Question

1. rhombus ghij with vertices g(-1, -3), h(3, -4), i(4, -8), and j(0, -7); (a) translation: (x, y) → (x + 4, y + 10) (b) rotation: 90° counterclockwise g(__, ) h(, ) i(, ) j(, __)

Explanation:

Part (a) Translation

Step1: Translate \( G(-1, -3) \)

Use the translation rule \( (x, y) \to (x + 4, y + 10) \). For \( G(-1, -3) \), \( x=-1,y=-3 \). So \( x + 4=-1 + 4 = 3 \), \( y + 10=-3 + 10 = 7 \). Thus \( G'(3,7) \).

Step2: Translate \( H(3, -4) \)

For \( H(3, -4) \), \( x = 3,y=-4 \). \( x + 4=3 + 4 = 7 \), \( y + 10=-4 + 10 = 6 \). So \( H'(7,6) \).

Step3: Translate \( I(4, -8) \)

For \( I(4, -8) \), \( x = 4,y=-8 \). \( x + 4=4 + 4 = 8 \), \( y + 10=-8 + 10 = 2 \). So \( I'(8,2) \).

Step4: Translate \( J(0, -7) \)

For \( J(0, -7) \), \( x = 0,y=-7 \). \( x + 4=0 + 4 = 4 \), \( y + 10=-7 + 10 = 3 \). So \( J'(4,3) \).

Part (b) Rotation (90° counterclockwise)

The rule for 90° counterclockwise rotation is \( (x, y) \to (-y, x) \). We first use the original vertices (we can also use the translated ones, but here we use original for clarity, if we use translated, we apply rotation on translated points). Let's use original vertices \( G(-1,-3),H(3,-4),I(4,-8),J(0,-7) \):

Step1: Rotate \( G(-1, -3) \)

Using \( (x,y)\to(-y,x) \), \( x=-1,y=-3 \). So \( -y = -(-3)=3 \), \( x=-1 \). Thus \( G'(-3,-1) \)? Wait, no, wait: Wait, 90° counterclockwise rotation rule is \( (x,y)\to(-y,x) \). Wait, let's recheck: For a point \( (x,y) \), 90° counterclockwise rotation around origin is \( (x,y)\to(-y,x) \). So for \( G(-1,-3) \): \( x=-1,y=-3 \). Then \( -y=-(-3)=3 \), \( x=-1 \)? No, wait, no: Wait, \( (x,y) \) rotated 90° counterclockwise is \( (-y,x) \). So \( x=-1,y=-3 \): \( -y = -(-3)=3 \), \( x=-1 \)? Wait, no, that's wrong. Wait, let's take a simple point, say \( (1,0) \) rotated 90° counterclockwise is \( (0,1) \). Using the rule \( (x,y)\to(-y,x) \): \( x=1,y=0 \), \( -y = 0 \), \( x = 1 \)? No, that's \( (0,1) \) is \( (-y,x) \) where \( x=1,y=0 \): \( -y = 0 \), \( x = 1 \)? No, \( (0,1) \) is \( ( - 0,1)=(0,1) \). Wait, another example: \( (0,1) \) rotated 90° counterclockwise is \( (-1,0) \). Using rule \( (x,y)\to(-y,x) \): \( x=0,y=1 \), \( -y=-1 \), \( x=0 \), so \( (-1,0) \), which is correct. So the rule is correct: \( (x,y)\to(-y,x) \).

So for \( G(-1,-3) \): \( x=-1,y=-3 \). Then \( -y=-(-3)=3 \), \( x=-1 \). So \( G'(-3,-1) \)? Wait, no, wait: Wait, \( (x,y)=(-1,-3) \), so \( -y = -(-3)=3 \), \( x=-1 \), so the new point is \( (3, -1) \)? Wait, I think I messed up the order. Wait, no: The correct rule is \( (x,y) \) rotated 90° counterclockwise is \( (-y, x) \). So \( x=-1,y=-3 \): \( -y = 3 \), \( x=-1 \)? No, that would be \( (3, -1) \)? Wait, no, let's use vector approach. A point \( (x,y) \) can be represented as a vector. Rotating 90° counterclockwise is equivalent to multiplying the vector by the rotation matrix \(

$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$

\). So \(

$$\begin{pmatrix}x'\\y'\end{pmatrix}$$

=

$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$

$$\begin{pmatrix}x\\y\end{pmatrix}$$

=

$$\begin{pmatrix}-y\\x\end{pmatrix}$$

\). So for \( (x,y)=(-1,-3) \), \( x'=-y=-(-3)=3 \), \( y'=x=-1 \). So the rotated point is \( (3, -1) \).

Step2: Rotate \( H(3, -4) \)

\( x=3,y=-4 \). \( x'=-y=-(-4)=4 \), \( y'=x=3 \). So \( H'(4,3) \).

Step3: Rotate \( I(4, -8) \)

\( x=4,y=-8 \). \( x'=-y=-(-8)=8 \), \( y'=x=4 \). So \( I'(8,4) \).

Step4: Rotate \( J(0, -7) \)

\( x=0,y=-7 \). \( x'=-y=-(-7)=7 \), \( y'=x=0 \). So \( J'(7,0) \).

Wait, but the problem says "Rotation: 90° counterclockwise" – it's not clear if we rotate the original rhombus or the translated one. The problem has (a) and (b) as separate? Wait, the original problem: (a) is translation, (b) is rotation. So (a) and (b) are two separate transformations: (a) translate the original r…

Answer:

(Part a - Translation):
\( G'(3, 7) \)
\( H'(7, 6) \)
\( I'(8, 2) \)
\( J'(4, 3) \)