Question

Use calculus to find the area a of the triangle with the given vertices. (0, 0), (6, 2), (4, 8)

Answer 1

To find the area of the triangle with the given vertices (0, 0), (6, 2), and (4, 8), we can use calculus. We first find the equation of the line passing through two points, then calculate the height of the triangle. Finally, we use the formula for the area of a triangle to find the result.

Step-by-step explanation:

To find the area of a triangle with the given vertices (0, 0), (6, 2), and (4, 8), we can use calculus. We can calculate the area by treating one of the sides of the triangle as the base and finding the height. Let's take the side connecting the points (0, 0) and (6, 2) as the base.

First, find the equation of the line passing through the two points using the slope-intercept form (y = mx + b). We have:

y = (2/6)x + b

Substituting one of the points, (0, 0), into the equation, we can solve for b:

0 = (2/6)(0) + b

b = 0

So the equation of the line is y = (1/3)x.

Now, let's find the area of the triangle using calculus. The base of the triangle is 6 units, and the height of the triangle is the vertical distance between the line y = (1/3)x and the point (4, 8). We can find the height by subtracting the y-coordinate of the point from the value of y at x = 4:

height = (1/3)(4) - 8 = 4/3 - 8 = -20/3

However, we only want the absolute value of the height, so we take the positive value: height = 20/3.

Finally, we can calculate the area of the triangle:

area = (1/2) * base * height = (1/2) * 6 * 20/3 = 10/3 * 6 = 60/3 = 20 square units.

Answer 2

The area of the triangle with vertices
`(x_1,y_1),(x_2,y_2),(x_3,y_3)` is

`(1)/(2)\left|\begin{array}{ccc}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{array}\right|`.

Inserting numerical values, the are of the triangle with vertices (0, 0), (6, 2), (4, 8) is

`\Delta = (1)/(2)\left|\begin{array}{ccc}1&0&0\\1&6&2\\1&4&8\end{array}\right |\\ \Delta = 0.5(48-8)\\ \Delta = 20 \;square\; units`