QUESTION IMAGE
Question
- triangle cde with vertices c(-2, -2), d(1, -4), and e(-3, -4):
a) dilation with scale factor of 4 centered (-1, -3)
b) reflection in the line y = -x
Part a) Dilation with scale factor 4 centered at \((-1, -3)\)
Step 1: Recall the dilation formula
The formula for dilating a point \((x, y)\) with center \((h, k)\) and scale factor \(r\) is:
\((x', y') = (h + r(x - h), k + r(y - k))\)
Step 2: Dilate point \(C(-2, -2)\)
Substitute \(x = -2\), \(y = -2\), \(h = -1\), \(k = -3\), \(r = 4\):
\(x' = -1 + 4(-2 - (-1)) = -1 + 4(-1) = -1 - 4 = -5\)
\(y' = -3 + 4(-2 - (-3)) = -3 + 4(1) = -3 + 4 = 1\)
So, \(C' = (-5, 1)\)
Step 3: Dilate point \(D(1, -4)\)
Substitute \(x = 1\), \(y = -4\), \(h = -1\), \(k = -3\), \(r = 4\):
\(x' = -1 + 4(1 - (-1)) = -1 + 4(2) = -1 + 8 = 7\)
\(y' = -3 + 4(-4 - (-3)) = -3 + 4(-1) = -3 - 4 = -7\)
So, \(D' = (7, -7)\)
Step 4: Dilate point \(E(-3, -4)\)
Substitute \(x = -3\), \(y = -4\), \(h = -1\), \(k = -3\), \(r = 4\):
\(x' = -1 + 4(-3 - (-1)) = -1 + 4(-2) = -1 - 8 = -9\)
\(y' = -3 + 4(-4 - (-3)) = -3 + 4(-1) = -3 - 4 = -7\)
So, \(E' = (-9, -7)\)
Part b) Reflection in the line \(y = -x\)
Step 1: Recall the reflection formula
The formula for reflecting a point \((x, y)\) over the line \(y = -x\) is:
\((x', y') = (-y, -x)\)
Step 2: Reflect point \(C(-2, -2)\)
Substitute \(x = -2\), \(y = -2\):
\(x' = -(-2) = 2\), \(y' = -(-2) = 2\)
So, \(C'' = (2, 2)\)
Step 3: Reflect point \(D(1, -4)\)
Substitute \(x = 1\), \(y = -4\):
\(x' = -(-4) = 4\), \(y' = -1\)
So, \(D'' = (4, -1)\)
Step 4: Reflect point \(E(-3, -4)\)
Substitute \(x = -3\), \(y = -4\):
\(x' = -(-4) = 4\), \(y' = -(-3) = 3\)
So, \(E'' = (4, 3)\)
Final Answers
a) Dilated vertices: \(C'(-5, 1)\), \(D'(7, -7)\), \(E'(-9, -7)\)
b) Reflected vertices: \(C''(2, 2)\), \(D''(4, -1)\), \(E''(4, 3)\)
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Part a) Dilation with scale factor 4 centered at \((-1, -3)\)
Step 1: Recall the dilation formula
The formula for dilating a point \((x, y)\) with center \((h, k)\) and scale factor \(r\) is:
\((x', y') = (h + r(x - h), k + r(y - k))\)
Step 2: Dilate point \(C(-2, -2)\)
Substitute \(x = -2\), \(y = -2\), \(h = -1\), \(k = -3\), \(r = 4\):
\(x' = -1 + 4(-2 - (-1)) = -1 + 4(-1) = -1 - 4 = -5\)
\(y' = -3 + 4(-2 - (-3)) = -3 + 4(1) = -3 + 4 = 1\)
So, \(C' = (-5, 1)\)
Step 3: Dilate point \(D(1, -4)\)
Substitute \(x = 1\), \(y = -4\), \(h = -1\), \(k = -3\), \(r = 4\):
\(x' = -1 + 4(1 - (-1)) = -1 + 4(2) = -1 + 8 = 7\)
\(y' = -3 + 4(-4 - (-3)) = -3 + 4(-1) = -3 - 4 = -7\)
So, \(D' = (7, -7)\)
Step 4: Dilate point \(E(-3, -4)\)
Substitute \(x = -3\), \(y = -4\), \(h = -1\), \(k = -3\), \(r = 4\):
\(x' = -1 + 4(-3 - (-1)) = -1 + 4(-2) = -1 - 8 = -9\)
\(y' = -3 + 4(-4 - (-3)) = -3 + 4(-1) = -3 - 4 = -7\)
So, \(E' = (-9, -7)\)
Part b) Reflection in the line \(y = -x\)
Step 1: Recall the reflection formula
The formula for reflecting a point \((x, y)\) over the line \(y = -x\) is:
\((x', y') = (-y, -x)\)
Step 2: Reflect point \(C(-2, -2)\)
Substitute \(x = -2\), \(y = -2\):
\(x' = -(-2) = 2\), \(y' = -(-2) = 2\)
So, \(C'' = (2, 2)\)
Step 3: Reflect point \(D(1, -4)\)
Substitute \(x = 1\), \(y = -4\):
\(x' = -(-4) = 4\), \(y' = -1\)
So, \(D'' = (4, -1)\)
Step 4: Reflect point \(E(-3, -4)\)
Substitute \(x = -3\), \(y = -4\):
\(x' = -(-4) = 4\), \(y' = -(-3) = 3\)
So, \(E'' = (4, 3)\)
Final Answers
a) Dilated vertices: \(C'(-5, 1)\), \(D'(7, -7)\), \(E'(-9, -7)\)
b) Reflected vertices: \(C''(2, 2)\), \(D''(4, -1)\), \(E''(4, 3)\)