The big square made with the Tangram pieces is now inside a Cartesian coordinate system.[br][br]The vertices of the square are [math]A\equiv\left(0,0\right)[/math], [math]B\equiv\left(8,0\right)[/math], [math]C\equiv\left(8,8\right)[/math] and [math]D\equiv\left(0,8\right)[/math].[br][br]You can move the square without changing its size by dragging its sides, or you can change its size by dragging the vertices [math]A[/math] and [math]B[/math].[br]
If [math]A\equiv\left(0,0\right)[/math] and [math]B\equiv\left(8,0\right)[/math], what is the equation of line [math]AC[/math]?
[math]\ell_{AC}:y=x[/math]
If we translate the square 1 unit up, what is the equation of line [math]AC[/math]?
Also the line is translated 1 unit up, therefore [math]\ell_{AC}:y=x+1[/math]
If [math]A\equiv\left(0,0\right)[/math] and [math]B\equiv\left(6,0\right)[/math], what is the equation of line [math]AC[/math]?
[math]\ell_{AC}:y=x[/math]
Considering what you have discovered by answering the previous questions, what can you say about:[br]- how does the equation of line [math]AC[/math] change when the square is translated [br]- how does the equation of line [math]AC[/math] change when the square is dilated/compressed (no rotation)?
If [math]A\equiv\left(0,0\right)[/math] and [math]B\equiv\left(8,0\right)[/math], what is the equation of line [math]HJ[/math]?
[math]\ell_{HJ}:y=x+4[/math]
Compare the equations of the lines [math]AC[/math] and [math]HJ[/math].[br]What can you say about their relative position?
Explain your previous answer.
The two lines are parallel because they have the same slope.
If [math]A\equiv\left(0,0\right)[/math] and [math]B\equiv\left(8,0\right)[/math], what is the equation of line [math]DB[/math]?
[math]\ell_{DB}:y=-x+8[/math]
Compare the equations of the lines [math]AC[/math] and [math]DB[/math].[br]What can you say about their relative position?
How can you decide if two lines are perpendicular, by comparing their equations?