Show that the points ( -4, 9, 6), (-1, 6, 6), (0, 7, 10) form a right-angled isosceles triangle.?

Answer 1

The distance between (-4,9,6) and (-1,6,6) =
square root of[ (-4-(-1))^2 + (9-6)^2 + (6-6)^2 ] =
square root of[ 3^2 + 3^2 + 0 ] =
square root of[ 18 ]

Similarly the distance between (-1,6,6) and (0,7,10) =
square root of[ 1 + 1 + 16 ] =
square root of[ 18 ]


Similarly the distance between (-4,9,6) and (0,7,10) =
square root of[ 16 + 4 + 16 ] =
square root of[ 36 ]

Thus , we can see that the length of the first two sides are the same proving that the triangle is isosceles.

By hypotenuse rule,
side1^2 + side^2 = hypotenuse^2
Therefore (square root of[ 18 ])^2 + (square root of[18])^2)
= 18 +18 = 36 = (square root of[36])^2
Thus the hypotenuse rule is satisfied proving the triangle is also right angled.

Answer 2

You could also use multivariable / vector calculus:

Let A = (-4, 9, 6), B = (-1, 6, 6), C = (0, 7, 10)
Let v1 be the vector (line segment) between A and B; that is,
v1 = A - B = (-4, 9, 6) - (-1, 6, 6) = (-3, 3, 0)
Let v2 be the vector between A and C; that is,
v2 = A - C = (-4, 9,6 ) - (0, 7, 10) = (-4, 2, -4)
Let v3 be the vector between B and C; that is,
v3 = B - C = (-1, 6, 6) - (0, 7, 10) = (-1, -1, -4)

We can then compute the angle between each pair of vectors v1, v2, v3 using the dot product, represented < _ , _ >. The dot product of two vectors, say

x = (a1, a2, a3, ...) and y = (b1, b2, b3, ...)

is equal to the scalar value of

= (a1 * b1 + a2 * b2 + ...).

As a rule, the dot product of two vectors equals the product of the magnitude of each vector, multiplied by the cosine of the angle between them; that is, for x and y above

= IIxII * IIyII * cos(-),

where (-) is the angle between x and y, and IIxII, IIyII is the magnitude of x and y, respectively. With some algebra, we get

(-) = cos^-1 ( / (IIxII * IIyII)),

where cos^-1 represents the inverse cosine function. To compute the magnitude of a vector, say x from above, it is simply

IIxII = sqrt(a1^2 + a2^2 + ...)

So

(-) of v1 and v2 = cos^-1 ((-3 * -4 + 3 * 2 + 0 * -4) / (((-3)^2 + (3)^2 + (0)^2) * ((-4)^2 + (2)^2 + (-4)^2))) =
= cos^-1 ((12 + 6 + 0) / ((9 + 9 + 0) * (16 + 4 + 16) = cos^-1 (18 / (18 * 36) =
= cos^-1 (1 / 36)

Similarly,

(-) of v1 and v3 = cos^-1 (0 / 324) = cos^-1 (0)
(-) of v2 and v3 = cos^-1 (18 / 648) = cos^-1 (1/36)

Thus, the (-) of v1 and v2 equals the (-) of v2 and v3; we have at least an isosceles triangle. Since the (-) of v1 and v3 equals the inverse cosine of zero, and the inverse cosine of zero equals 90 degrees, we conclude we have a right isosceles triangle.

Answer 3

Label the three points
A(-4, 9, 6)
B(-1, 6, 6)
C(0, 7, 10)

AB = √(3^2 + 3^2 + 0^2) = √(18)
BC = √(1^2 + 1^2 + 4^2) = √(18)
AC = √(4^2 + 2^2 + 4^2) = √(36)

AB² + BC² = 18 + 18 = 36 = AC²

To sides of the triangle are equal, so it is an isosceles triangle. The Pythagorean Theorem is satisfied, so it is also a right triangle. This makes it a right-angled isosceles triangle.

Answer 4

say,
A( -4, 9, 6),
B(-1, 6, 6),
C(0, 7, 10)
now calculate the dist between these points.
AB = ( -4, 9, 6) - (-1, 6, 6)
= (-3,3,0)
now, sqrt -3^2+3^2
AB =sqrt 18
similarly,find out BC,CA

use the pythogoras theorem tht, (hypotenuse)^2 = opp side ^2 + adj side ^2

then u can prove it.

hope this helps

Answer 5

S1^2 = (3^2 + 3^2 + 0^2) = 18
S2^2 = (1^2 + 1^2 + 4^2) = 18
S3^2 = (4^2 + 2^2 + 4^2) = 36
QED

Answer 6

Are you sure of correctness of the problem ? Why each point is having three coordinates ? Hope you don't mean three dimensional ..... and talking about a triangle .....

Answer 7

umm this is very simple if you do it on graph paper.. it should have 2 equal sides for it to be isosceles.