prove that (6,-4),(-2,-4),(2,10) are the vertices of a right angled isosceles triangle?

Answer 1

The distance from (6,-4) to (-2,-4) is 8.

The distance from (6, -4) to (2, 10) is sqrt(14^2 + 4^2) = sqrt(212) = 2sqrt(53).

The distance from (-2,4) to (2,10) is sqrt(4^2 + 14^2) = 2sqrt(53).

You can already see that the triangle is isosceles because two of the sides are the same length.

Now we can use the pythagorean theorem to check to see if it is also a right triangle. If the side lengths satisfy a^2 + b^2 = c^2, then it is a right triangle.

[2sqrt(53)]^2 + [2sqrt(53)]^2 = 8^2
212 + 212 = 64
424 = 64
THIS IS NOT TRUE ===> The triangle is isosceles but NOT right.

Graph it and you will see that it is clearly NOT a right triangle.

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To clear up any confusion, here is the mistake in papyrus' answer:

"Gradient of AC = (10-(6))/(10-(-4)) = -2/7"
The gradient (or slope) of AC is (10-(-4))/(2-6) = -7/2
Thus the two sides have opposite slopes, but not opposite reciprocal slopes. They are not perpendicular.

Answer 2

The line containing (6, -4) and (-2, -4) is horizontal, with a slope (gradient) of zero.

The line containing (6, -4) and (2, 10) has a negative slope... it's going down to the right.
m = Δy / Δx = (-4 - 10) / (6 - 2) = -14 / 4 = -3½.

The line containing (-2, -4) and (2, 10) has a positive slope... it goes up to the right.
m = Δy / Δx = (-4 - 10) / (-2 - 2) = (-14) / (-4) = +3½.

Since the second two lines aren't vertical, they're not perpendicular to the horizontal line y = 4.
If the second two lines are to be perpendicular, their slopes would be negative reciprocals of one another. They're not. (One has a slope of -7 / 2, the other +7 / 2... opposites, but not reciprocals.)

On the bright side, it is, at least, an isosceles triangle. The distance between (2, 10) to (6, 4) = √[(2 - 6)² + (10 - 4)²] = √[(-4)² + 6²] = √[16 + 36] = √52 = 2√13.

The distance between (2, 10) to (-2, 4) = √[(2 - (-2))² + (10 - 4)²] = √[(4)² + 6²] = √[16 + 36] = √52 = 2√13.

Since two of the sides are congruent, it is an isosceles triangle... but it's not a right triangle.

Answer 3

Calculate distance between points:

A (6, -4) B (-2, -4) C (2, 10)

AB = 8
AC = sqrt (4^2 + 14^2) = sqrt (16 + 196) = sqrt 212 = 2 sqrt 53
BC = sqrt (4^2 + 14^2) = 2 sqrt 53

The triangle is clearly isosceles, as AC and BC are congruent.

If the triangle is right, then 2(2 sqrt 53)^2 = 8^2
2*4*53 = 64
53 = 8

The triangle is NOT a right triangle.

Answer 4

(6,-4) and (-2,-4)
D = sqrt((-2 - 6)^2 + (-4 - (-4))^2)
D = sqrt((-2 + (-6))^2 + 0^2)
D = sqrt((-8)^2)
D = sqrt(64)
D = 8

(6,-4) and (2,10)
D = sqrt((2 - 6)^2 + (10 - (-4))^2)
D = sqrt((-4)^2 + (10 + 4)^2)
D = sqrt(16 + 14^2)
D = sqrt(16 + 196)
D = sqrt(212)
D = sqrt(4 * 53)
D = 2sqrt(53)

(-2,-4) and (2,10)
D = sqrt((2 - (-2))^2 + (10 - (-4))^2)
D = sqrt((4)^2 + (10 + 4)^2)
D = sqrt(16 + 14^2)
D = sqrt(16 + 196)
D = sqrt(212)
D = 2sqrt(53)

a^2 + b^2 = c^2
a = b
a^2 + a^2 = c^2
2a^2 = c^2
a^2 = (c^2)/2
a = sqrt((c^2)/2)

a = sqrt((8^2)/2)
a = sqrt(64/2)
a = sqrt(32)
a = sqrt(16 * 2)
a = 4sqrt(2)

for this to be an isosceles right triangle

2a^2 = c^2
c = sqrt(2a^2)
c = a * sqrt(2)

c = sqrt(212) * sqrt(2)
c = sqrt(424)
c = 2sqrt(106)

Its an isoscles triangle, but it isn't an isosceles right triangle.

Answer 5

Calculate the distances between the points and the angles. Two points are equidistant and one of the angles is 90 degrees. Therefore, these points are vertices of a right angled isosceles triangle.

Answer 6

Let A(6,-4), B(-2,-4), C(2,10)
Gradient of AC = (10-(6))/(10-(-4)) = -2/7
Gradient of AB = (10-(-4))/(2-(-2)) = 7/2
Gradient of AC * Gradient of AB = -1

Therefore, triangle ABC is a right-angled triangle (AC is perpendicular to AB).

Answer 7

Find the length of each sides. If any two sides are equal, then the triangle is isosceles. Or calculate the angle of each corner. If any two are equal, then the triangle is isosceles.

Answer 8

hmmm

lets try:

sqrt((6--2)^2+(-4--4)^2) = sqrt(64) = 8

sqrt((-2-2)^2+(-4-10)^2) = sqrt(16 + 196) = sqrt(212)

sqrt((6-2)^2+(-4-10)^2) = sqrt(16 + 196) = sqrt(212)

sqrt(212)^2+sqrt(212)^2=8^2
sqrt(424)=sqrt(4096)
incorrect its not a right triangle

Answer 9

I shall show the steps.
Find the length of sides

and then check.

Answer 10

state it in the form of a question