Solutions to In Class Exercises on Analytic Geometry

1. First let E be a point on Line BC such that C-B-E. Then the measure of angle ABE is ½ the measure of the arc AB or 50 degrees. This gives the measure of ABC as 130 degrees. Triangle ABC is isosceles by assumption. This says that angles BAC and ACB are both 25 degrees. Since angle BAC is 50 degrees, arc BD is 50 degrees. This gives us all the angles

angle ABD = 105

angle DBC = 25

angle ACB = 25

angle BAC = 25

angle ADB = 50

2. To find the Euler line we will 1) find the centroid, 2) find the circumcenter, and then 3) find the equation of the line through theses two points.

1. Let A = (0,0), B = (8,16) , and C = (10,2). The centroid is 2/3 of the distance from A along the median to side BC. The midpoint of BC is (9,9) = ((8+10)/2,(16+2)/2). The centroid is then given by .

2. The slope of BC is ((16-2)/(8-10))= -7. The slope of a line perpendicular to BC is thus 1/7. The midpoint of BC is (9,9). The perpendicular bisector of BC is thus or . The midpoint of AC is (5,1). The slope of AC is 1/5 so the slope of the line perpendicular to AC is –5. The equation of the perpendicular bisector of AC is (y-1)= -5(x-5) or y = -5x+26. The intersection of these two lines is which is the circumcenter.

3. The line through and (6,6) has slope . Thus the line is .

3. The two ways to show that the quadrilateral is a rectangle are1) the opposite sides are parallel and the adjacent sides are perpendicular and 2) the diagonals are of equal length and bisect each other.

1. The slopes of the sides are 5/7 and –7/5 so we have parallels and perpendiculars.

2. The lengths of the diagonals are each and the midpoint of each diagonal is (2,7). Since they have the same midpoints, they bisect each other.