Question

In what quadrant would the image of point A (2,-1) be located if it is rotated 135° clockwise around point (1, -3)? Enter the numeric

value of the quadrant

Answer 1

The new point A location is (1.69, -51267)

The quadrant is 4th quadrant

Explanation:

Here we have a point rotating about another point, therefore we have;

Let point (1, -3) be = P

Therefore the length of line AP is given by the following relation;

`AP =√((2-1)^2+(-1-(-3))^2) = √(5)`

With angle

`Actan ((-1 -(-3))/(2-1)) = Actan( 2) = 63^(\circ) \ due \ North \ of \ East`

Rotating through 135° clockwise gives;

63° - 135° = 72° Due south of east from P

The coordinates is then given as follows;

y coordinates = -3 - sin 72×√5 = -5.1267

The x coordinates will be 1 + cos 72×√5 = 1.69

The coordinate of the new point A location = (1.69, -51267)

Therefore the quadrant remains the 4th quadrant.

Answer 2

`A' = (1.707,-5.121)`

The new point belongs to the fourth quadrant.

Explanation:

The relative vector is:

`\vec r = (2-1,-1+3)`

`\vec r = (1, 2)`

The length of the vector is:

`\|\vec r\| = \sqrt{1^(2)+2^(2)}`

`\|\vec r\| = √(5)`

The angle with respect to the horizontal (+x direction) is:

`\theta = \tan^(-1) (2)/(1)`

`\theta \approx 63.435^(\textdegree)`

The new angle with respect to the horizontal (+x direction) is:

`\theta' = 63.435^(\textdegree)-135^(\textdegree)`

`\theta' = -71.565^(\textdegree)`

The new point is:

`A' = (1 + √(5)\cdot \cos (-71.565^(\textdegree)),-3 + √(5)\cdot \sin(-71.565^(\textdegree)))`

`A' = (1.707,-5.121)`

The new point belongs to the fourth quadrant.