Answer :
Answer:
[tex]A' = (1.707,-5.121)[/tex]
The new point belongs to the fourth quadrant.
Step-by-step explanation:
The relative vector is:
[tex]\vec r = (2-1,-1+3)[/tex]
[tex]\vec r = (1, 2)[/tex]
The length of the vector is:
[tex]\|\vec r\| = \sqrt{1^{2}+2^{2}}[/tex]
[tex]\|\vec r\| = \sqrt{5}[/tex]
The angle with respect to the horizontal (+x direction) is:
[tex]\theta = \tan^{-1} \frac{2}{1}[/tex]
[tex]\theta \approx 63.435^{\textdegree}[/tex]
The new angle with respect to the horizontal (+x direction) is:
[tex]\theta' = 63.435^{\textdegree}-135^{\textdegree}[/tex]
[tex]\theta' = -71.565^{\textdegree}[/tex]
The new point is:
[tex]A' = (1 + \sqrt{5}\cdot \cos (-71.565^{\textdegree}),-3 + \sqrt{5}\cdot \sin(-71.565^{\textdegree}))[/tex]
[tex]A' = (1.707,-5.121)[/tex]
The new point belongs to the fourth quadrant.
Answer:
The new point A location is (1.69, -51267)
The quadrant is 4th quadrant
Step-by-step 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;
[tex]AP =\sqrt{(2-1)^2+(-1-(-3))^2} = \sqrt{5}[/tex]
With angle
[tex]Actan (\frac{-1 -(-3)}{2-1}) = Actan( 2) = 63^{\circ} \ due \ North \ of \ East[/tex]
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.