Question

Drag the point A to the location indicated in each scenario to complete each statement.

Answer 1

The graph from which the position of the point A can determined following

the multiplication with a scalar is attached.

Responses:

• If A is in quadrant I and is multiplied by a negative scalar, c, then c·A is in quadrant III

• If A is in quadrant II and is multiplied by a positive scalar, c, then c·A is in quadrant II

• If A is in quadrant II and is multiplied by a negative scalar, c, then c·A is in quadrant IV

• If A is in quadrant III and is multiplied by a negative scalar, c, then c·A is in quadrant I

Methods by which the above responses are obtained

Background information;

The question relates to the coordinate system with the abscissa represent the real number and the ordinate representing the imaginary number.

Solution:

If A is in quadrant I; A = a + b·i

When multiplied by a negative scalar, c, we get;

c·A = c·a + c·b·i

Therefore;

c·a is negative

c·b is negative

• c·A = c·a + c·b·i is in the quadrant III (third quadrant)

If A is quadrant II, we have;

A = -a + b·i

When multiplied by a positive scalar c, we have;

c·A = c·(-a) + c·b·i = -c·a + c·b·i

-c·a is negative

c·b·i is positive

Therefore;

• c·A = -c·a + c·b·i is in quadrant II

Multiplying A by negative scalar if A is in quadrant II, we have;

c·A = -c·a + c·b·i

-c·a is positive

c·b·i is negative

Therefore;

c·A = -c·a + c·b·i is in quadrant IV

If A is in quadrant III, we have;

A = a + b·i

a is negative

b is negative

Multiplying A with a negative scalar c gives;

c·A = c·a + c·b·i

c·a is positive

c·b  is positive

Therefore;

• c·A = c·a + c·b·i is in quadrant I

Learn more about real and imaginary numbers here;

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