Please help I need it please please please
2 answers:
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<span>A(2, -2), B(4, -2), C(1, -3), D(5, -3).
A'(1, 2), B'(1, 4), C'(2, 1), D'(2, 5)
That looks like a rotation 90 degrees around the origin then a translation.
Let's do the rotation ABCD->A"B"C"D"
The rotation is (x,y)->(-y,x)
</span>A"(2, 2), B"(2,4), C"(3,1), D"(3,5).
Compare to
A'(1, 2), B'(1, 4), C'(2, 1), D'(2, 5)
we need to translate one unit to the left.
Answer: 90 degree rotation around the origin then translation by (-1,0)
A'(1, 2), B'(1, 4), C'(2, 1), D'(2, 5)
That looks like a rotation 90 degrees around the origin then a translation.
Let's do the rotation ABCD->A"B"C"D"
The rotation is (x,y)->(-y,x)
</span>A"(2, 2), B"(2,4), C"(3,1), D"(3,5).
Compare to
A'(1, 2), B'(1, 4), C'(2, 1), D'(2, 5)
we need to translate one unit to the left.
Answer: 90 degree rotation around the origin then translation by (-1,0)
One way to do it is with calculus. The distance between any point 
on the line to the origin is given by
Now, both
and 
attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.
Solving for
, you find a critical point of 
.
Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.
You have
so indeed, a minimum occurs at.
The minimum distance is then
Now, both
Solving for
Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.
You have
so indeed, a minimum occurs at
The minimum distance is then