Subjective Type

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

Solution

For parallelogram $$ABCD$$ the area is
$$S = \left | (x_B - x_A) . (y_D - y_A) - (y_B - y_A) . (x_D - x_A) \right |$$
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 $$\left | AB \right |$$ and the altitude $$\left | DH \right |$$from vertex $$D$$ to point $$H$$ on side $$AB$$ (that is,$$DH \perp AB)$$.
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:
$$A[U_A = 0, B_B = 0]$$
$$B[U_B = X_B - X_A, V_B = Y_B - Y_A]$$
$$C[U_C = X_C - X_A, V_C = Y_C - Y_A]$$
$$D[U_D = X_D - X_A, V_D = Y_D - Y_A]$$
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 $$p$$:
$$\left | AB \right | = \sqrt{U_{B}^{2} + V_{B}^{2}}$$
The length of altitude $$\left | DH \right |$$ can be expressed as $$\left | AD \right | . sin (\angle BAD)$$.
The length $$AD$$ is the length of vector $$q$$:
$$\left | AD \right | = \sqrt{U_{D}^{2} + V_{D}^{2}}$$
Angle $$\angle BAD$$ can be determined by using two expressions for the scalar (dot) product of vectors $$p$$ and $$q$$:
$$(p . q) = U_B . U_D + V_B . V_D = \left | P \right | . \left |q \right | . cos (\angle BAD)$$
from which
$$cos^2 (\angle BAD) = \frac{(U_B . U_D + V_B . V_D)^2}{(U_{B}^{2} + V_{B}^{2}) . (U_{D}^{2} + V_{D}^{2})}$$
$$cos^2 (\angle BAD) = 1 - cos^2 (\angle BAD)= $$
$$= 1 - \frac{(U_B . U_D + V_B . V_D)^2}{(U_{B}^{2} + V_{B}^{2}) . (U_{D}^{2} + V_{D}^{2})} =$$
$$= \frac{(U_B . V_D - V_B . U_D)^2}{(U_{B}^{2} + V_{B}^{2}) . (U_{D}^{2} + V_{D}^{2})}$$
Now we know all components to calculate the area:
Base $$\left | AB \right | = \sqrt{U_{B}^{2} + V_{B}^{2}}$$
Altitude $$\left | DH \right | = \sqrt{U_{D}^{2} + V_{D}^{2}} . \frac{\left | U_A . V_D - V_A . U_D \right |}{\sqrt{(U_{B}^{2} + V_{B}^{2}) . (U_{D}^{2} + V_{D}^{2})}}$$
The area is their product:
$$\left | AB \right | . \left | DH \right | = \left | U_B . V_D - V_B . U_D \right |$$
In terms of original coordinates, it looks like this:
$$S = \left | (x_B - x_A) . (y_D - y_A) - (y_B - y_A) . (x_D - x_A) \right |$$

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