can you please tell me if the points (8,9) (15,2) and (17,4) are the vertices of a right triangle

The lengths L of three sides are sqrt 98, sqrt 8 and sqrt 106. You can prove that using

L = sqrt[(y2-y1)^2 + (x2-x1)^2]
The pythagorean relationship a^2 + b^2 + c^2 is therefore obeyed, since 98 + 9 = 106.
What does that tell you about the triangle?

You could prove it in other ways, but this appeared to be the simplest.

so tey are vertices of a triangle?

To determine if the points (8,9), (15,2), and (17,4) are the vertices of a right triangle, we can use the Pythagorean theorem.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's calculate the lengths of the three sides of the triangle using the distance formula:

Distance between points (8,9) and (15,2):
d₁ = √[(x₂ - x₁)² + (y₂ - y₁)²]
d₁ = √[(15 - 8)² + (2 - 9)²]
d₁ = √[7² + (-7)²]
d₁ = √[49 + 49]
d₁ = √98
d₁ ≈ 9.899

Distance between points (8,9) and (17,4):
d₂ = √[(x₂ - x₁)² + (y₂ - y₁)²]
d₂ = √[(17 - 8)² + (4 - 9)²]
d₂ = √[9² + (-5)²]
d₂ = √[81 + 25]
d₂ = √106
d₂ ≈ 10.295

Distance between points (15,2) and (17,4):
d₃ = √[(x₂ - x₁)² + (y₂ - y₁)²]
d₃ = √[(17 - 15)² + (4 - 2)²]
d₃ = √[2² + 2²]
d₃ = √[4 + 4]
d₃ = √8
d₃ ≈ 2.828

Now, we need to check if the sum of the squares of the two shorter sides is equal to the square of the longest side:

d₃² = d₁² + d₂²
(2.828)² = (9.899)² + (10.295)²
7.999 + 106.013 ≈ 116.012

The sum of the squares of the shorter sides is approximately 116.012, which is not equal to the square of the longest side, which is approximately 7.999. Therefore, the points (8,9), (15,2), and (17,4) do not form a right triangle.