Question

△abc △abc
a(4,3) a(8,6)
b(4,0) b(8,0)
c(0,0) c(0,0)
1 how are the coordinates related?
2 you can? the coordinates of
△abc by? to find the coordinates
of △abc.

Explanation:

Step1: Analyze the x - coordinates

For point \(A(4,3)\) and \(A'(8,0)\), \(4\times2 = 8\). For point \(B(4,0)\) and \(B'(8,0)\), \(4\times2=8\).

Step2: Analyze the y - coordinates

For point \(A(4,3)\) and \(A'(8,0)\), \(3\times0 = 0\). But if we consider the non - zero y - coordinate transformation (ignoring the \(y = 0\) case for \(B\) and \(B'\) as a special case of the transformation rule). If we assume a linear transformation for non - zero \(y\) values (since \(C(0,0)\) and \(C'(0,0)\) is a fixed point).
We can see that for non - zero \(x\) values (e.g., \(x = 4\) in \(\triangle ABC\) and \(x = 8\) in \(\triangle A'B'C'\)), we multiply the \(x\) - coordinate of \(\triangle ABC\) by \(2\) and the \(y\) - coordinate of \(\triangle ABC\) by \(0\) (a degenerate transformation in the \(y\) - direction). But if we consider the non - vertical transformation (since \(B\) and \(B'\) have \(y = 0\)), we can say that we can transform the coordinates of \(\triangle ABC\) by a transformation.

We note that for \(x\) - coordinates: \(x_{A'}=2x_A\), \(x_{B'}=2x_B\), \(x_{C'}=2x_C\) (since \(x_C = 0\), \(x_{C'}=0\)). For \(y\) - coordinates: \(y_{A'}=0\times y_A\) (a special case, but if we consider the non - zero \(x\) transformation as a scaling and \(y\) as a projection).

We can say that you can transform (a combination of scaling in \(x\) and projection in \(y\)) the coordinates of \(\triangle ABC\). If we consider the non - zero \(x\) values (since \(C\) is a fixed point), we can say that you can scale the \(x\) - coordinates of \(\triangle ABC\) by \(2\) (for non - zero \(x\)) and set \(y\) - coordinates (except for \(C\)) according to the transformation rule (in this case, \(y = 0\) for \(A'\) and \(B'\) from \(A\) and \(B\) with non - zero \(y\) in \(A\)). But if we consider the overall pattern (ignoring the \(y\) - value change for \(A\) as a degenerate case of the transformation relative to \(x\) - scaling), we can say that you can scale the \(x\) - coordinates of \(\triangle ABC\) by \(2\) (since \(4\times2=8\) for \(x\) of \(A\) and \(B\))

Answer:

You can scale the coordinates of \(\triangle ABC\) by \(2\) (for \(x\) - coordinates, and a special transformation for \(y\) - coordinates) to find the coordinates of \(\triangle A'B'C'\)