How could the distance formula and slope be used to classify triangles and quadrilaterals in the coordinate plane?
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Answer:
See below
Step-by-step explanation:
I think we had a question similar to this before. Again, let's figure out the vertical and horizontal distances figured out. The distance from C at x=8 to D at x=-5 is 13 units while the distance from C at y=-2 to D at y=9 is 11 units. (8+5=13 and 2+9=11, even though some numbers are negative, we're looking at their value in those calculations)
Next, we have to divide each distance by 4 so we can apply it to the ratio. 13/4= and 11/4=
. Next, we need to read the question carefully. It's asking us to place the point in the ratio <em>3</em> to <em>1</em> from <em>C</em> to <em>D</em>. The point has to be closer to endpoint D because of this. Let's take each of our fractions, multiply them by 3, then add them towards the direction of endpoint D to get our answer (sorry if that sounds confusing):
Therefore, our point that partitions CD into a 3:1 ratio is ().
I'm not sure if there was more to #5 judging by how part B was cut off. From what I can understand of part B, however, I believe that Beatriz started from endpoint D and moved towards C, the wrong direction. She found the coordinates for a 1:3 ratio point.
Also, for #6, since a square is a 2-dimensional object, the answer needs to be written showing that. The answer for #6 is 9 units^2.