find the equation for the diagonal BD

B

bumblebee123

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question:

ABCD is a square. A is the point (0,5) and C is the point ( 8, -7 ). find the equation for the diagonal BD. give your answer in the form ax + by = c

I attempted this question by trying to find the coordinates of D, which I found must be ( 0, -7 ) and B, which must be ( 8,5)

I then tried to find the gradient of the line BD = ( -7 - 5 ) / ( 0 -8 ) = 3/2

I found that line equation must now be y = 3/2 x - 7

to change it into the form the question, asked I multiplied by 2. This gave me 2y = 3x - 14
3x - 2y = 14

however, this is the wrong answer. Can anyone help?
 
bumblebee123 said:
question:

ABCD is a square. A is the point (0,5) and C is the point ( 8, -7 ). find the equation for the diagonal BD. give your answer in the form ax + by = c

I attempted this question by trying to find the coordinates of D, which I found must be ( 0, -7 ) and B, which must be ( 8,5)

I then tried to find the gradient of the line BD = ( -7 - 5 ) / ( 0 -8 ) = 3/2

I found that line equation must now be y = 3/2 x - 7

to change it into the form the question, asked I multiplied by 2. This gave me 2y = 3x - 14
3x - 2y = 14

however, this is the wrong answer. Can anyone help?
Click to expand...
Have you drawn these points and joined them up? Is it a square? No, it's a rectangle of side lengths 12 and 8.

The square won't have a horizontal bottom, it will be at an angle.

Ok so a different approach is needed

Can you state two properties about the diagonals of a square?
 
bumblebee123 said:
question: ABCD is a square. A is the point (0,5) and C is the point ( 8, -7 ). find the equation for the diagonal BD. give your answer in the form ax + by = c
I attempted this question by trying to find the coordinates of D, which I found must be ( 0, -7 ) and B, which must be ( 8,5)
I then tried to find the gradient of the line BD = ( -7 - 5 ) / ( 0 -8 ) = 3/2
I found that line equation must now be y = 3/2 x - 7
to change it into the form the question, asked I multiplied by 2. This gave me 2y = 3x - 14
3x - 2y = 14 however, this is the wrong answer.
Click to expand...
Your mistake is in the coordinates for the points \(\displaystyle B~\&~D\).
If \(\displaystyle \overline{AC}\) is a diagonal of a square has slope \(\displaystyle \frac{-3}{2}\),
then the other diagonal would have slope \(\displaystyle \frac{2}{3}\).
Thus the line \(\displaystyle \overleftrightarrow {BD}\) has equation \(\displaystyle 2x-3y=C\).
In order to find \(\displaystyle C\) we need a point on \(\displaystyle \overline{BD}\).
Well its midpoint is \(\displaystyle (4,-1)\). WHY?
What is the value of \(\displaystyle \bf{C}~?\)
 
bumblebee123 said:
at c ( the intercept ) x = 0
Click to expand...
No indeed. \(\displaystyle \large{c=2(4)-3(-1)=11}\)
So the line \(\displaystyle \overleftrightarrow {BD}\) is \(\displaystyle 2x-3y=11\)
What are the coordinates of \(\displaystyle B~\&~D~?\) HINT: They are not what you think!

 
bumblebee123 said:
  • The diagonals are of equal length.
  • All the diagonals are perpendicular to one another.
Click to expand...
Good. Maybe I should have asked for 3 properties. The other relevant one here is that the diagonals bisect each other, in other words they have the same midpoint.
So you can find the gradient of BD after you find the gradient of AC (using your second property)
Find the midpoint of AC which is also lies on BD.
Then you have enough info to find the equation of BD.
 
bumblebee123 said:
i honestly have no idea
Click to expand...
Harry_the_cat said:
Good. Maybe I should have asked for 3 properties. The other relevant one here is that the diagonals bisect each other, in other words they have the same midpoint. So you can find the gradient of BD after you find the gradient of AC (using your second property)
Click to expand...
Hi Bee & Cat, can either of you find the coordinates of \(\displaystyle B~\&~D\). Have a look here.
Look Here to find the coordinates.
 
Hi pka,

I'd just use vectors.

Let M be the midpoint (4, -1)

AM=(4 -6) …… (pretend this is written vertically. I'm too lazy atm to look up how to do it in LaTex)
Therefore MB = (6 4) since AM and MB have same length and are perpendicular
So Point B is (10, 3)

Similarly
MD = ( -6 -4)
So Point D is (-2 -5)

Not sure what your first link was doing
 
Harry_the_cat said:
... AM=(4 -6) …… (pretend this is written vertically. I'm too lazy atm to look up how to do it in LaTex) ...
Click to expand...
Markbot to the rescue!

Code: 
[tex]\boldsymbol{AM} = \vec{AM} = \left< 4, -6 \right> = \begin{bmatrix} 4 \\ -6 \end{bmatrix}[/tex]

\(\displaystyle \boldsymbol{AM} = \vec{AM} = \left< 4, -6 \right> = \begin{bmatrix} 4 \\ -6 \end{bmatrix}\)

?
 
After spending a literal hour trying different methods, I think I've done it!

the information I have: A ( 0, 5) C(8, -7 )

AC midpoint = ( 4, -1 )

BD is the other diagonal, it's perpendicular to AC and has the same midpoint.

AC gradient = -3/2

AC line equation: y = -3/2 x + 5

As BD is perpendicular, its gradient will be 2/3

BD equation: y = 2/3x + c

2x - 3y = c

to find C, the midpoint coordinates can be used ( it has the same midpoint as AC )

c = ( 2* 4 ) - ( 3*-1)
c= 8 - - 3
c = 11

full BD equation is therefore: 2x - 3y = 11
 
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