Finding Vertices and Area of Parallelogram

The question asks for the area of a parallelogram with given vertices A (1, 2), B (4, y), C (x, 6), and D (3, 5) taken in order. To find the area, we first need to determine the unknown coordinates, x and y.

In a parallelogram, the diagonals bisect each other. This means the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

Calculating Unknown Coordinates (x and y)

Let's find the midpoint of AC and BD using the midpoint formula: Midpoint of a segment with endpoints $ (x_1, y_1) $ and $ (x_2, y_2) $ is $ (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) $.

• Midpoint of AC with A(1, 2) and C(x, 6): $$ \left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{1+x}{2}, 4\right) $$
• Midpoint of BD with B(4, y) and D(3, 5): $$ \left(\frac{4+3}{2}, \frac{y+5}{2}\right) = \left(\frac{7}{2}, \frac{y+5}{2}\right) $$

Since the midpoints are the same, we can equate the corresponding coordinates:

• Equating x-coordinates: $$ \frac{1+x}{2} = \frac{7}{2} $$ $$ 1+x = 7 $$ $$ x = 7 - 1 $$ $$ x = 6 $$
• Equating y-coordinates: $$ 4 = \frac{y+5}{2} $$ $$ 4 \times 2 = y+5 $$ $$ 8 = y+5 $$ $$ y = 8 - 5 $$ $$ y = 3 $$

So, the vertices of the parallelogram are A(1, 2), B(4, 3), C(6, 6), and D(3, 5).

Calculating the Area of the Parallelogram

Now that we have all the coordinates, we can calculate the area of the parallelogram. One way is to use the coordinates directly. The area of a parallelogram with vertices $ (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) $ taken in order can be calculated using the determinant of two adjacent vectors, for example, $\vec{AB}$ and $\vec{AD}$.

• Vector $\vec{AB} = B - A = (4-1, 3-2) = (3, 1)$
• Vector $\vec{AD} = D - A = (3-1, 5-2) = (2, 3)$

The area of the parallelogram formed by vectors $(x_1, y_1)$ and $(x_2, y_2)$ is given by the absolute value of the determinant $ |x_1 y_2 - x_2 y_1| $.

Area = $ |(3)(3) - (1)(2)| $

Area = $ |9 - 2| $

Area = $ |7| $

Area = 7 square units.

Alternatively, we can find the area of one of the triangles formed by a diagonal (e.g., triangle ABC) and double it, since the diagonal divides the parallelogram into two congruent triangles.

Using vertices A(1, 2), B(4, 3), C(6, 6), the area of triangle ABC is:

Area(ABC) = $ \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $

Area(ABC) = $ \frac{1}{2} |1(3 - 6) + 4(6 - 2) + 6(2 - 3)| $

Area(ABC) = $ \frac{1}{2} |1(-3) + 4(4) + 6(-1)| $

Area(ABC) = $ \frac{1}{2} |-3 + 16 - 6| $

Area(ABC) = $ \frac{1}{2} |7| $

Area(ABC) = $ \frac{7}{2} $ square units.

The area of the parallelogram is twice the area of triangle ABC.

Area of Parallelogram = $ 2 \times \text{Area(ABC)} = 2 \times \frac{7}{2} = 7 $ square units.

Both methods give the same area, 7 square units.

Summary of Calculations

The area of the parallelogram is 7 square units.

Revision Table: Parallelogram Properties

Additional Information: Area Calculation Methods

There are several ways to calculate the area of a parallelogram in coordinate geometry:

• Using Vectors: If two adjacent sides are represented by vectors $\vec{u} = (x_1, y_1)$ and $\vec{v} = (x_2, y_2)$, the area is $|x_1 y_2 - x_2 y_1|$. This is the magnitude of the 2D cross product (or determinant).
• Using Triangles: A diagonal divides a parallelogram into two congruent triangles. Calculate the area of one triangle using the determinant formula ($ \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $) and double it.
• Using Shoelace Formula: The Shoelace formula can be used for any polygon. For a parallelogram with vertices $ (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) $ in order, the area is $ \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| $.

All these methods yield the same correct area for the given parallelogram.