SOLUTION: DETERMINE THE POINT A(x,y) SO THAT THE POINTS A(x,y),B(O,3),C(1,0),D(7,2) WILL BE THE VERTICES OF A PARALLELOGRAM.

Algebra ->  Points-lines-and-rays -> SOLUTION: DETERMINE THE POINT A(x,y) SO THAT THE POINTS A(x,y),B(O,3),C(1,0),D(7,2) WILL BE THE VERTICES OF A PARALLELOGRAM.       Log On


   

Question 64973: DETERMINE THE POINT A(x,y) SO THAT THE POINTS A(x,y),B(O,3),C(1,0),D(7,2) WILL BE THE VERTICES OF A PARALLELOGRAM.
Answer by cristiana(10) About Me  (Show Source):
You can put this solution on YOUR website!
Let's plot the given vertices (B,C and D). The graph would look like this:


Now, we know that a parallelogram has its sides parallel and congruent two by two. In order for our points to shape a parallelogram we need to have BA || CD, BA = CD and BC || AD, BC = AD. Therefore, the line segment passing through points A,B needs to be parallel and congruent to the line segment passing through points C,D. Similarly, the line segment passing through points B,C needs to be parallel and congruent to the line segment passing through points A,D.

We also know that two parallel lines have the same slope.

Since we need to find the coordinates X,Y for point A we will most likely need a system of equations in x, y.

Let's focus on lines AB and CD. These line segments have to be parallel and congruent.
The equation of the line going through point D: y-yD = m(x-xD); y = m(x-7) + 2
The equation of the line going through point C: y-yC = m(x-xC); y = m(x-1)
These want these two equations to be valid for the same line segment, CD, and to have the same slope m.
Therefore, we have m(x-7) + 2 = m(x-1) which leads to m=1/3

Now, BA needs to be parallel to CD, which translates into mBA=mCD=1/3. We can now write the point-slope form of the equation going through point B and having the slope m=1/3:
y-yB=1/3(x-xB) which is y = 1/3x + 3.
This will be the first equation of our system.

To get to the second equation, we use the fact that AB needs to be congruent to CD.
Let's calculate CD as the distance between points C and D:
CD+=+sqrt%28%28xD-xC%29%5E2%2B%28yD-yC%29%5E2%29
CD+=+2sqrt%2810%29
Let's calculate AB now:
AB+=+sqrt%28%28xB-xA%29%5E2%2B%28yB-yA%29%5E2%29
sqrt%28%28y-3%29%5E2%2Bx%5E2%29+=+2sqrt%2810%29
This will be the second equation of our system.
Now we have to solve the system:
y = 1/3x + 3
sqrt%28%28y-3%29%5E2%2Bx%5E2%29+=+2sqrt%2810%29
We will get two solution sets: {x=6, y=5} and {x=-6, y=1}.
If you check both by trying to plot them into our graph you'll easily notice that only {x=6, y=5} makes our ABCD a parallelogram.

Hope this helped.
Cristiana