Is the square root of 9 a real number
1 answer:
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V1=(-2,4)=(x1,y1)→x1=-2, y1=4
V2=(4,0)=(x2,y2)→x2=4, y2=0
V3=(2,-3)=(x3,y3)→x3=2, y3=-3
V4=(x4,y4)→x4=?, y4=?
V1-V2
dx=x2-x1=4-(-2)=4+2→dx=6
dy=y2-y1=0-4→dy=-4
V4-V3
dx=x3-x4→6=2-x4
Solving for x4:
6=2-x4→6-2=2-x4-2→4=-x4→(-1)(4=-x4)→-4=x4→x4=-4
dy=y3-y4→-4=-3-y4
Solving for y4:
-4=-3-y4→-4+3=-3-y4+3→-1=-y4→(-1)(-1=-y4)→1=y4→y4=1
V4=(x4, y4)→V4=(-4, 1)
Answer: The coordinates of the fourth vertex are (-4,1)
V2=(4,0)=(x2,y2)→x2=4, y2=0
V3=(2,-3)=(x3,y3)→x3=2, y3=-3
V4=(x4,y4)→x4=?, y4=?
V1-V2
dx=x2-x1=4-(-2)=4+2→dx=6
dy=y2-y1=0-4→dy=-4
V4-V3
dx=x3-x4→6=2-x4
Solving for x4:
6=2-x4→6-2=2-x4-2→4=-x4→(-1)(4=-x4)→-4=x4→x4=-4
dy=y3-y4→-4=-3-y4
Solving for y4:
-4=-3-y4→-4+3=-3-y4+3→-1=-y4→(-1)(-1=-y4)→1=y4→y4=1
V4=(x4, y4)→V4=(-4, 1)
Answer: The coordinates of the fourth vertex are (-4,1)