Question -
Answer -
Given:
The mid-points of thesides of a triangle ABC is given as (-2, 3, 5), (4, -1, 7) and (6, 5, 3).
By using the sectionformula,
We know the mid-pointdivides side in the ratio of 1:1.
The coordinates of Cis given by,
P(-2, 3, 5) ismid-point of A(x1, y1, z1) and B(x2,y2, z2)
Then,
Q(4, -1, 7) ismid-point of B(x2, y2, z2) and C(x3,y3, z3)
Then,
R(6, 5, 3) ismid-point of A(x1, y1, z1) and C(x3,y3, z3)
Then,
Now solving for ‘x’terms
x1 + x2 =-4……………………(4)
x2 + x3 =8………………………(5)
x1 + x3 =12……………………(6)
By adding equation(4), (5), (6)
x1 + x2 +x2 + x3 + x1 + x3 =8 + 12 – 4
2x1 +2x2 + 2x3 = 16
2(x1 +x2 + x3) = 16
x1 + x2 +x3 = 8………………………(7)
Now, subtract equation(4), (5) and (6) from equation (7) separately:
x1 + x2 +x3 – x1 – x2 = 8 – (-4)
x3 =12
x1 + x2 +x3 – x2 – x3 = 8 – 8
x1 = 0
x1 + x2 +x3 – x1 – x3 = 8 – 12
x2 =-4
Now solving for ‘y’terms
y1 + y2 =6……………………(8)
y2 + y3 =-2……………………(9)
y1 + y3 =10……………………(10)
By adding equation(8), (9) and (10) we get,
y1 + y2 +y2 + y3 + y1 + y3 =10 + 6 – 2
2y1 +2y2 + 2y3 = 14
2(y1 +y2 + y3) = 14
y1 + y2 +y3 = 7………………………(11)
Now, subtract equation(8), (9) and (10) from equation (11) separately:
y1 + y2 +y3 – y1 – y2 = 7 – 6
y3 = 1
y1 + y2 +y3 – y2 – y3 = 7 – (-2)
y1 = 9
y1 + y2 +y3 – y1 – y3 = 7 – 10
y2 =-3
Now solving for ‘z’terms
z1 + z2 =10……………………(12)
z2 + z3 =14……………………(13)
z1 + z3 =6……………………(14)
By adding equation(12), (13) and (14) we get,
z1 + z2 +z2 + z3 + z1 + z3 =6 + 14 + 10
2z1 +2z2 + 2z3 = 30
2(z1 +z2 + z3) = 30
z1 + z2 +z3 = 15………………………(15)
Now, subtract equation(8), (9) and (10) from equation (11) separately:
z1 + z2 +z3 – z1 – z2 = 15 – 10
z3 = 5
z1 + z2 +z3 – z2 – z3 = 15 – 14
z1 = 1
z1 + z2 +z3 – z1 – z3 = 15 – 6
z2 = 9
∴Thevertices of sidesof a triangle ABC are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5).