Given:

The mid-points of the sides of a triangle ABC is given as (-2, 3, 5), (4, -1, 7) and (6, 5, 3).

By using the section formula,

We know the mid-point divides side in the ratio of 1:1.

The coordinates of C is given by,

P(-2, 3, 5) is mid-point of A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2})

Then,

Q(4, -1, 7) is mid-point of B(x_{2}, y_{2}, z_{2}) and C(x_{3}, y_{3}, z_{3})

Then,

R(6, 5, 3) is mid-point of A(x_{1}, y_{1}, z_{1}) and C(x_{3}, y_{3}, z_{3})

Then,

Now solving for ‘x’ terms

x_{1} + x_{2} = -4……………………(4)

x_{2} + x_{3} = 8………………………(5)

x_{1} + x_{3} = 12……………………(6)

By adding equation (4), (5), (6)

x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3} = 8 + 12 – 4

2x_{1} + 2x_{2} + 2x_{3} = 16

2(x_{1} + x_{2} + x_{3}) = 16

x_{1} + x_{2} + x_{3} = 8………………………(7)

Now, subtract equation (4), (5) and (6) from equation (7) separately:

x_{1} + x_{2} + x_{3} – x_{1} – x_{2} = 8 – (-4)

x_{3} = 12

x_{1} + x_{2} + x_{3} – x_{2} – x_{3} = 8 – 8

x_{1} = 0

x_{1} + x_{2} + x_{3} – x_{1} – x_{3} = 8 – 12

x_{2} = -4

Now solving for ‘y’ terms

y_{1} + y_{2} = 6……………………(8)

y_{2} + y_{3} = -2……………………(9)

y_{1} + y_{3} = 10……………………(10)

By adding equation (8), (9) and (10) we get,

y_{1} + y_{2} + y_{2} + y_{3} + y_{1} + y_{3} = 10 + 6 – 2

2y_{1} + 2y_{2} + 2y_{3} = 14

2(y_{1} + y_{2} + y_{3}) = 14

y_{1} + y_{2} + y_{3} = 7………………………(11)

Now, subtract equation (8), (9) and (10) from equation (11) separately:

y_{1} + y_{2} + y_{3} – y_{1} – y_{2} = 7 – 6

y_{3} = 1

y_{1} + y_{2} + y_{3} – y_{2} – y_{3} = 7 – (-2)

y_{1} = 9

y_{1} + y_{2} + y_{3} – y_{1} – y_{3} = 7 – 10

y_{2} = -3

Now solving for ‘z’ terms

z_{1} + z_{2} = 10……………………(12)

z_{2} + z_{3} = 14……………………(13)

z_{1} + z_{3} = 6……………………(14)

By adding equation (12), (13) and (14) we get,

z_{1} + z_{2} + z_{2} + z_{3} + z_{1} + z_{3} = 6 + 14 + 10

2z_{1} + 2z_{2} + 2z_{3} = 30

2(z_{1} + z_{2} + z_{3}) = 30

z_{1} + z_{2} + z_{3} = 15………………………(15)

Now, subtract equation (8), (9) and (10) from equation (11) separately:

z_{1} + z_{2} + z_{3} – z_{1} – z_{2} = 15 – 10

z_{3} = 5

z_{1} + z_{2} + z_{3} – z_{2} – z_{3} = 15 – 14

z_{1} = 1

z_{1} + z_{2} + z_{3} – z_{1} – z_{3} = 15 – 6

z_{2} = 9

Thevertices of sides of a triangle ABC are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5).

Answered by Aaryan | 1 year agoA(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Show that the three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.