Question

In Exercises \(53-56,\) use integration to find the area of the figure having the given vertices. $$ (2,-3),(4,6),(6,1) $$

Short answer

The area of the triangle is 14 square units.

Step-by-step solution

Identify the vertices

The given vertices of the triangle are (2,-3), (4,6) and (6,1). They shall be labeled as points A, B, and C respectively for easy referencing.

Utilize the area formula

Using the formula for the area of a triangle from the coordinates of its vertices: \( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1- y_2) |\), we substitute the coordinates into this formula and finish up with \( \frac{1}{2} | 2(6 - 1) + 4(1 + 3) + 6(-3 - 6) | \).

Evaluate the expression

Perform the calculations in the bars before taking the absolute value, for the area to be positive. Evaluate the expression \( \frac{1}{2} |2(5) + 4(4) + 6(-9)| \) to obtain \( \frac{1}{2} |10 + 16 - 54| = \frac{1}{2} |-28| \).

Final Calculation

To find the area of the triangle, calculate \(\frac{1}{2} |-28| = 14\).

Key concepts

Integration

Integration is a fundamental tool in calculus that allows us to calculate the area under a curve, the accumulated quantity, and various other applications. In the context of finding the area of a shape such as a triangle, integration can sometimes be used to sum up infinitesimally small sections to find the total area. Especially when dealing with irregular shapes or curves, integration becomes a powerful method to compute areas that might not be easily calculable through simple geometric formulas.

However, the exercise we are considering deals with a triangle determined by its vertices. Normally, for such straightforward geometric shapes whose vertices are known, integration is not the immediate tool we reach for. Instead, we apply direct formulas that leverage the coordinates of the vertices themselves. But it's important to understand that the concept of integration is fundamentally about adding up small pieces to find a whole - and this is true whether we are summing small sections of area under a curve or considering the implications of coordinates on the area of a triangle.

Coordinates of Vertices

Coordinates of vertices are the specific points plotted on a two-dimensional plane that define the corners or 'vertices' of a geometric shape like a triangle or polygon. Each vertex is given by an ordered pair \( (x,y) \) in a Cartesian coordinate system, where 'x' represents the horizontal position and 'y' the vertical position.

In finding the area of a triangle using its vertices, the coordinates provide the necessary information to determine the length of the sides and hence calculate the area. The formula that utilizes the coordinates of vertices to find the area of a triangle is derived from the concept of a determinant in linear algebra and takes into account the position of each vertex relative to the others. This formula is often preferred because it offers a direct and efficient way to arrive at the area without the need for graphing the shape or performing integration:
\( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1- y_2)| \) where \( x_1, x_2, x_3 \) and \( y_1, y_2, y_3 \) are the coordinates of the vertices. By simply plugging in the coordinates, the area can be calculated succinctly.

Absolute Value

The absolute value of a number is essentially its distance from zero on the number line, regardless of direction. For any real number 'a', its absolute value is denoted by |a| and is defined as \( |a| = a \) if 'a' is positive or zero and \( |a| = -a \) if 'a' is negative. Hence, the absolute value is always a non-negative quantity.

In the context of area calculations, particularly for a triangle using the coordinates of its vertices, the absolute value ensures that the calculated area is always positive. Since the magnitude of area cannot physically be negative, any calculation of area resulting in a negative value simply reflects the order in which the vertices were processed. By taking the absolute value of the resulting determinant expression, you negate the possibility of an erroneous negative area and ensure the final answer represents a real, measurable area.

During the final stage of calculating the triangle's area in the given problem, we see the absolute value in action: \( \frac{1}{2} |-28| \) ensures that the negative result from the determinant is converted to a positive 28, which, when halved, yields the true area of the triangle, 14 square units.