Find the 4th Vertex of a Parallelogram (Grade 10 Math)

Course Site - Grade 10 Academic Math (MPM2D)

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key words: FIN300, FIN 300, FIN401, FIN 401, QMS 102, QMS 101, QMS10, ADMS 3530, ADMS3530, ADMS 4501, ADMS 4502, RYERSON UNIVERSITY, YORK UNIVERSITY, QUEENS UNIVERSITY, COMM 121, COMM121, COMM122, COMM 122, MAT133, MAT 133, MCV4U, MHF4U, MPM2D, MPM1D, MAT 134, MAT 135, calculus and linear algebra, MISSISSAUGA, TORONTO, calculus, advanced functions, grade 12, grade 11, high school, COMM 298, UBC, ACC 100, Ryerson, AMF 102, Waterloo university, STATS 1024, CALC 1000, Western University

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[veautiiiiifuuuulll]

In the parallelogram ABCD,
AD=BC
AB=CD
B(8, 1) C(11, 5)
Taking coordinates of B as (x1, y1)
and C as (x2, y2)
Difference between the x coordinates = x2 - x1
= 11-8
=3
Difference between the y coordinates = y2 - y1
=5-1
= 4
A = (2, 3)
D= (2+3, 3+4)
D=( 5, 7)

[thedivinedilemmacreationwo]

Thank you sir. I understood clearly ! Thank you 😊

[z.m]

Can i use the same method for a square???

[ntsikiezasawe]

"Better simplify now so initially you'll have less simplifying to do later" GIRL MATH

[KTKMaths]

Guess we can find this with midpoint way, easier than this! As we know, diagonals of parallelogram are bisecting each other. So, Midpoint of AC = Midpoint of BD method, we can solve quickly and get the solution sir.

[casketpaint]

Why was the 1 negative ? Wasnt the 6 supposed to stay as negative ?

[kandgray]

Why is it so complicated. Im having a mental breakdown

[Katthelego]

Great job sir but your method was way too long I got the answer in the first 2 minutes