McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
4. Indirect Proof
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Exercise 51 Page 444

To find the circumcenter, find the intersection of two perpendicular bisectors of two sides of the triangle.

(1 25,2 35)

Practice makes perfect

Let's start by graphing the triangle using the given coordinates.

To find the circumcenter, we will first find the equations of perpendicular bisectors. Finally, to find the circumcenter, we will calculate the intersection of the bisectors. Let's do it one at a time.

Finding Equations of Perpendicular Bisectors

To find the equations of perpendicular bisectors, we first need to find the midpoints of at least two arbitrary sides of the triangle â–ł ABC. In our case, let's choose sides AB and BC.

After calculating the midpoints, we will find the slopes of the lines containing the sides and the slopes of perpendicular lines. Then, we will use the slopes and the midpoints to find the equations of the perpendicular bisectors.

Finding Midpoints

Let's find the midpoints of two sides of the triangle, AB and BC. To do so, we can use the Midpoint Formula.
Side Points M(x_1+x_2/2,y_1+y_2/2) Midpoint
AB ( 4,0), ( -2,4) U(4+( -2)/2,0+ 4/2) U(1,2)
BC ( -2,4), ( 0,6) V(-2+ 0/2,4+ 6/2) V(-1,5)

We can now add these midpoints to our graph.

Slopes of the Sides

Let's find the slopes of the lines that contain the segments AB and BC, respectively. The slope of a line can be calculated using the Slope Formula. To do so, we will use the coordinates of the endpoints of the given sides.

Side Points m=y_2-y_1/x_2-x_1 Slope
AB ( 4,0), ( -2,4) m_1 = 4- 0/-2- 4 m_1 = -2/3
BC ( -2,4), ( 0,6) m_2 = 6- 4/0-( -2) m_2=1

Slopes of the Perpendicular Bisectors

Now, we can find the slopes of lines perpendicular to the segment AB and to the segment BC, respectively. By the Slopes of Perpendicular Lines Theorem, we know that non-vertical lines are perpendicular if and only if the product of their slopes is -1. s_1* s_2 = -1 We can use the slopes of the sides that we found to calculate the slopes of lines perpendicular to the them.

Side Slope s_1* s_2 = -1 Perpendicular Line's Slope
AB m_1 = -1/6 -2/3* a_1 = -1 a_1 = 3/2
BC m_2 = 3 1* a_2 = -1 a_2 = -1

Equations

We have already found that the slopes of the bisectors are 32 and -1. We will now find the equations of the bisectors using midpoints U(1,2) and V(-1,5). To do so, we can apply the point-slope form of the equation of a line. An equation in point-slope form follows a specific format. y- y_1= m(x- x_1) In this format, m represents the slope and the point ( x_1, y_1) is the point that the line passes through. To find the equations, we will use a table.

Midpoint Slope Point-slope Form Equation
U( 1, 2) a_1 = 3/2 y- 2= 3/2(x- 1) y = 3/2x+1/2
V( -1, 5) a_2 = -1 y- 5= -1(x-( -1)) y = - x+4

Let's now add both bisectors to our graph.

Finding the Circumcenter

The triangle's circumcenter is the point at which the perpendicular bisectors intersect. We can find this point of intersection by representing the equations of the bisectors as a system of equations. y= 32x+ 12 y=- x+4 We can solve this system of equations using using the Substitution Method. Note that y is already isolated in both equations. We can substitute the value of y from the first equation into the second equation.
y= 32x+ 12 y=- x+4
y= 32x+ 12 32x+ 12=- x+4
â–Ľ
(II): Simplify
y= 32x+ 12 32x+x+ 12=4
y= 32x+ 12 3x2+x+ 12=4
y= 32x+ 12 3x2+ 2x2+ 12=4
y= 32x+ 12 5x2+ 12=4
y= 32x+ 12 5x2=4- 12
y= 32x+ 12 5x2= 82- 12
y= 32x+ 12 5x2= 72
y= 32x+ 12 52x= 72
y= 32x+ 12 x= 72( 25)
y= 32x+ 12 x= 1410
y= 32x+ 12 x= 75
Finally, let's substitute x= 75 into the first equation.
y= 32x+ 12 x= 75
y= 32( 75)+ 12 x= 75
â–Ľ
(I): Simplify
y= 2110+ 12 x= 75
y= 2110+ 510 x= 75
y= 2610 x= 75
y= 135 x= 75

(I), (II): Write fraction as a mixed number

y=2 35 x=1 25
We found that the solution to the system of equations is the point (1 25, 2 35). Therefore, the circumcenter is located at (1 25,2 35).