Solve: 1/3(9k + 12) = 15
2 answers:
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Answer:
Step-by-step explanation:
The question is as following:
The verticies of a triangle on the coordinate plane are
A(0, 0), B(2, 0) and C(0, 2).
What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?
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Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).
IF the triangle is dilated by a factor of k about the origin, then
(x,y) → (kx , ky)
that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.
It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.
If a figure dilated by a factor of 1/3 about the origin
So,
<u>So, The coordinates of the triangle A'B'C' are:</u>