using the section formula show that the points A(1,0),B(5,3),C(2,7)D(-2,4) are the vertices of a parallogram taken in order
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Let A (-7 , 6) B (8 , 5) and C (a , b) are the three vertices of the triangle.
centroid of the triangle = (1 , 3)
Centroid of the triangle = (x1 + x2 + x3)/3 , (y1 + y2 + y3)/3
(1 , 3) = (-7 + 8 + a)/3 ,(6 + 5 + b)/3
(1 , 3) = (1 + a)/3 ,(11 + b)/3
Equating x and y coordinates
(1+a)/3 = 1 (11+b)/3 = 3
1 + a = 3 11 + b = 9
a = 3 – 1 b = 9 -11
a = 2 b = -2
Therefore the missing vertex of the triangle is (2,-2)
centroid of the triangle = (1 , 3)
Centroid of the triangle = (x1 + x2 + x3)/3 , (y1 + y2 + y3)/3
(1 , 3) = (-7 + 8 + a)/3 ,(6 + 5 + b)/3
(1 , 3) = (1 + a)/3 ,(11 + b)/3
Equating x and y coordinates
(1+a)/3 = 1 (11+b)/3 = 3
1 + a = 3 11 + b = 9
a = 3 – 1 b = 9 -11
a = 2 b = -2
Therefore the missing vertex of the triangle is (2,-2)
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