A(0,6) B(2,1) C(7,3) are the 3 possible vertices of a square. Find the coordinate of D, the fouth vertex of the square.
You can find the gradients of the lines AB and BC, get the equations of DC and AD, and then solve the two line equations simultaneously to get the coordinates of D.
I hope that helps.
ILoveMaths07.
D is (5,8) There is two ways i can think to solve this problem. 1 way is to use midpoints and the other way is to find the slope of AB and BC and use that info to find the equation of line AD and the equation of line CD. After that equal equation AD and CD to see where they intersect and that will be point D.
Oh, there's an easier way... Consider all the points as position vectors.
$\displaystyle \vec {OA} = 6\vec {j}$
$\displaystyle \vec {OB} = 2\vec {i} + \vec {j} $
$\displaystyle \vec {OC} = 7\vec {i} + 3 \vec {j} $
Then, $\displaystyle \vec {BC} = \vec {OC} - \vec {OB} $
$\displaystyle \Rightarrow \vec {BC} = (7\vec{i} + 3 \vec {j}) - (2\vec {i} + \vec {j}) = 5\vec{i} + 2\vec{j}.$
Therefore, $\displaystyle \vec {OD} - \vec {OA}$, or $\displaystyle \vec{AD}$, will also equal $\displaystyle 5\vec {i} + 2 \vec {j} $ (Because $\displaystyle BC = AD$; $\displaystyle ABCD$ is a square)
So $\displaystyle \vec {OD} = \vec{OA} + 5 \vec {i} + 2\vec {j} = 6\vec{j} + 5 \vec {i} + 2\vec {j} = 5 \vec {i} + 8 \vec {j}. $.
So the coordinates of $\displaystyle D$ are $\displaystyle (5, 8)$.
I hope that helps.
ILoveMaths07.
Use vectors:
$\displaystyle \overrightarrow{OB} = (2,1)$
and:
$\displaystyle \overrightarrow{BA} = (-2,5)$
$\displaystyle \overrightarrow{BC} = (5, 2)$
then
$\displaystyle \overrightarrow{OD} = \overrightarrow{OB} + \overrightarrow{BA} + \overrightarrow{BC} = (2,1) + (-2,5) + (5,2) = (5,8)$