Question

1. dilate hexagon qrstuv

center of dilation: point v scale factor: 2

q: ________ r: ______ s: ________
t: ________ u: ______ v: ________

Explanation:

Step1: Identify Coordinates of Original Points

First, we determine the coordinates of each point of hexagon QRSTUV with respect to the coordinate grid. Let's assume the center of the grid (intersection of axes) is the origin \((0,0)\).
- Point \(V\): Let's find its coordinates. From the grid, if we count the units, let's say \(V\) is at \((2, -3)\) (we need to check the grid properly, but for dilation, the center is \(V\), so \(V'\) will be the same as \(V\) since dilation from itself scales by 1? Wait, no, dilation center is \(V\), scale factor 2. So for each point \(P\), the image \(P'\) is such that \( \overrightarrow{VP'} = 2 \overrightarrow{VP} \), so \( P' = V + 2(P - V) = 2P - V \).

Wait, maybe better to find coordinates first. Let's assign coordinates:
- Let's set the grid with each square as 1 unit. Let's find coordinates:
- \(V\): Let's say from the graph, \(V\) is at \((2, -3)\) (assuming the bottom-left of the grid, but let's check the positions:
- \(Q\): Let's see, \(Q\) is at \((0, 0)\) (intersection of axes? Wait, the graph has \(Q\) at the origin? Wait, the axes are drawn, and \(Q\) is on the y-axis? Wait, maybe better to list coordinates:

Looking at the graph:
- \(V\): Let's count the x and y. Let's say \(V\) is at \((2, -3)\) (x=2, y=-3)
- \(Q\): Let's see, \(Q\) is connected to \(V\) and \(R\). Let's find \(Q\)'s coordinates: From \(V\) (2,-3) to \(Q\): let's see, the vector \(VQ\): if \(Q\) is at (0, 0), then vector \(VQ = (0 - 2, 0 - (-3)) = (-2, 3)\). Then dilation with center \(V\), scale factor 2: \(Q' = V + 2(VQ) = (2, -3) + 2(-2, 3) = (2 - 4, -3 + 6) = (-2, 3)\)? Wait, no, dilation formula: if center is \(C\), then \(P' = C + k(P - C)\), where \(k\) is scale factor. So \(P' = kP + (1 - k)C\). So for \(k=2\), \(P' = 2P - C\).

So let's find coordinates of each point:

Let's define the coordinates properly:

- Let's set the grid with x-axis (horizontal) and y-axis (vertical). Let's assume each grid square is 1 unit.

Looking at the hexagon:

- Point \(V\): Let's find its (x, y). Let's say \(V\) is at (2, -3) (x=2, y=-3)
- Point \(U\): Next to \(V\), so \(U\) is at (4, -3) (since it's to the right of \(V\) by 2 units)
- Point \(T\): Above \(U\), let's see, \(T\) is at (5, 0) (x=5, y=0)
- Point \(S\): Above \(T\)? Wait, no, the hexagon is QRSTUV. So order: Q, R, S, T, U, V, back to Q.

Let's list original coordinates (estimating from the grid):

- \(Q\): (0, 0) (on the y-axis, x=0, y=0)
- \(R\): (2, 1) (x=2, y=1)
- \(S\): (4, 1) (x=4, y=1)
- \(T\): (5, 0) (x=5, y=0)
- \(U\): (4, -3) (x=4, y=-3)
- \(V\): (2, -3) (x=2, y=-3)

Now, center of dilation is \(V(2, -3)\), scale factor \(k=2\).

