Two vertices of a right triangle are (0,1) and (0,6). The area of the triangle is 10 square units. Which point could be the third vertex of the triangle?

let's consider (0,1) and (0,6) to be the base,

the the base is 5 units.
let the height be h
giving us a nice isosceles triangle

(1/2)(5)h = 10
5h = 20
h = 4

so a possible third point is (4, 3.5)

As a matter of fact, we could now take any point along the vertical line x = 4
Since x = 4 is parallel to the y-axis, the distance between them is always 4 and the base of our triangle is always 5, giving us an area of
(1/2)(4)(5) = 10

A third point looks like (4, k), where k is any real number.

Forgot to mention that we could also have placed out third point on the left side of the y-axis,

e.g. (-4,8) would work

To find the third vertex of the right triangle, we need to understand the given information and use it to determine the possible coordinates.

Given that two vertices of the right triangle are (0,1) and (0,6), we can tell that the triangle is a right triangle with the right angle at the vertex (0,1).

To determine the third vertex, we can consider the characteristics of a right triangle: the two legs are perpendicular, and the length of one leg multiplied by the length of the other leg and divided by 2 gives the area of the triangle.

Thus, the length of the vertical leg is 6 - 1 = 5 units, and the length of the horizontal leg is 0 units. Therefore, the area of the triangle is 5 * 0 / 2 = 0 square units, which does not match the given area of 10 square units.

Since the area does not match, it means that there is no valid third vertex for the right triangle with the given coordinates.