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.