Question

determine whether the given points (8,4) (5,7)and (-1,1) are vertices of right triangle.

Answer 1

Answer:

Step-by-step explanation:

(8,4),(5,7),(-1,1)

By using distance formula,

AB =√(x2-x1)^+(y2-y1)^

=√(5-8)^+(7-4)^

=√17+33

=√50

=2√25 sq. Units

BC=√(x2-x1)^+(y2-y1)^

=√{(-1)-5}^+(1-7)^

=√72

=2√36 sq. Units

CA=√(x2-x1)^+(y2-y1)^

=√{(-1)-8}^+(1-4)^

=√81+9

=2√45 sq. Units

Therefore, given pts.are not the vertices of right angled triangle.

Answer 2

The given points do not form a right triangle.

To determine whether the given points (8, 4), (5, 7), and (-1, 1) are vertices of a right triangle, we can use the Pythagorean theorem.

First, we need to find the distances between each pair of points:

Distance between (8, 4) and (5, 7):

d1 = $√((8-5)^2 + (4-7)^2) = √(3^2 + (-3)^2) = √18$

Distance between (5, 7) and (-1, 1):

d2 = $√((5-(-1))^2 + (7-1)^2) = √(6^2 + 6^2) = √72$

Distance between (-1, 1) and (8, 4):

d3 = $√((-1-8)^2 + (1-4)^2) = √((-9)^2 + (-3)^2) = √90$

Now, we can check if any of these distances satisfy the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the longest side.

If $d1^2 + d2^2 = d3^2 or d1^2 + d3^2 = d2^2 or d2^2 + d3^2 = d1^2$, then the given points are vertices of a right triangle.

Plugging in the distances we found, we get:

$d1^2 + d2^2 = (√18)^2 + (√72)^2 = 18 + 72 = 90 ≠ (√90)^2 = d3^2$

$d1^2 + d3^2 = (√18)^2 + (√90)^2 = 18 + 90 = 108 ≠ (√72)^2 = d2^2$

$d2^2 + d3^2 = (√72)^2 + (√90)^2 = 72 + 90 = 162 ≠ (√18)^2 = d1^2$

None of the above equations hold true, so the given points do not form a right triangle.

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