Jake+B

Jake B Hour 3 Second Quarter Project

Triangle IJH is a scalene acute triangle. In Order to find the midpoints of segment IJ and IH I have to find the average x and y values so i used the formula X1 + X2 / 2, Y1 + Y2 / 2 I: (0, 6) J: (-2, -1) H: (5, 0) Midpoint of segment IJ X= 0 + -2 / 2 = -1, Y= 6 + -1 / 2 = 2.5 K= (-1, 2.5) Midpoint of segment IH X= 0 +5 / 2 = 2.5, Y= 6 + 0 / 2 = 3 L= (2.5, 3) In order to find the slope of a line I used the formula, Slope = (Y2 - Y1) / (X2 - X1) The slope of segment IJ is (-1 - 6) / (-2 - 0) -7 / -2 = 3.5 the y-intercept is 6 so the final slope is Y = 3.5X + 6 The slope of segment IH is (6 - 0) / (0 - 5) 6 / -5 = -1.2 the y-intercept is 6 so the final slope is Y = -1.2X + 6 This is how I found the Perpendicular bisectors to segment IJ and IH. For Segment IJ I first had to find the opposite reciprocal which is Y= -1/3.5 Then I needed to add the midpoint K to the equation in order to find the Y-intercept. This is the equation- 2.5 = -1/3.5(-1) + b 2.5 = .2857 + b -.2857 -.2857 2.2143 = b K: Y = -1/3.5 + 2.2143 First I find the opposite reciprocal of segment IH which is Y = 1/1.2X + b Next I add the midpoint L to the equation so that I can find the Y-intercept. 3 = 1/1.2(2.5) + b 3 = 2.083 + b -2.083 -2.083 0.917 = b Therefore the perpindicular bisector is Y = 1/1.2X + 0.917 The perpendicular bisector of segment IJ is known as K: Y = -0.29X + 2.21 The perpendicular bisector of segment IH is known as L: Y = 0.83X + 0.92 The point where these two lines meet is called the circumcenter. On the diagram below Slope k is acually L and Slope j is acually K.

To find the circumcenter I need to find the point where K and L meet. To do this I must solve for X -0.29X + 2.21 = 0.83X + 0.92 +0.29X +0.29X 2.21 = 1.12X + 0.92 -.92 -0.92 1.29 = 1.12X 1.29/1.12 1.12/1.12 1.15 = X Now I can plug the x into the Y equations to find the Y coordinate. Y = -0.29(1.15) + 2.21 Y = -0.33 + 2.21 Y = 1.88 So my coordinates for N is (1.15, 1.88) Next I need to find the distance between a vertice of my triangle and the circumcenter-N I will find the distance from I to N this distance will be the radius of the circle. To find this distance I have to use the distance formula. N (1.15,1.88) I (0,6) D = Square root of (6 - 1.88)^2 + (0 - 1.15)^2 D = Square root of 4.12^2 + -1.15^ D = Square root of 16.97 + -1.32 D = Square root of 15.65 Now I Can Construct a circle with N being the midpoint and the radius being the square root of 15.65. Geometer sketchpad shows this circle.