given W(-1,5), X(6,-2) an Y(-2,-7) find the co-ordinate of the point if WXYZ is a parallelogram

Question

it is a vectors question needed to be solved by using the dot product

Answer 1

We are given 3 vertices of the parallelogram, so the coords of the fourth vertex can be found.

GEOMETRIC SOLUTION

WY and XZ are parallel, and WX and YZ are parallel.

The slope of WY=(5-(-7))/(-1-(-2))=12=slope of XZ.

The slope of WX=(5-(-2))/(-1-6)=7/(-7)=-1=slope of YZ.

Knowing the slopes of XZ and YZ we can work out their equations.

XZ=y+2=12(x-6), y=12x-72-2=12x-74.

YZ=y+7=-(x+2), y=-x-2-7=-x-9.

The lines meet at Z, so 12x-74=-x-9, 13x=65, x=5, so y=-5-9=-14⇒Z(5,-14).

VECTOR SOLUTION

Let position vectors w=<-1,5>, x=<6,-2>, y=<-2,-7>, z=<a,b>.

WY=y-w=<-1,-12>, WX=x-w=<7,-7>, YZ=z-y=<a+2,b+7>, XZ=z-x=<a-6,b+2>.

XZ=WY=√((-1)²+(-12)²)=√145, YZ=WX=√(7²+(-7)²)=√98=7√2. Angle between parallel vectors=0 and cos(0)=1.

By definition of dot-product, WX.YZ=|WX||YZ|cos(0)=(WX)(YZ)=WX²=98; WY.XZ=|WY||XZ|cos(0)=WY²=145;

WX.YZ=<7,-7>.<a+2,b+7>=7a-7b-35=98, a-b-5=14, a-b=19; WY.XZ=<-1,-12>.<a-6,b+2>=-a-12b-18=145, a+12b=-163.

So we have the system: a-b=19 and a+12b=-163, that is, substituting for a: 19+b+12b=-163, 13b=-182, b=-14⇒a+14=19, a=5.

So z=<5,-14>, making Z(5,-14). This confirms the geometric solution.