WGU C992 Analytic Geometry Graph and Permitted
Analytic Methods Task 2 |Latest Update with
Complete Solution
Kelsi Offerman
College Geometry – C992
April 1, 2025
Task 2: Analytic Geometry Graph and Permitted Analytic Methods
1. Compute..the..coordinates..for..point..M,..showing..and..explaining..all..work.
In..the..diagram..(shown..at..the..bottom..of..this..document),..we..are..told..that..M..is..the..midpoint..of..KN...Point..K..is..located
at..(1,1)..and..Point..N..is..at..(2,1)...Since..M..is..the..midpoint,..we..can..use.the..midpoint..formula,. . (𝑥1+𝑥2. . ,𝑦1+𝑦2 )...I..use..Point
2 2
K..as..(x1,..y1)..and..Point..N..as..(x2,y2).
1..+..2. 1..+..
.( , 1
2 )
3 2
( .., 2
2 )
2
(1.5,1)
Therefore,..using..the..midpoint..formula,..Point..M..is..located..at..(1.5,1).
2. Demonstrate..that..∆𝑺𝑲𝑷.. is..an..isosceles..right..triangle..by..doing..the..following:
a. Demonstrate..that..the..triangle..is..isosceles,..showing..and..explaining..all..work.
An..isosceles..triangle..has..two..equal..sides...I..will..use..the..distance..formula..to..prove..that..two..sides..are..equal..in
length...The..distance..formula..is..√(𝑥2.. −..𝑥1)2.. +..(𝑦2.. −..𝑦1)2...Point..S..is..located..at..(1,0),..Point..K..is..located..at..(1,1),..and..
Point..P..is..located..at..(1.5,0.5).
The..distance..of..side..KP..is..√(1.5..−..1)2.. +..(0.5..−..1)2 The..distance..of..side..SP..is..√(1.5..−..1)2.. +..(0.5..−..0)2
√(0.5)2..+..(−0.5)2 √(0.5)2..+..(0.5)2
√0.25..+..0.25 √0.25..+..0.25
√0.5.. ≈.. 0.70 √0.5.. ≈.. 0.70
The..distances..of..sides..KP..and..SP..are..both..√0.5..≈.. 0.70..meaning..they..are.equal..in..length...Since..these..are.equal..in..len
gth,..the..triangle..is..isosceles.
b. Demonstrate..that..the..triangle..is..a..right..triangle,..showing..and..explaining..all..work.
Analytic Methods Task 2 |Latest Update with
Complete Solution
Kelsi Offerman
College Geometry – C992
April 1, 2025
Task 2: Analytic Geometry Graph and Permitted Analytic Methods
1. Compute..the..coordinates..for..point..M,..showing..and..explaining..all..work.
In..the..diagram..(shown..at..the..bottom..of..this..document),..we..are..told..that..M..is..the..midpoint..of..KN...Point..K..is..located
at..(1,1)..and..Point..N..is..at..(2,1)...Since..M..is..the..midpoint,..we..can..use.the..midpoint..formula,. . (𝑥1+𝑥2. . ,𝑦1+𝑦2 )...I..use..Point
2 2
K..as..(x1,..y1)..and..Point..N..as..(x2,y2).
1..+..2. 1..+..
.( , 1
2 )
3 2
( .., 2
2 )
2
(1.5,1)
Therefore,..using..the..midpoint..formula,..Point..M..is..located..at..(1.5,1).
2. Demonstrate..that..∆𝑺𝑲𝑷.. is..an..isosceles..right..triangle..by..doing..the..following:
a. Demonstrate..that..the..triangle..is..isosceles,..showing..and..explaining..all..work.
An..isosceles..triangle..has..two..equal..sides...I..will..use..the..distance..formula..to..prove..that..two..sides..are..equal..in
length...The..distance..formula..is..√(𝑥2.. −..𝑥1)2.. +..(𝑦2.. −..𝑦1)2...Point..S..is..located..at..(1,0),..Point..K..is..located..at..(1,1),..and..
Point..P..is..located..at..(1.5,0.5).
The..distance..of..side..KP..is..√(1.5..−..1)2.. +..(0.5..−..1)2 The..distance..of..side..SP..is..√(1.5..−..1)2.. +..(0.5..−..0)2
√(0.5)2..+..(−0.5)2 √(0.5)2..+..(0.5)2
√0.25..+..0.25 √0.25..+..0.25
√0.5.. ≈.. 0.70 √0.5.. ≈.. 0.70
The..distances..of..sides..KP..and..SP..are..both..√0.5..≈.. 0.70..meaning..they..are.equal..in..length...Since..these..are.equal..in..len
gth,..the..triangle..is..isosceles.
b. Demonstrate..that..the..triangle..is..a..right..triangle,..showing..and..explaining..all..work.
