Question

1. if quadrilateral abcd was rotated 270 degrees counterclockwise about point d, what would be the coordinates of b?

(4, -1)
(-1, 4)
(-2, 3)
(3, -2)

Explanation:

Step1: Find coordinates of D and B

From the graph, \( D=(1,1) \), \( B=(4,3) \).

Step2: Translate to make D the origin

Subtract \( D \)'s coordinates from \( B \): \( (4 - 1, 3 - 1)=(3,2) \).

Step3: Apply 270° counterclockwise rotation

The rule for 270° counterclockwise rotation about the origin is \( (x,y)\to(y,-x) \). So \( (3,2)\to(2,-3) \).

Step4: Translate back (add D's coordinates)

Add \( D \)'s coordinates: \( (2 + 1, -3 + 1)=(3,-2) \). Wait, no, wait. Wait, maybe I made a mistake. Wait, let's re - do the rotation. Wait, 270° counterclockwise is equivalent to 90° clockwise. The formula for 270° counterclockwise rotation about a point \( (a,b) \) for a point \( (x,y) \) is: first, translate the point so that \( (a,b) \) is the origin: \( (x - a,y - b) \). Then rotate 270° counterclockwise: the rotation of a point \( (x',y') \) 270° counterclockwise about the origin is \( (y',-x') \). Then translate back: \( (y' + a,-x' + b) \).

So for \( D=(1,1) \) and \( B=(4,3) \):

1. Translate: \( (4 - 1,3 - 1)=(3,2) \)
2. Rotate 270° counterclockwise: \( (2,-3) \) (since \( (x',y')=(3,2) \), after rotation \( (y',-x')=(2,-3) \))
3. Translate back: \( (2+1,-3 + 1)=(3,-2) \)? Wait, but let's check the options. Wait, maybe I messed up the rotation direction. Wait, 270° counterclockwise is the same as 90° clockwise. The formula for 90° clockwise rotation is \( (x,y)\to(y,-x) \). Wait, no, 90° clockwise: \( (x,y)\to(y,-x) \), 180°: \( (-x,-y) \), 270° counterclockwise (or 90° counterclockwise is \( (-y,x) \), 270° counterclockwise is \( (y,-x) \)? Wait, no, let's recall the rotation matrices.

The rotation matrix for \( \theta \) degrees counterclockwise is \(

$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

\). For \( \theta = 270^\circ \), \( \cos270^\circ = 0 \), \( \sin270^\circ=- 1 \). So the matrix is \(

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

\). So applying to a point \( (x,y) \), we get \( (y,-x) \).

Wait, so for the translated point \( (3,2) \), applying the rotation matrix: \( x' = 2 \), \( y'=-3 \). Then translate back: \( x' + 1=3 \), \( y' + 1=-2 \). So \( (3,-2) \), which is one of the options.

Wait, but let's check the coordinates again. Let's find the vector from D to B: \( \overrightarrow{DB}=(4 - 1,3 - 1)=(3,2) \). When we rotate this vector 270° counterclockwise about the origin, the new vector \( \overrightarrow{DB'} \) should be \( (2,-3) \) (using the rotation matrix). Then the coordinates of \( B' \) are \( D+\overrightarrow{DB'}=(1 + 2,1+(-3))=(3,-2) \).

Answer:

\( (3,-2) \)