1. sum of coordinates

Points A = ( 3, 9) B = (1 , 1) c = ( 5, 3 ) and D = (a , b) lie in the first quadrant and are the vertices of quadrilateral ABCD. The quadrilateral formed by joining the midpoints of ab, bc, cd, and da is a square. What is the sum of the coordinates of point D?

I solved the problem but someone said that I could've solved it easier and faster by using vectors. I looked up vectors on google and I dont understand it that well. Could someone take a look at my work and show me where i could apply vectors.

Thanks

Vicky.

2. Hi

Let I be the midpoint of [AB], J the midpoint of [BC], K the midpoint of [CD] and L the midpoint of [DA]
From the data we have I(2,5) ; J(3,2) ; K((a+5)/2,(b+3)/2) ; L((a+3)/2,(b+9)/2)
Using vectors
$\displaystyle \vec{IJ} (1,-3)$

$\displaystyle \vec{JK} \left(\frac{a-1}{2},\frac{b-1}{2} \right)$

[IJ] and [JK] being two sides of a square, they are perpendicular
The dot product of $\displaystyle \vec{IJ}$ and $\displaystyle \vec{JK}$ is equal to 0

$\displaystyle \frac{a-1}{2}-3\:\frac{b-1}{2} = 0$

$\displaystyle \frac{a-1}{2}=3\:\frac{b-1}{2}$ [1]

The lengths of [IJ] and [JK] are equal

$\displaystyle 10 = \left(\frac{a-1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2$

Using [1]
$\displaystyle 10 = 10\:\frac{(b-1)^2}{4}$

$\displaystyle (b-1)^2 = 4$

b-1=2 => b=3 => a=7 => D(7,3)
b-1=-2 => b=-1 => a=-5 => D'(-5,-1)

There are 2 solutions depending if you consider ABCD' as a quadrilateral (probably not)

3. Originally Posted by running-gag
Hi

Let I be the midpoint of [AB], J the midpoint of [BC], K the midpoint of [CD] and L the midpoint of [DA]
From the data we have I(2,5) ; J(3,2) ; K((a+5)/2,(b+3)/2) ; L((a+3)/2,(b+9)/2)
Using vectors
$\displaystyle \vec{IJ} (1,-3)$

$\displaystyle \vec{JK} \left(\frac{a-1}{2},\frac{b-1}{2} \right)$

[IJ] and [JK] being two sides of a square, they are perpendicular
The dot product of $\displaystyle \vec{IJ}$ and $\displaystyle \vec{JK}$ is equal to 0

$\displaystyle \frac{a-1}{2}-3\:\frac{b-1}{2} = 0$

$\displaystyle \frac{a-1}{2}=3\:\frac{b-1}{2}$ [1]

The lengths of [IJ] and [JK] are equal

$\displaystyle 10 = \left(\frac{a-1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2$

Using [1]
$\displaystyle 10 = 10\:\frac{(b-1)^2}{4}$

$\displaystyle (b-1)^2 = 4$

b-1=2 => b=3 => a=7 => D(7,3)
b-1=-2 => b=-1 => a=-5 => D'(-5,-1)

There are 2 solutions depending if you consider ABCD' as a quadrilateral (probably not)