Chapter 11.5, Problem 16IP

a.

To determine

To write the coordinate of the vertex after each rotation:

(1) ${90}^0$ clockwise about the origin

(2) ${180}^0$ about the origin.

(3) ${270}^0$ clockwise about the origin

Answer to Problem 16IP

(1) $(x,y)\to (y,-x)$

(2) $(x,y)\to (-x,-y)$

(3) $(x,y)\to (-y,x)$

Explanation of Solution

Given:

A figure with one of its vertices as $(x,y)$

Calculation:

1. ${90}^0$ clockwise about the origin:

When rotating vertex clockwise by ${90}^0$ the $x$ and $y$ coordinates interchanges and a negative sign is added to the $x$ coordinate.

  $(x,y)\to (y,-x)$

So, the coordinates of the vertex will become $(y,-x).$

2. ${180}^0$ about the origin:

When rotating vertex clockwise by 180° the x and $y$ coordinate changes to $-x\text{and}-y$ .

  $(x,y)\to (-x,-y)$

So, the coordinates of the vertex will become $(-x,-y).$

3. ${270}^0$clockwise about the origin:

When rotating vertex clockwise by ${270}^0$ the $x$ and $y$ coordinate interchanges and a negative sign is added to the $y$ coordinate.

  $(x,y)\to (-y,x)$

So, the coordinates of the vertex will become $(-y,x).$

Conclusion:

Therefore,

When vertex is rotated by ${90}^0$ (clockwise) about the origin, the coordinates becomes $(y,-x)$ .

When vertex is rotated by ${180}^0$ about the origin, the coordinates becomes $(-x,-y)$ .

When vertex is rotated by ${270}^0$ (clockwise) about the origin, the coordinates becomes $(-y,x)$ .

b.

To determine

To determine the Quadrant in which the vertex of the image will be located after the original figure is rotated ${180}^0$ about the origin.

Answer to Problem 16IP

Quadrant IV

Explanation of Solution

Given:

A vertex in Quadrant II.

Let the vertex be $P(-x,y)$ in Quadrant II.

Now on rotating the figure by ${180}^0$ about the origin, the vertex becomes $P(x,-y)$ .

Because when rotating clockwise by ${180}^0$ the $x$ and $y$ coordinate changes to $-x$ and $-y$ .

  $\therefore (x,y)\to (-x,-y)$

Now, vertex with coordinates $(x,-y)$ are found in Quadrant IV.

So, $P(x,-y)$ will be located in Quadrant IV.

Conclusion:

Therefore, the corresponding vertex $P(x,-y)$ is located in Quadrant IV.