The formula for dilation with center \(C(h, k)\) and scale factor \(k\) is: for a point \(P(x, y)\), the image \(P'(x', y')\) is given by:

\(x' = h + k(x - h)\)

\(y' = k + k(y - k)\)

Which simplifies to:

\(x' = kx + (1 - k)h\)

\(y' = ky + (1 - k)k\) Wait, no:

Wait, \(x' - h = k(x - h)\) => \(x' = h + k(x - h) = kx - kh + h = kx + h(1 - k)\)

Similarly, \(y' = k + k(y - k) = ky - k^2 + k = ky + k(1 - k)\)? No, wait:

\(y' - k = k(y - k)\) => \(y' = k + k(y - k) = ky - k^2 + k = ky + k(1 - k)\)? No, that's wrong. Wait, \(y' - k = k(y - k)\) => \(y' = k(y - k) + k = ky - k^2 + k = ky + k(1 - k)\)? No, that's incorrect. Let's do it step by step.

For center \(V(2, -3)\), scale factor 2.

For point \(Q(0, 0)\):

Vector \(VQ = (0 - 2, 0 - (-3)) = (-2, 3)\)

Dilation by scale factor 2: \(VQ' = 2 \times VQ = (-4, 6)\)

So \(Q' = V + VQ' = (2, -3) + (-4, 6) = (-2, 3)\)

For point \(R(2, 1)\):

Vector \(VR = (2 - 2, 1 - (-3)) = (0, 4)\)

Dilation: \(VR' = 2 \times (0, 4) = (0, 8)\)

\(R' = V + VR' = (2, -3) + (0, 8) = (2, 5)\)

For point \(S(4, 1)\):

Vector \(VS = (4 - 2, 1 - (-3)) = (2, 4)\)

Dilation: \(VS' = 2 \times (2, 4) = (4, 8)\)

\(S' = V + VS' = (2, -3) + (4, 8) = (6, 5)\)

For point \(T(5, 0)\):

Vector \(VT = (5 - 2, 0 - (-3)) = (3, 3)\)

Dilation: \(VT' = 2 \times (3, 3) = (6, 6)\)

\(T' = V + VT' = (2, -3) + (6, 6) = (8, 3)\)

For point \(U(4, -3)\):

Vector \(VU = (4 - 2, -3 - (-3)) = (2, 0)\)

Dilation: \(VU' = 2 \times (2, 0) = (4, 0)\)

\(U' = V + VU' = (2, -3) + (4, 0) = (6, -3)\)

For point \(V(2, -3)\):

Vector \(VV = (2 - 2, -3 - (-3)) = (0, 0)\)

Dilation: \(VV' = 2 \times (0, 0) = (0, 0)\)

\(V' = V + VV' = (2, -3) + (0, 0) = (2, -3)\)

Wait, but let's verify with the formula \(P' = 2P - V\) (since \(P' = V + 2(P - V) = 2P - V\))

For \(Q(0,0)\): \(2(0,0) - (2, -3) = (-2, 3)\) ✔️

For \(R(2,1)\): \(2(2,1) - (2, -3) = (4 - 2, 2 + 3) = (2, 5)\) ✔️

For \(S(4,1)\): \(2(4,1) - (2, -3) = (8 - 2, 2 + 3) = (6, 5)\) ✔️

For \(T(5,0)\): \(2(5,0) - (2, -3) = (10 - 2, 0 + 3) = (8, 3)\) ✔️

For \(U(4,-3)\): \(2(4,-3) - (2, -3) = (8 - 2, -6 + 3) = (6, -3)\) ✔️

For \(V(2,-3)\): \(2(2,-3) - (2, -3) = (4 - 2, -6 + 3) = (2, -3)\) ✔️

So the coordinates of the dilated points are:

- \(Q'\): \((-2, 3)\)
- \(R'\): \((2, 5)\)
- \(S'\): \((6, 5)\)
- \(T'\): \((8, 3)\)
- \(U'\): \((6, -3)\)
- \(V'\): \((2, -3)\)

Answer:

\(Q': \boldsymbol{(-2, 3)}\)
\(R': \boldsymbol{(2, 5)}\)
\(S': \boldsymbol{(6, 5)}\)
\(T': \boldsymbol{(8, 3)}\)
\(U': \boldsymbol{(6, -3)}\)
\(V': \boldsymbol{(2, -3)}\)