Verify that the area of the triangle with vertices (4, 6), (7, 10) and (1, -2) remains invariant under the translation of axes when the origin is shifted to the point (-2, 1).

Given points (4, 6), (7, 10), and (1, -2).


To show: The area of a triangle is invariant to shifting of origin.


The area of triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is


= [x1(y2 – y3) + x2(y3 -y1) + x3(y1 – y2)]


Hence, the area of given triangle = [4(10-(-2)) + 7(-2-6) + 1(6-10)]


= [48 – 56 - 4]


= [-12]


= -6


we takes modulus value of -6 i.e. 6 since the area cannot be negative.


Origin shifted to point (-2, 1), the new coordinates of the triangle are (6, 5), (9, 9), and (3, -3) obtained from subtracting a point (-2, 1).


Hence, the new area of triangle = [6(9-(-3)) + 9(-3-5) + 3(5-9)]


= [72 – 72 + (-12)]


= [-12]


= -6


we takes modulus value of -6 i.e. 6 sq. sq. units since the area cannot be negative.


Since the area of the triangle before and after the translation after shifting of origin remains same, i.e. 6 sq. units, therefore we can say that the area of a triangle is invariant to shifting of origin.


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