator. So 3*2 = 6 and 8*1 = 8. That's why, you have 6/8 on the third line. It is easier to solve equations without fractions, so to get rid of the denominator (8), mutiply both sides by 8 to have the fourth line then simplify. Result is the last line. Earlier, you had:

`y = -x +7`

So, you have 2 equations, 2 unknowns. Solve that by substitution.

`8 (-x +7) = 3x - 10`

`-8x + 56 = 3x -10`

`-8x -3x = -10 -56`

`-11x = -66`

`(-11x)/11 = -66/11`

`x = 6`

From y = -x + 7. Plug-in 6 in place of x to solve for y.

y = -6 + 7

y = 1.

So point Z is (6,1).

You can check your answer by getting the length of line WX and line YZ using distance formula `d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)` . They must be equal. You'll have `4sqrt(2).`

In parallelogram ,diagonal bisect each other.Just this is sufficient to find fourth vertex of parallelogram.

W(-2,-2),X(-6,2),Y(2,5) .

Let Z( x,y)

Find the mid point of the diagonal WY .Let it be `(x_1,y_1)` ,where

`x_1=(-2+2)/2 ,y_1=(-2+5)/2`

`(x_1,y_1)=(0,3/2)`

But this is also mid point of diagonal XZ.There fore

`x_1=(x-6)/2 ,y_1=(y+2)/2`

`0=(x-6)/2,3/2=(y+2)/2`

`x-6=0,y+2=3`

`x=6,y=3-2=1`

`(x,y)=(6,1)`

Thus fourth vertex Z(6,1)