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 find𝒀 Tℎe
!𝟏"𝑿 𝟐
Coordinates Of Point M.
𝟏 "𝒀 𝟐
Midpoint Formula: ( , )
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.
!𝟏"𝑿 𝟐 𝒀 𝟏 "𝒀 𝟐
Midpoint Formula: ( , )
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
S, K, And P. For Points
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
Line Lengtℎ
SegmentOf 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 find𝒀 Tℎe
!𝟏"𝑿 𝟐
Coordinates Of Point M.
𝟏 "𝒀 𝟐
Midpoint Formula: ( , )
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.
!𝟏"𝑿 𝟐 𝒀 𝟏 "𝒀 𝟐
Midpoint Formula: ( , )
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
S, K, And P. For Points
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
Line Lengtℎ
SegmentOf 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.
