Question

what are the coordinates of the vertices of the image of trapezoid pqrt after a translation along the vector (2, -6) followed by a dilation centered at the origin with scale factor of \\(\frac{1}{2}\\)?

Explanation:

First, we need to find the original coordinates of the vertices of trapezoid \( PQRT \) from the graph. Let's assume the grid has each square with side length 1.

Looking at the graph:

• \( P \): Let's say the coordinates are \( (-2, 4) \) (since it's 2 units left of the origin on the x - axis and 4 units up on the y - axis)
• \( Q \): Coordinates are \( (0, 4) \) (on the y - axis, 4 units up)
• \( R \): Coordinates are \( (2, 2) \) (2 units right of the origin on the x - axis and 2 units up on the y - axis)
• \( T \): Coordinates are \( (-4, 2) \) (4 units left of the origin on the x - axis and 2 units up on the y - axis)

Step 1: Apply the translation vector \( (2,-6) \)

The translation rule for a point \( (x,y) \) is \( (x + 2,y-6) \)

• For \( P(-2,4) \):

\( x=-2 + 2=0 \), \( y = 4-6=-2 \). So the translated point \( P_1=(0,-2) \)

• For \( Q(0,4) \):

\( x = 0+2 = 2 \), \( y=4 - 6=-2 \). So the translated point \( Q_1=(2,-2) \)

• For \( R(2,2) \):

\( x=2 + 2=4 \), \( y=2-6=-4 \). So the translated point \( R_1=(4,-4) \)

• For \( T(-4,2) \):

\( x=-4 + 2=-2 \), \( y=2-6=-4 \). So the translated point \( T_1=(-2,-4) \)

Step 2: Apply the dilation with scale factor \( \frac{1}{2} \) centered at the origin

The dilation rule for a point \( (x,y) \) centered at the origin with scale factor \( k \) is \( (kx,ky) \)

• For \( P_1(0,-2) \):

\( x=\frac{1}{2}\times0 = 0 \), \( y=\frac{1}{2}\times(-2)=-1 \). So the dilated point \( P'=(0,-1) \)

• For \( Q_1(2,-2) \):

\( x=\frac{1}{2}\times2 = 1 \), \( y=\frac{1}{2}\times(-2)=-1 \). So the dilated point \( Q'=(1,-1) \)

• For \( R_1(4,-4) \):

\( x=\frac{1}{2}\times4 = 2 \), \( y=\frac{1}{2}\times(-4)=-2 \). So the dilated point \( R'=(2,-2) \)

• For \( T_1(-2,-4) \):

\( x=\frac{1}{2}\times(-2)=-1 \), \( y=\frac{1}{2}\times(-4)=-2 \). So the dilated point \( T'=(-1,-2) \)

Answer:

The coordinates of the vertices of the image are \( P'(0,-1) \), \( Q'(1,-1) \), \( R'(2,-2) \), \( T'(-1,-2) \)