Question

How do you find the area of a parallelogram with vertices?

Answer 1

For parallelogram `ABCD` the area is
`S = |(x_B-x_A)*(y_D-y_A)-(y_B-y_A)*(x_D-x_A)|`

Explanation:

Let's assume that our parallelogram `ABCD` is defined by the coordinates of its four vertices - `[x_A,y_A]`, `[x_B,y_B]`, `[x_C,y_C]`, `[x_D,y_D]`.

To determine the area of our parallelogram, we need the length of its base `|AB|` and the altitude `|DH|` from vertex `D` to point `H` on side `AB` (that is, `DH_|_AB`).

First of all, to simplify the task, let's move it to a position when its vertex `A` coincides with the origin of coordinates. The area will be the same, but calculations will be easier.
So, we will perform the following transformation of coordinates:
`U=x-x_A`
`V=y-y_A`

Then the (`U,V`) coordinates of all vertices will be:
`A[U_A=0,V_B=0]`
`B[U_B=x_B-x_A,V_B=y_B-y_A]`
`C[U_C=x_C-x_A,V_C=y_C-y_A]`
`D[U_D=x_D-x_A,V_D=y_D-y_A]`

Our parallelogram now is defined by two vectors:
`p=(U_B,V_B)` and `q=(U_D,V_D)`

Determine the length of base `AB` as the length of vector `p`:
`|AB|=sqrt(U_B^2+V_B^2)`

The length of altitude `|DH|` can be expressed as `|AD|*sin(/_BAD)`.
The length `AD` is the length of vector `q`:
`|AD|=sqrt(U_D^2+V_D^2)`
Angle `/_BAD` can be determined by using two expressions for the scalar (dot) product of vectors `p` and `q`:
`(p*q)=U_B*U_D+V_B*V_D=|p|*|q|*cos(/_BAD)`
from which
`cos^2(/_BAD)=(U_B*U_D+V_B*V_D)^2/[(U_B^2+V_B^2)*(U_D^2+V_D^2)]`
`sin^2(/_BAD)=1-cos^2(/_BAD)=`
`=1-(U_B*U_D+V_B*V_D)^2/[(U_B^2+V_B^2)*(U_D^2+V_D^2)]=`
`=(U_B*V_D-V_B*U_D)^2 / [(U_B^2+V_B^2)*(U_D^2+V_D^2)]`

Now we know all components to calculate the area:
Base `|AB|=sqrt(U_B^2+V_B^2)`:
Altitude `|DH|=sqrt(U_D^2+V_D^2)*|U_A*V_D-V_A*U_D| / sqrt[(U_B^2+V_B^2)*(U_D^2+V_D^2)]`

The area is their product:
`S = |AB|*|DH|=|U_B*V_D-V_B*U_D|`

In terms of original coordinates, it looks like this:
`S = |(x_B-x_A)*(y_D-y_A)-(y_B-y_A)*(x_D-x_A)|`

Answer 2

another discussion

Explanation:

Geometric proof
Considering the figure

we can easily establish the formula for calculation of the area of a parallelogram ABCD, when any three vertices (say A,B,D) are known.

Since diagonal BD bisects the parallelogram into two congruent triangle.
The area of the parallelogram ABCD
= 2 area of triangle ABD
=2[ area of trapezium BAPQ +area of trap BQRD - area of trap DAPR]
=2[`1/2(AP+BQ)PQ+1/2(BQ+DR)QR-1/2(AP+DR)PR]`
= `(Y_A+Y_B)(X_B-X_A )+(Y_B+Y_D)(X_D-X_B)-(Y_A+Y_D)(X_D-X_A)`
=`Y_AX_B+cancel (Y_BX_B)-cancel(Y_AX_A)-Y_BX_A +Y_BX_D+cancel(Y_DX_D)-cancel(Y_BX_B)-Y_AX_D-cancel(Y_DX_D)+cancel(Y_AX_A)+Y_DX_A`

=`Y_A(X_B_X_D)+Y_B(X_D-XA)+Y_D(X_A-X_B)`

This formula will give the area of the parallelogram .
Proof considering vector
It can also be established considering `vec(AB)` and`vec(AD)`
Now
Position vector of point A w.r,t the origin O, `vec(OA)= X_Ahati +Y_Ahatj`
Position vector of point B w.r,t the origin O, `vec(OB)= X_Bhati +Y_Bhatj`
Position vector of point D w.r,t the origin O, `vec(OD)= X_Dhati +Y_Dhatj`

Now
Area of the Parallelogram ABCD
`= Base (AD)*Height (BE)=AD*h`
`=AD*ABsintheta=|vec(AD)Xvec(AB)|`

Again
`vec(AD)=vec(OD)-vec(OA)=(X_D-X_A)hati+(Y_D-Y_A)hatj`
`vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj`
`vec(AD)`X`vec(AB)=[(X_D-X_A)(Y_B-Y_A)-(X_B-X_A)(Y_D-Y_A)]hatk`
Area = `|vec(AD)`X`vec(AB)|`
=`|Y_BX_D-Y_BX_A-Y_AX_D+cancel(Y_AX_A)-Y_DX_B +Y_DX_A+Y_AX_B-cancel(Y_AX_A)|`
=`|Y_BX_D-Y_BX_A-Y_AX_D-Y_DX_B +Y_DX_A+Y_AX_B|`
=`|Y_BX_D-Y_BX_A-Y_AX_D-Y_DX_B +Y_DX_A+Y_AX_B|`
=`|Y_A(X_B_X_D)+Y_B(X_D-XA)+Y_D(X_A-X_B)|`
Thus we have the same formula