SOLUTION: If. A(-8;6),B,C and D(3;9) are vertices of a rhombus. The equation for AC is 3y=-x+10 Show that BD is 3x-Y=0 2. Calculate. The co-ordinates of k the intersection of the diago

Algebra ->  Parallelograms -> SOLUTION: If. A(-8;6),B,C and D(3;9) are vertices of a rhombus. The equation for AC is 3y=-x+10 Show that BD is 3x-Y=0 2. Calculate. The co-ordinates of k the intersection of the diago      Log On


   



Question 885759: If. A(-8;6),B,C and D(3;9) are vertices of a rhombus. The equation for AC is 3y=-x+10
Show that BD is 3x-Y=0
2. Calculate. The co-ordinates of k the intersection of the diagonals of the rhombus ABCD

Found 2 solutions by richwmiller, Edwin McCravy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
2)
3x-y=0
y=3x
3y=-x+10
3*3x=10-x
9x=10
10x=10
x=1
y=3

Remember y is not the same as Y

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The green line is the line 3y=-x+10

As you can see from the graphs below, there are two possible 
solutions for the rhombus.

   

There is a mistake in your problem, for as you see the line BD cannot
possibly pass through the origin, yet the problem states:

"Show that BD is 3x-y=0"

But 3x-y=0 passes through the origin (0,0), so there is no way that
3x-y=0 could be the equation of BD. 

Let's find what the equation of BD really is.  

BD is parallel to AC and has the same slope.

AC has the equation  3y%22%22=%22%22-x%2B10

Solving for y,  y%22%22=%22%22expr%28-1%2F3%29x%2B10

thus the slope of AC and BD is -1%2F3

And BD passes through D(3,9)

Using the point-slope formula:

y-y%5B1%5D%22%22=%22%22m%28x-x%5B1%5D%29

y-9%22%22=%22%22expr%28-1%2F3%29%28x-3%29

Multiply through by 3

3y-27%22%22=%22%22{{-{x-3)}}} 

3y-27%22%22=%22%22{{-x+3}}}

3y%2Bx%22%22=%22%2224

This is correct equation for BD, not 3y-x=0.  

Now we must find the coordinates of the K, the intersection of the
diagonals.  Since the diagonals of a parallellogram bisect each other,
we only need to find the midpoint of one of the diagonals.  We will
need to find the coordinates of C, and then K will be the midpoint of
diagonal CD. 

Since this is a rhombus, all sides must be equal.  So we find side 
AD using the distance formula:

d%22%22=%22%22+sqrt%28+%28x%5B2%5D-x%5B1%5D%29%5E2+%2B+%28y%5B2%5D-y%5B1%5D%29%5E2%29 

AD%22%22=%22%22+sqrt%28+%289-6%29%5E2+%2B+%283-%28-8%29%29%5E2%29
AD%22%22=%22%22+sqrt%28+%283%29%5E2+%2B+%283%2B8%29%5E2%29
AD%22%22=%22%22+sqrt%28+9+%2B+%2811%29%5E2%29
AD%22%22=%22%22+sqrt%28+9+%2B+121%29
AD%22%22=%22%22+sqrt%28130%29

So the length of all 4 sides is %22%22=%22%22+sqrt%28+%28x%5B2%5D-x%5B1%5D%29%5E2+%2B+%28y%5B2%5D-y%5B1%5D%29%5E2%29 
sqrt%28130%29%22%22=%22%22+sqrt%28+%28p-%28-8%29%29%5E2+%2B+%28q-6%29%5E2%29
sqrt%28130%29%22%22=%22%22+sqrt%28+%28p%2B8%29%5E2+%2B+%28q-6%29%5E2%29 

Square both sides:

130%22%22=%22%22%28p%2B8%29%5E2+%2B+%28q-6%29%5E2
 
Now we must solve this system of equations to find p and q,
the coordinates of C

system%283q+=+-p%2B10%2C+130=%28p%2B8%29%5E2+%2B+%28q-6%29%5E2%29
Solve the first equation for p, p = 10-3q
Substitute in the second equation,

130%22%22=%22%22%28p%2B8%29%5E2+%2B+%28q-6%29%5E2
130%22%22=%22%22%2810-3q%2B8%29%5E2+%2B+%28q-6%29%5E2
130%22%22=%22%22%2818-3q%29%5E2+%2B+%28q-6%29%5E2
130%22%22=%22%229q%5E2-108q%2B324+%2B+q%5E2-12q%2B36
130%22%22=%22%2210q%5E2-120q%2B360
%220%22%22%22=%22%2210q%5E2-120q%2B230
Divide through by 10
%220%22%22%22=%22%22q%5E2-12q%2B23
Solve by the quadratic formula and get
q%22%22=%22%226+%2B-+sqrt%2813%29

using the +, substitute in
p%22%22=%22%2210-3q
p%22%22=%22%22+10-3%286%2Bsqrt%2813%29%29+%5D

p%22%22=%22%2210-18-3sqrt%2813%29%29
p%22%22=%22%22-8-3sqrt%2813%29%29

So point C in the graph on the right above is

C%28matrix%281%2C3%2C++-8-3sqrt%2813%29%2C+%22%2C%22%2C+6%2Bsqrt%2813%29+%29%29

using the -, substitute in
p%22%22=%22%2210-3q
p%22%22=%22%22+10-3%286-sqrt%2813%29%29+%5D

p%22%22=%22%2210-18%2B3sqrt%2813%29%29
p%22%22=%22%22-8%2B3sqrt%2813%29%29

So point C in the graph on the left above is

C%28matrix%281%2C3%2C++-8%2B3sqrt%2813%29%2C+%22%2C%22%2C+6-sqrt%2813%29+%29%29

Now we will find the midpoint of CD which will be K,
where the two diagonals intersect.

We will find K for the graph on the left.

For the graph on the left

C%28matrix%281%2C3%2C++-8%2B3sqrt%2813%29%2C+%22%2C%22%2C+6-sqrt%2813%29+%29%29 and D(3,9)

Using the midpoint formula:

Midpoint = 

Midpoint = 

Midpoint = 

For the graph on the right

C%28matrix%281%2C3%2C++-8-3sqrt%2813%29%2C+%22%2C%22%2C+6%2Bsqrt%2813%29+%29%29 and D(3,9)

Using the midpoint formula:

Midpoint = 

Midpoint = 

Midpoint = 

Edwin