Verifying Right Triangle Vertices And Finding Equidistant Points On The Y Axis

To demonstrate that the points (-2, 3), (8, 3), and (6, 7) form the vertices of a right-angled triangle, we will use the distance formula and the Pythagorean theorem. The distance formula helps us calculate the lengths of the sides of the triangle, and the Pythagorean theorem (a² + b² = c²) will help us verify if the triangle satisfies the condition for a right-angled triangle. The core idea here is to calculate the distances between each pair of points and then check if the square of the longest side equals the sum of the squares of the other two sides.

First, let's denote the points as A(-2, 3), B(8, 3), and C(6, 7). We will calculate the distances between each pair of points using the distance formula, which is given by:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

1. Distance between A(-2, 3) and B(8, 3) (AB):

   AB = √[(8 - (-2))² + (3 - 3)²] = √[(10)² + (0)²] = √100 = 10

2. Distance between B(8, 3) and C(6, 7) (BC):

   BC = √[(6 - 8)² + (7 - 3)²] = √[(-2)² + (4)²] = √(4 + 16) = √20

3. Distance between C(6, 7) and A(-2, 3) (CA):

   CA = √[(-2 - 6)² + (3 - 7)²] = √[(-8)² + (-4)²] = √(64 + 16) = √80

Now that we have the lengths of the sides, AB = 10, BC = √20, and CA = √80, we will check if the Pythagorean theorem holds true. The longest side is AB with a length of 10, so we will check if AB² = BC² + CA².

AB² = 10² = 100

BC² = (√20)² = 20

CA² = (√80)² = 80

Now, let's see if the sum of BC² and CA² equals AB²:

BC² + CA² = 20 + 80 = 100

Since AB² = BC² + CA² (100 = 100), the Pythagorean theorem holds true. Therefore, the points A(-2, 3), B(8, 3), and C(6, 7) are the vertices of a right-angled triangle. Specifically, the right angle is at vertex C, as AB is the hypotenuse.

In conclusion, by applying the distance formula to find the lengths of the sides and verifying the Pythagorean theorem, we have successfully shown that the given points form a right-angled triangle. This method is a standard approach in coordinate geometry for verifying the properties of geometric shapes using their coordinates.

To find a point on the y-axis that is equidistant from two given points, we need to understand the properties of points on the y-axis and the concept of equidistance. A point on the y-axis has an x-coordinate of 0. Therefore, we can represent the point we are looking for as (0, y). The term "equidistant" means that the distance from this point to the first given point is equal to the distance from this point to the second given point. We will use the distance formula to set up an equation and solve for the y-coordinate.

Let the point on the y-axis be P(0, y), and the two given points be A(5, -2) and B(-3, 2). We want to find the y-coordinate such that the distance PA is equal to the distance PB.

Using the distance formula:

PA = √[(5 - 0)² + (-2 - y)²] = √(25 + (y + 2)²)

PB = √[(-3 - 0)² + (2 - y)²] = √(9 + (2 - y)²)

Since PA = PB, we can set the distances equal to each other:

√(25 + (y + 2)²) = √(9 + (2 - y)²)

To eliminate the square roots, we square both sides of the equation:

25 + (y + 2)² = 9 + (2 - y)²

Now, we expand the squared terms:

25 + (y² + 4y + 4) = 9 + (4 - 4y + y²)

Next, we simplify and rearrange the equation:

25 + y² + 4y + 4 = 9 + 4 - 4y + y²

Combine like terms:

y² terms cancel out:

29 + 4y = 13 - 4y

Now, we solve for y:

4y + 4y = 13 - 29

8y = -16

y = -16 / 8

y = -2

Therefore, the point on the y-axis that is equidistant from the points (5, -2) and (-3, 2) is (0, -2).

In summary, we used the distance formula to set up an equation based on the condition of equidistance. By solving this equation, we found the y-coordinate of the point on the y-axis. This approach is a fundamental technique in coordinate geometry for solving problems involving distances and points in a coordinate plane.

In this article, we addressed two distinct problems in coordinate geometry. The first problem involved verifying that three given points form the vertices of a right-angled triangle. We achieved this by applying the distance formula to calculate the lengths of the triangle's sides and then using the Pythagorean theorem to confirm the right-angle condition. This approach demonstrates a practical application of fundamental geometric principles in a coordinate system.

The second problem focused on finding a point on the y-axis that is equidistant from two given points. Here, we again utilized the distance formula to express the distances and set up an equation based on the equidistance condition. Solving this equation yielded the y-coordinate of the desired point. This problem highlights how algebraic techniques can be combined with geometric concepts to solve for unknown points satisfying specific distance-related criteria.

Both problems illustrate the power and versatility of coordinate geometry in solving geometric problems. By leveraging the distance formula and algebraic manipulations, we can effectively analyze and solve a wide range of geometric questions. These techniques are crucial in various fields, including mathematics, physics, engineering, and computer graphics, where understanding spatial relationships and distances is essential.