Question

Вершины треугольника АВС заданы координатами А (1,-2,0),В(1,-4,2),С(3,2,0) Найдите длину медианы СМ

Answer 1

$M(x_M;y_M;z_M)$ - координаты точки М - середины отрезка АВ.
$x_M=\dfrac{1+1}{2}=1;\ y_M=\dfrac{-2-4}{2}=-3;\ z_M=\dfrac{0+2}{2}=1.\\ M(1;-3;1) \\ CM=\sqrt{(1-3)^2+(-3-2)^2+(1-0)^2} = \sqrt{30}$

Answer 2

Calculation of the Length of Median SM in Triangle ABC

To find the length of the median SM in triangle ABC, we can use the distance formula. The median SM is the line segment connecting the midpoint of side AC to vertex B.

The coordinates of the vertices of triangle ABC are: A(1, -2, 0) B(1, -4, 2) C(3, 2, 0)

To find the midpoint of side AC, we can use the formula:

Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2, (z1 + z2) / 2)

Let's calculate the midpoint of side AC:

Midpoint of AC = ((1 + 3) / 2, (-2 + 2) / 2, (0 + 0) / 2) = (2, 0, 0)

Now that we have the coordinates of the midpoint, we can calculate the length of the median SM using the distance formula:

Distance = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

Let's calculate the length of the median SM:

Distance = √((1 - 2)^2 + (-4 - 0)^2 + (2 - 0)^2) = √((-1)^2 + (-4)^2 + 2^2) = √(1 + 16 + 4) = √21

Therefore, the length of the median SM in triangle ABC is √21.