Question

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triangle def has coordinates d(4, -1), e(5, 2), and f(1, 2). determine the coordinates of the vertices of the image after each transformation.

1. reflection in x - axis
2. reflection in y - axis
3. translation along the vector (1, 0)
4. translation along the vector (-3, 1)
5. rotation 180° about the origin
6. rotation 270° counterclockwise about the origin

Explanation:

Step1: Reflection in x - axis

When reflecting a point \((x,y)\) over the \(x\) - axis, the rule is \((x,y)\to(x, - y)\).
For \(D(4,-1)\): \(D'(4,1)\)
For \(E(5,2)\): \(E'(5,-2)\)
For \(F(1,2)\): \(F'(1,-2)\)

Step2: Reflection in y - axis

When reflecting a point \((x,y)\) over the \(y\) - axis, the rule is \((x,y)\to(-x,y)\).
For \(D(4,-1)\): \(D'(-4,-1)\)
For \(E(5,2)\): \(E'(-5,2)\)
For \(F(1,2)\): \(F'(-1,2)\)

Step3: Translation along the vector \((1,0)\)

When translating a point \((x,y)\) along the vector \((a,b)\), the rule is \((x,y)\to(x + a,y + b)\). Here \(a = 1,b=0\)
For \(D(4,-1)\): \(D'(4 + 1,-1+0)=(5,-1)\)
For \(E(5,2)\): \(E'(5 + 1,2+0)=(6,2)\)
For \(F(1,2)\): \(F'(1 + 1,2+0)=(2,2)\)

Step4: Translation along the vector \((-3,1)\)

Using the rule \((x,y)\to(x + a,y + b)\) with \(a=-3,b = 1\)
For \(D(4,-1)\): \(D'(4-3,-1 + 1)=(1,0)\)
For \(E(5,2)\): \(E'(5-3,2 + 1)=(2,3)\)
For \(F(1,2)\): \(F'(1-3,2 + 1)=(-2,3)\)

Step5: Rotation \(180^{\circ}\) about the origin

When rotating a point \((x,y)\) \(180^{\circ}\) about the origin, the rule is \((x,y)\to(-x,-y)\)
For \(D(4,-1)\): \(D'(-4,1)\)
For \(E(5,2)\): \(E'(-5,-2)\)
For \(F(1,2)\): \(F'(-1,-2)\)

Step6: Rotation \(270^{\circ}\) counter - clockwise about the origin

When rotating a point \((x,y)\) \(270^{\circ}\) counter - clockwise about the origin, the rule is \((x,y)\to(y,-x)\)
For \(D(4,-1)\): \(D'(-1,-4)\)
For \(E(5,2)\): \(E'(2,-5)\)
For \(F(1,2)\): \(F'(2,-1)\)

Answer:

1. \(D'(4,1),E'(5,-2),F'(1,-2)\)
2. \(D'(-4,-1),E'(-5,2),F'(-1,2)\)
3. \(D'(5,-1),E'(6,2),F'(2,2)\)
4. \(D'(1,0),E'(2,3),F'(-2,3)\)
5. \(D'(-4,1),E'(-5,-2),F'(-1,-2)\)
6. \(D'(-1,-4),E'(2,-5),F'(2,-1)\)