C972
College Geometry
SDM1 TASK 2 – Passed
Analytic Geometry
Western Governors University
, SDM1 Task 2- Analy0c Geometry
A1. I first iden0fied that point M is the midpoint of line segment KN, according to the chart
that was given with the graph. Since I need to find the coordinates for point M and M is
the midpoint on line segment KN, I used the midpoint formula to find the coordinates of
point M.
! "𝑿 𝒀 "𝒀
Midpoint formula: ( 𝟏 𝟐 , 𝟏 𝟐 )
𝟐 𝟐
Using the line segment KN, I used the points of point K (1,1) and point N (2,1) to plug
these coordinates into the midpoint formula. (The coordinates of points K and N came
from the chart that was given with the graph.)
I then allocated 𝑋& , 𝑌& and 𝑋' , 𝑌' to the points in line segment KN.
𝐾 = (1,1)
(𝑋& , 𝑌& ) = (1,1)
𝑋& = 1
𝑌& = 1
𝑁 = (2,1)
(𝑋' , 𝑌' ) = (2,1)
𝑋' = 2
𝑌' = 1
Then, I plugged these numbers into the midpoint formula and simplified to find the
coordinates for point M.
! "𝑿 𝒀 "𝒀
Midpoint formula: ( 𝟏 𝟐 𝟐 , 𝟏 𝟐 )
𝟐
&"' &"&
𝑀=- '
, '
.
&"' ( &"& '
I evaluated - '
. = ', and evaluated - '
. = '.
( ' ( '
Now I have 𝑀 = -' , '. or -'. = 1.5, and -'. = 1, giving me the coordinates of M to be
𝑀 = (1.5,1).
, A2a. Demonstrate that ∆𝑆𝐾𝑃 is an isosceles right triangle.
To demonstrate that ∆𝑆𝐾𝑃 is an isosceles triangle, I first iden0fied the three-line
segments of ∆𝑆𝐾𝑃 to be SK, KP, and PS.
To find the length of these three-line segments, I used the
distance formula: 4(𝑿𝟐 − 𝑿𝟏 )𝟐 + (𝒀𝟐 − 𝒀𝟏 )𝟐
According to the chart that was given with the graph, I know the coordinates for points
S, K, and P.
𝑆 = (1,0) 𝐾 = (1,1) 𝑃 = (1.5, 0.5)
Star4ng with line segment SK:
𝑆 = (1,0)
(𝑋& , 𝑌& ) = (1,0)
𝑋& = 1
𝑌& = 0
𝐾 = (1,1)
(𝑋' , 𝑌' ) = (1,1)
𝑋' = 1
𝑌' = 1
Now I plug these numbers into the distance formula and simplify to find the length of
line segment SK.
distance formula: 4(𝑿𝟐 − 𝑿𝟏 )𝟐 + (𝒀𝟐 − 𝒀𝟏 )𝟐
𝑆𝐾 = 4(1 − 1)' + (1 − 0)'
I evaluated (1 − 1)' = (0)' , and evaluated (1 − 0)' = (1)' to now have
𝑆𝐾 = 4(0)' + (1)' .
Then evaluated (0)' = 0 , and (1)' = 1 to now have √0 + 1.
Now add (0 + 1) = 1 to have 𝑆𝐾 = √1 which equals 1.
𝑺𝑲 = 𝟏
The length of line segment 𝑆𝐾 = 1.
College Geometry
SDM1 TASK 2 – Passed
Analytic Geometry
Western Governors University
, SDM1 Task 2- Analy0c Geometry
A1. I first iden0fied that point M is the midpoint of line segment KN, according to the chart
that was given with the graph. Since I need to find the coordinates for point M and M is
the midpoint on line segment KN, I used the midpoint formula to find the coordinates of
point M.
! "𝑿 𝒀 "𝒀
Midpoint formula: ( 𝟏 𝟐 , 𝟏 𝟐 )
𝟐 𝟐
Using the line segment KN, I used the points of point K (1,1) and point N (2,1) to plug
these coordinates into the midpoint formula. (The coordinates of points K and N came
from the chart that was given with the graph.)
I then allocated 𝑋& , 𝑌& and 𝑋' , 𝑌' to the points in line segment KN.
𝐾 = (1,1)
(𝑋& , 𝑌& ) = (1,1)
𝑋& = 1
𝑌& = 1
𝑁 = (2,1)
(𝑋' , 𝑌' ) = (2,1)
𝑋' = 2
𝑌' = 1
Then, I plugged these numbers into the midpoint formula and simplified to find the
coordinates for point M.
! "𝑿 𝒀 "𝒀
Midpoint formula: ( 𝟏 𝟐 𝟐 , 𝟏 𝟐 )
𝟐
&"' &"&
𝑀=- '
, '
.
&"' ( &"& '
I evaluated - '
. = ', and evaluated - '
. = '.
( ' ( '
Now I have 𝑀 = -' , '. or -'. = 1.5, and -'. = 1, giving me the coordinates of M to be
𝑀 = (1.5,1).
, A2a. Demonstrate that ∆𝑆𝐾𝑃 is an isosceles right triangle.
To demonstrate that ∆𝑆𝐾𝑃 is an isosceles triangle, I first iden0fied the three-line
segments of ∆𝑆𝐾𝑃 to be SK, KP, and PS.
To find the length of these three-line segments, I used the
distance formula: 4(𝑿𝟐 − 𝑿𝟏 )𝟐 + (𝒀𝟐 − 𝒀𝟏 )𝟐
According to the chart that was given with the graph, I know the coordinates for points
S, K, and P.
𝑆 = (1,0) 𝐾 = (1,1) 𝑃 = (1.5, 0.5)
Star4ng with line segment SK:
𝑆 = (1,0)
(𝑋& , 𝑌& ) = (1,0)
𝑋& = 1
𝑌& = 0
𝐾 = (1,1)
(𝑋' , 𝑌' ) = (1,1)
𝑋' = 1
𝑌' = 1
Now I plug these numbers into the distance formula and simplify to find the length of
line segment SK.
distance formula: 4(𝑿𝟐 − 𝑿𝟏 )𝟐 + (𝒀𝟐 − 𝒀𝟏 )𝟐
𝑆𝐾 = 4(1 − 1)' + (1 − 0)'
I evaluated (1 − 1)' = (0)' , and evaluated (1 − 0)' = (1)' to now have
𝑆𝐾 = 4(0)' + (1)' .
Then evaluated (0)' = 0 , and (1)' = 1 to now have √0 + 1.
Now add (0 + 1) = 1 to have 𝑆𝐾 = √1 which equals 1.
𝑺𝑲 = 𝟏
The length of line segment 𝑆𝐾 = 1.
