Respuesta :
Answer:
[tex]\overline{PQ} = 3[/tex]
[tex]\overline{QR} = 3\sqrt{5}[/tex]
[tex]\overline{PQ} = 6[/tex]
PQR is a right-angled triangle.
PQR is NOT an isosceles triangle.
Step-by-step explanation:
P(2, -1, 0), Q(4, 1, 1), R(4, -5, 4) are coordinates on a (x,y,z) space.
we can easily find the lengths of each side using the good ol' distance formula.:
[tex]\overline{AB}=\sqrt{(x_1 - x_2)^2+(y_1 - y_2)^2+(z_1 - z_2)^2}[/tex]
To find the distance PQ
[tex]\overline{PQ}=\sqrt{(2 - 4)^2+(-1 - 1)^2+(0 - 1)^2}[/tex]
[tex]\overline{PQ}=\sqrt{(-2)^2+(-2)^2+(-1)^2}[/tex]
[tex]\overline{PQ}=\sqrt{9}[/tex]
[tex]\overline{PQ}=3[/tex]
To find the distance QR
[tex]\overline{QR}=\sqrt{(4 - 4)^2+(1 - (-5))^2+(1 - 4)^2}[/tex]
[tex]\overline{QR}=\sqrt{45}[/tex]
[tex]\overline{QR}=\sqrt{9\times5}[/tex]
[tex]\overline{QR}=\sqrt{9}\times\sqrt{5}[/tex]
[tex]\overline{QR}=3\sqrt{5}[/tex]
To find the distance RP
[tex]\overline{PQ}=\sqrt{(4 - 2)^2+(-5 - (-1))^2+(4 - 0)^2}[/tex]
[tex]\overline{PQ}=6[/tex]
Now we have the side lengths of the triangle. And since no two side-lengths are equal, we can confidently say that this is NOT an isosceles triangle.
Our next task is to make sure whether this is a right-angled triangle or not.
Although there are multiple ways for this. One easy way to find this out is to use the Pythagorean Theorem. and see whether this equation is true?
[tex](\overline{QR})^2 = (\overline{PQ})^2 + (\overline{RP})^2[/tex]
to put it in words, what we're trying to find is: does the square of the larger side (QR) equal the sum of the squares of the smaller sides?
if this is true, then we've found that the triangle PQR is indeed a right-angled triangle.
[tex](3\sqrt{5})^2 = (3)^2 + (6)^2[/tex]
[tex]45 = 9+36[/tex]
[tex]45 = 45[/tex]
thus, we've proved that the triangle is right-angled!
side note:
An alternative way to find whether the triangle PQR is right-angled or not is to use the cosine law. (it also looks a bit like the Pythagorean theorem too if you look closely)
[tex](\overline{QR})^2 = (\overline{PQ})^2 + (\overline{RP})^2 - 2(\overline{PQ})(\overline{RP})\cos{\theta}[/tex]
you can solve the above equation for theta and check whether [tex]\theta[/tex] equals to 90 or not.
if [tex]\theta = 90[/tex], then the triangle PQR is right angle! (and it WILL BE since we've already found this out using Pythagorean theorem)
It is not an isosceles triangle.
For an isosceles triangle two sides must have same length.
We will use the distance formula to find the lengths.
[tex]PQ=\sqrt{\left(4-2\right)^{2}+\left(1+1\right)^{2}+\left(1-0\right)^{2}}=3[/tex]
[tex]QR=\sqrt{\left(4-4\right)^{2}+\left(-5-1\right)^{2}+\left(4-1\right)^{2}}=3\sqrt{5}[/tex]
[tex]PR=\sqrt{\left(4-2\right)^{2}+\left(-5+1\right)^{2}+\left(4-0\right)^{2}}=6[/tex]
Since no sides are of the same length, so it is not an isosceles triangle.
Learn more: https://brainly.com/question/12356025
