Question

Coordinates of the point An are (n^2,2n) and of Bn are (n^2,-2n). What is the area of the quadrilateral with vertices
A1B1BnAn ?

Answer 1

Answer on Question #33354 – Math – Analytic Geometry

Question

Coordinates of the point $A_n$ are $(n^2, 2n)$ and of $B_n$ are $(n^2, -2n)$.

What is the area of the quadrilateral with vertices $A_1B_1A_nB_n$?

Solution

Suppose that $n$ is a natural number.

Coordinates of point $A_1$ is $(1^2, 2 \cdot 1)$, that is $(1, 2)$.

$$B_1 (1^2, -2 \cdot 1), \text{ that is } (1, -2).$$ $$A_n (n^2, 2n)$$ $$B_n (n^2, -2n)$$

If $n = 1$ than $A_1B_1$ is a line segment and its area is 0.

If $n > 1$ than $A_1B_1A_nB_n$ is trapezoid.

Area of trapezoid is given by the formula

$$S = \frac{(a + b)}{2} h,$$

where $S$ is area, $a$ and $b$ are bases of trapezoid, and $h$ is height (altitude).

$$h = x_{A_n} - x_{A_1} = n^2 - 1$$ $$a = y_{A_1} - y_{B_1} = 2 - (-2) = 4$$ $$b = y_{A_n} - y_{B_n} = 2n - (-2n) = 4n$$

So,

$$S = \frac{(a + b)}{2} h$$ $$S = \frac{(4 + 4n)}{2} (n^2 - 1)$$ $$S = 2(n + 1)(n^2 - 1)$$

**Answer:**

$$S = \begin{cases} 2(n + 1)(n^2 - 1), & \text{if } n > 1, \\ 0, & \text{if } n = 1. \end{cases}$$

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