Question

thursday

1. find the coordinates of the image,

given $(x, y) \to (x - 1, y + 3)$

1. find the area (honors try to do

without a calculator):
$\frac{8}{5}$ yd
$\frac{3}{2}$ yd
3.find the coordinates of the image,
given a $90^\circ$ clockwise rotation.

1. find the coordinates after the

reflection over $y = -1$

Explanation:

Problem 1:

Step1: Identify original coordinates

Original points: $J(-3,1)$, $K(-4,-1)$, $L(2,-1)$, $M(-1,2)$

Step2: Apply transformation $(x,y)\to(x-1,y+3)$

• $J': (-3-1, 1+3)=(-4,4)$
• $K': (-4-1, -1+3)=(-5,2)$
• $L': (2-1, -1+3)=(1,2)$
• $M': (-1-1, 2+3)=(-2,5)$

Problem 2:

Step1: Recall triangle area formula

Area = $\frac{1}{2}\times \text{base}\times\text{height}$

Step2: Substitute given values

Base = $\frac{3}{2}$ yd, Height = $\frac{8}{5}$ yd
Area = $\frac{1}{2}\times\frac{3}{2}\times\frac{8}{5}$

Step3: Calculate the product

$\frac{1}{2}\times\frac{3}{2}\times\frac{8}{5}=\frac{3\times8}{2\times2\times5}=\frac{24}{20}=\frac{6}{5}$

Problem 3:

Step1: Identify original coordinates

Original points: $P(-3,-3)$, $Q(-2,0)$, $N(-3,2)$, $O(0,1)$

Step2: Apply 90° clockwise rotation rule $(x,y)\to(y,-x)$

• $P': (-3, 3)$
• $Q': (0, 2)$
• $N': (2, 3)$
• $O': (1, 0)$

Problem 4:

Step1: Identify original coordinates

Original points: $M(-2,-3)$, $I(-2,2)$, $N(0,3)$, $L(4,-2)$

Step2: Apply reflection over $y=-1$ rule: for a point $(x,y)$, new $y$-coordinate is $-1-(y-(-1))=-2-y$

• $M': (-2, -2-(-3))=(-2,1)$
• $I': (-2, -2-2)=(-2,-4)$
• $N': (0, -2-3)=(0,-5)$
• $L': (4, -2-(-2))=(4,0)$

Answer:

1. Image coordinates: $J'(-4,4)$, $K'(-5,2)$, $L'(1,2)$, $M'(-2,5)$
2. Area: $\frac{6}{5}$ square yards
3. Rotated coordinates: $P'(-3,3)$, $Q'(0,2)$, $N'(2,3)$, $O'(1,0)$
4. Reflected coordinates: $M'(-2,1)$, $I'(-2,-4)$, $N'(0,-5)$, $L'(4,0)$