Question

Name:
Chapter 5 Section 1
More Practice Finding the Coordinates of the point of Concurrency
1. The vertices of AABC are A(1,6), B(5,4), C(5,-2). Find the coordinates of the circumcenter.
a) Graph and label the triangle
b) Find the midpoint of each side of the triangle
Midpoint AB = 5, 1)
Midpointbc =
し32)
Midpointac = [1, 2]
c) Find the slope of each side of the triangle
MAB =
MBC =
MAC =
d) Find the slope of each perpendicular bisector
IMAB =
IMBc =
IMAC =
e) Use the midpoint and the perpendicular slope to accurately draw each perpendicu
bisector on the triangle.

Answer 1

The triangle is formed the points on a x–y plane and the perpendicular

bisectors are given from the slope and midpoints of the lines.

Responses:

a) The required labelled graph is attached

b) Midpoint of AB = (3, 5)

Midpoint of BC = (5, 1)

Midpoint of AC = (3, 4)

c) $m_{AB} = -\dfrac{1}{2}$

$m_{BC} = \infty$

$m_{AC} = -2$

$d) \hspace{0.15 cm}Im_{AB} =2$

$Im_{BC} = 0$

$Im_{AC} = 0.5$

e) The graph of the ΔABC showing the perpendicular bisectors of the sides AB, BC, and AC, created with MS Excel is attached

Which method can be used to solve the given questions?

The given vertices of the triangle, ΔABC = A(1, 6), B(5, 4), C(5, -2)

(a) Please find attached the graph showing the labeled triangle created with MS Excel.

(b) The midpoints are given as follows;

$Midpoint \ of \ AB = \mathbf{ \left(\dfrac{1 + 5}{2} , \, \dfrac{6 + 4}{2} \right)} = \underline{(3, \, 5)}$

$Midpoint \ of \ BC = \mathbf{ \left(\dfrac{5 + 5}{2} , \, \dfrac{-2 + 4}{2} \right)} = \underline{(5, \, 1)}$

$Midpoint \ of \ AC = \left(\dfrac{1 + 5}{2} , \, \dfrac{6 - 2}{2} \right) =\underline{ (3, \, 4)}$

(c) The slopes are;

$\mathbf{ m_{AB}} = \dfrac{6 - 4}{1 - 5} = \underline{-\dfrac{1}{2}}$

$\mathbf{ m_{BC}}= \dfrac{4 - (-2)}{5 - 5} =\underline{ \infty}$

$\mathbf{ m_{AC}}= \dfrac{6 - (-2)}{1 - 5} = -\dfrac{8}{4} = \underline{-2}x^{2}$

d) The slope of a line perpendicular to another line having slope m is $-\dfrac{1}{m}$, which gives;

$\mathbf{ Im_{AB}}= -\dfrac{1}{ -\dfrac{1}{2}} =\underline{2}$

$\mathbf{ Im_{BC} }= -\dfrac{1}{ \infty} = \underline{0}$

$\mathbf{ Im_{AC}}= -\dfrac{1}{-2} = \dfrac{1}{2} = \underline{0.5}$

e) Using the midpoint and the perpendicular slope, we have;

The equation of the perpendicular line to AB is; y - 5 = 2·(x - 3)

Which gives;

y = 2·x - 6 + 5 = 2·x - 1

At x = 0, y = -1

At x = 4, y = 7

The points on the perpendicular bisector to AB are;

(0, -1) and (4, 7)

The equation of the perpendicular line to BC is; y - 1 = 0·(x - 5) = 1

Which gives;

y = 1

The points on the perpendicular bisector to BC are;

(0, 1) and (8, 1)

The equation of the perpendicular line to AC is; y - 4 = 0.5·(x - 3)

Which gives;

y = 0.5·x - 1.5 + 4 = 0.5·x + 2.5

At x = 0, y = 2.5

At x = 6, y = 5.5

The points on the perpendicular bisector to AC are;

(0, 2.5) and (6, 5.5)

• Please find attached the drawing of the ΔABC showing the perpendicular bisectors created using the above points in MS Excel.

Learn more about plotting points on a graph here:

https://brainly.com/question/14321394

Answer 2

Answer:

  see the attachments for the graph, and a spreadsheet with midpoints and slopes

Step-by-step explanation:

We are given the coordinates of the vertices of a triangle, and asked to find the parameters of the perpendicular bisectors of the sides of the triangle. The perpendicular bisectors are to be plotted on the graph.

Given:

Coordinates A(1, 6), B(5, 4), C(5, -2)

__

Find:

a) Graph and label the triangle

b) Find the midpoint of each side of the triangle

c) Find the slope of each side of the triangle

d) Find the slope of each perpendicular bisector

e) Use the midpoint and the perpendicular slope to accurately draw each perpendicular bisector on the triangle

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Solution:

a)

See the attached graph for shaded triangle ABC.

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b)

The midpoint (M) of a segment AB will be ...

  M = (A+B)/2

For example, the midpoint of segment AB is ...

  D = ((1, 6) +(5, 4))/2 = (1+5, 6+4)/2 = (6, 10)/2 = (3, 5)

This repetitive arithmetic is carried out in the spreadsheet shown in the second attachment. The midpoints are ...

  D(3, 5) is midpoint of AB

  E(5, 1) is midpoint of BC

  F(3, 2) is midpoint of CA

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c)

The slope of a segment is found using the slope formula (or by counting grid squares). That formula is ...

  m = (y2 -y1)/(x2 -x1)

For segment AB, this is ...

  mAB = (4 -6)/(5 -1) = -2/4 = -1/2

The other slopes are calculated similarly in the spreadsheet. When the denominator is zero (a vertical line), the slope is "undefined."

  mBC = undefined

  mCA = -2

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d)

The slope of the perpendicular line is the opposite reciprocal of the slope of the segment.

  m⟂AB = -1/(-1/2) = 2

  m⟂BC = 0 . . . . . a horizontal line has 0 slope

  m⟂CA = -1/-2 = 1/2

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e)

The perpendicular bisectors of each of the sides of the triangle are shown in the first attachment. As the lesson title indicates, their point of concurrency is G(1, 1), the circumcenter of the triangle.