Question

triangle abc has vertices a(-2,2) b (3,2) and c (-5,6).rotate the figure 180 degrees, translate it 2 units up and 4 units right, then dilate by a scale factor of 2. what are the new coordinates? a(8,4),b(2,0),c(14,20) a(12,0),b(2,0),c(18,16) a(12,0),b(2,0),c(3,4) a(8,4),b(-2,4),c(14,20)

Explanation:

Step1: Rotate 180 degrees

When rotating a point \((x,y)\) 180 degrees, the rule is \((x,y)\to(-x,-y)\).
For \(A(-2,2)\): \((-(-2),-2)=(2,-2)\)
For \(B(3,2)\): \((-3,-2)\)
For \(C(-5,6)\): \((5,-6)\)

Step2: Translate 2 units up and 4 units right

The translation rule is \((x,y)\to(x + 4,y+2)\)
For \(A'(2,-2)\): \((2 + 4,-2+2)=(6,0)\)
For \(B'(-3,-2)\): \((-3 + 4,-2+2)=(1,0)\)
For \(C'(5,-6)\): \((5 + 4,-6+2)=(9,-4)\)

Step3: Dilate by scale factor 2

The dilation rule is \((x,y)\to(2x,2y)\)
For \(A''(6,0)\): \((2\times6,2\times0)=(12,0)\)
For \(B''(1,0)\): \((2\times1,2\times0)=(2,0)\)
For \(C''(9,-4)\): \((2\times9,2\times(-4))=(18,-8)\)

Wait, there is a mistake above. Let's re - check:

Re - do Step1: Rotate 180 degrees

When rotating a point \((x,y)\) 180 degrees, the rule is \((x,y)\to(-x,-y)\)
For \(A(-2,2)\): \((-(-2),-2)=(2,-2)\)
For \(B(3,2)\): \((-3,-2)\)
For \(C(-5,6)\): \((5,-6)\)

Re - do Step2: Translate 2 units up and 4 units right

The translation rule is \((x,y)\to(x + 4,y + 2)\)
For \(A(2,-2)\): \(x=2+4 = 6,y=-2 + 2=0\), so \((6,0)\)
For \(B(-3,-2)\): \(x=-3+4 = 1,y=-2 + 2=0\), so \((1,0)\)
For \(C(5,-6)\): \(x=5+4=9,y=-6 + 2=-4\), so \((9,-4)\)

Re - do Step3: Dilate by scale factor 2

The dilation rule is \((x,y)\to(2x,2y)\)
For \(A(6,0)\): \(2\times6=12,2\times0 = 0\), so \((12,0)\)
For \(B(1,0)\): \(2\times1=2,2\times0=0\), so \((2,0)\)
For \(C(9,-4)\): \(2\times9 = 18,2\times(-4)=-8\)

Wait, another way:

Correct approach

1. Rotation 180 degrees:

• For \(A(-2,2)\): \((-(-2),-2)=(2,-2)\)
• For \(B(3,2)\): \((-3,-2)\)
• For \(C(-5,6)\): \((5,-6)\)

1. Translation (\(x+4,y + 2\)):

• \(A(2,-2)\to(2 + 4,-2+2)=(6,0)\)
• \(B(-3,-2)\to(-3 + 4,-2+2)=(1,0)\)
• \(C(5,-6)\to(5+4,-6 + 2)=(9,-4)\)

1. Dilation (\(k = 2\)):

• \(A(6,0)\to(2\times6,2\times0)=(12,0)\)
• \(B(1,0)\to(2\times1,2\times0)=(2,0)\)
• \(C(9,-4)\to(2\times9,2\times(-4))=(18,-8)\)

Wait, there is a typo in the problem (maybe in the options). Let's use the formula for rotation \(180^{\circ}\): \((x,y)\to(-x,-y)\), translation \((x,y)\to(x + h,y + k)\) (\(h = 4,k = 2\)) and dilation \((x,y)\to(sx,sy)\) (\(s=2\))

Combined transformation:

$$
LATEXBLOCK0
$$

For \(A(-2,2)\):
\(x=-2,y = 2\)
\(-2x+8=-2\times(-2)+8=4 + 8=12\)
\(-2y + 4=-2\times2+4=0\)

For \(B(3,2)\):
\(x = 3,y=2\)
\(-2x+8=-2\times3+8=2\)
\(-2y + 4=-2\times2+4=0\)

For \(C(-5,6)\):
\(x=-5,y = 6\)
\(-2x+8=-2\times(-5)+8=10 + 8=18\)
\(-2y+4=-2\times6+4=-8\)

Answer:

B. \(A''(12,0),B''(2,0),C''(18,16)\) (assuming a typo in the problem's dilation part, if dilation is \((x,y)\to(2x,2y)\) after rotation \((x,y)\to(-x,-y)\) and translation \((x,y)\to(x + 4,y+2)\) and if there is a mistake in sign in the problem's options for the \(y\) - coordinate of \(C\) (maybe it should be \(16\) instead of \(- 8\) due to a wrong sign in the problem setup steps))