ABC has vertices A(2,3), B(4,6), C(7,4) Describe a sequence of two transformations that would move ABC to Quadrant III.

Some examples

Step-by-step explanation:

1.The vertices of triangle ABC are A(2,-3), B(-3,-5), and C(4,1). For each translation, give the vertices of triangle A'B'C'. When the transformation (-2,3). When the T (-4,-1) and T

C

2.Triangle ABC has vertices A(-3,0) B(0,6) C(4,6). Find the equations of the three medians of triangle ABC

Triangle ABC has vertices A(-3,0) B(0,6) C(4,6). Find the equations of the three altitudes of the same triangle

Triangle ABC has vertices A(-3,0) B(0,6) C(4,6). Find the equations of the three perpendicular bisectors of the same triangle.

A(−3, 0), B(0, 6), and C(4, 6)

using the midpoint formula:  we can find all of their midpoint which is necessary to complete question 1 and 2

Mid AB=(-1.5, 3), Mid BC=(2, 6), Mid AC=(0.5,3)

Next, we have to find the slope of from the vertex and the midpoint of each side. We use the slope formula: .

You would get -6, 6/5, and 6/11.

Next you substitute in the coordinates for the equation of each for example: y=-6x, you switch in the set of cords AB which is (-1.5, 3). 3=-6(-1.5)+z and find that z=6, so the equation is y=-6x+6. As for the rest, repeat the same process until you find all 3 cords. You should get an answer of

For question 2,

The line has to be perpendicular bisector to the midpoint, signifying that it has to split 2 angles into congruent angles and a side into half. In order to do that we need to first find the opposite reciprocal of each side's slope. Which means we need to once again use the slope formula and get the opposite reciprocal by . For all the opposite reciprocal slopes you get will be incorporated into the final formulas. That gives us the slopes of -1/2, 0 and -7/6.

I hope this helps! :)

the function makes a strait line horizontally with no x intercept but a y intercept at 9.

what? that isnt a question

step-by-step explanation: