Find the smallest value of k, such that 16k is a perfect cube.
1 answer:
You might be interested in
The answer is really none of the above, because all of these have divisions by difference in x, and there are perfectly fine rectangles where those will mean dividing by zero.
To do the problem the way they want us to, we note segments with the same slope are parallel (or collinear); segments with slopes which multiply to -1 are perpendicular.
Let's figure out what each choice is trying to tell us by translating it into parallel and perpendicular. Remember A is point 1, ...
A. ((y_4-y_3)/(x_4-x_3)= (y_3-y_2)/(x_3-x_2)) and ((y_4-y_3)/(x_4-x_3)×(y_3-y_2)/(x_3-x_2))=-1
translated, that's CD║BC and CD⊥BC - contradictory; second half right
B. (y_4-y_3)/(x_4-x_3)= (y_2-y_1)/(x_2-x_1) and ((y_4-y_3)/(x_4-x_3)×(y_2-y_1)/(x_2-x_1))=-1
CD ║ AB and CD ⊥ AB - only the parallelogram half is right
C. (y_4-y_3)/(x_4-x_3)= (y_2-y_1)/(x_2-x_1) and (y_4-y_3)/(x_4-x_3)×(y_3-y_2)/(x_3-x_2)=-1
CD║AB and CD⊥BC - that's TRUE
D. (y_4-y_3)/(x_4-x_3)= (y_3-y_1)/(x_3-x_1) and (y_4-y_3)/(x_4-x_3)×(y_2-y_1)/(x_2-x_1)=-1
CD║AC and CD⊥AB - nope
Answer: C
The way I'd get the proper expression is with the dot product.
A parallelogram is a rectangle when adjacent sides are perpendicular. It only takes one right angle to make them all right.
Two sides are perpendicular when the dot product of their direction vectors is zero. All the answers have so one side is CD. The other side must be BC or AD; AD would give
terms which don't appear among the answer, so let's go with BC.
The direction vectors are
For perpendicularity we need a zero dot product:
That's the correct expression, given it's a parallelogram, to demonstrate it's a rectangle. The divisions make it iffy.