Rachel+and+Shannon

Rachel S and Shannon G Hour 7 Second Quarter Project

1.We created an acute scalene triangle choosing random points on the graph. 2. We created midpoints for segment FE to create point H, and segment DE to create point G. 3.Then we calculated the slopes of segment FE and segment DE, then we created a perpendicular bisector for both points G and H, where those intersected, we plotted point I.  4. To create the systems of equations, we created equations from the graph and equaled the equations, solved for x. Then we put the answer in the first equation to find y.  5.We used the distance formula, D= (X1-X2)^2 + (Y1-Y2)^2, to calculate the distance between the circumcenter and the three points of the triangle 6. To create the circumcenter, we constructed a circle by using point I, and confirming that the circle intersected with the three plotted points of the original triangle

D(-3.00,-1.00)
E(3.00,3.00) F(-6.00,6.00)

**__2. Midpoints:__** F (-6.00, 6.00) __E +(3.00, 3.00)__ (-3.00, 9.00) / 2= **H (-1.50, 4.50)**

D (3.00, -1.00) __E +(3.00, 3.00)__ (6.00, 2.00) / 2= **G (3.00, 2.00)**

**__3. Slopes:__** D (-3.00, -1.00) __E - (3.00,3.00)__ (-6.00, -4.00)= **__-4.00/-6.00)__= .67**

E (3.00,3.00) __F – (-6.00, 6.00)__ (9.00, -3.00) = **__-3.00/9.00__ = -.33**

3x+9 = -1.5x+1 +1.5x +1.5x 4.5x+9 = 1 +-9 +-9 __4.5x__ = __-8__ 4.5 4.5 x = -1.77 = __-16/9__  3(-16/9)+9 = y y = 3.66 = __11/3__
 * __﻿4. Systems of Equations __**

__ ** 5. Distance Formula ** __ ** Point F ** D= (-6- (-1.78))^ 2 + (6-3.67)^2 (-6+ 1.78)^2 + (6-3.67)^2 (-4.22)^2 + (2.33)^2  __ 17.8084 + 5.4289 __

23.2373 = 4.82 = 2.19

** Point E ** D= (3- (-1.78))^2+ (3-3.67)^2 (3+ 1.78)^2 + (3-3.67)^2 (4.78)^2 + (-.67)^2  __ 22.9484 + .4489 __  23.2973 = 4.82 = 2.19

** Point D ** D= (-3- (-1.78))^2+ (-1-3.67)^2 (-3+ 1.787)^2 + (-1-3.67)^2 (-1.22)^2 + (-4.67)^2  __ 1.4884 + 21.8089 __  23.2973 = 4.82 = 2.19