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4.1 translations (pp. 173-180) graph quadrilateral abcd with vertices a…

Question

4.1 translations (pp. 173-180)
graph quadrilateral abcd with vertices a(1, -2), b(3, -1), c(0, 3), and d(-4, 1) and its image after the translation (x, y) → (x + 2, y - 2).
graph quadrilateral abcd. to find the coordinates of the vertices of the image, add 2 to the x-coordinates and subtract 2 from the y-coordinates of the vertices of the preimage. then graph the image.
(x, y) → (x + 2, y - 2)
a(1, -2) → a(3, -4)
b(3, -1) → b(5, -3)
c(0, 3) → c(2, 1)
d(-4, 1) → d(-2, -1)
graph △xyz with vertices x(2, 3), y(-3, 2), and z(-4, -3) and its image after the translation.

  1. (x, y) → (x, y + 2)
  2. (x, y) → (x - 3, y)
  3. (x, y) → (x + 3, y - 1)
  4. (x, y) → (x + 4, y + 1)

graph △pqr with vertices p(0, -4), q(1, 3), and r(2, -5) and its image after the composition.

  1. translation: (x, y) → (x + 1, y + 2)

translation: (x, y) → (x - 4, y + 1)

  1. translation: (x, y) → (x, y + 3)

translation: (x, y) → (x - 1, y + 1)
4.2 reflections (pp. 181-188)
graph △abc with vertices a(1, -1), b(3, 2), and c(4, -4) and its image after a reflection in the line y = x.
graph △abc and the line y = x. then use the coordinate rule for reflecting in the line y = x to find the coordinates of the vertices of the image.
(a, b) → (b, a)
a(1, -1) → a(-1, 1)
b(3, 2) → b(2, 3)
c(4, -4) → c(-4, 4)
graph the polygon and its image after a reflection in the given line.

  1. x = 4
  2. y = 3
  3. how many lines of symmetry does the figure have?

Explanation:

To solve these problems, we'll address each sub - question one by one.

1. Translation \((x,y)\to(x,y + 2)\) for \(\triangle XYZ\) with \(X(2,3)\), \(Y(-3,2)\), \(Z(-4,-3)\)
Step 1: Apply translation to \(X\)

For point \(X(2,3)\), using the rule \((x,y)\to(x,y + 2)\), we keep the \(x\) - coordinate the same and add 2 to the \(y\) - coordinate. So \(X(2,3)\to X'(2,3 + 2)=X'(2,5)\)

Step 2: Apply translation to \(Y\)

For point \(Y(-3,2)\), using the rule \((x,y)\to(x,y + 2)\), we have \(Y(-3,2)\to Y'(-3,2 + 2)=Y'(-3,4)\)

Step 3: Apply translation to \(Z\)

For point \(Z(-4,-3)\), using the rule \((x,y)\to(x,y + 2)\), we get \(Z(-4,-3)\to Z'(-4,-3 + 2)=Z'(-4,-1)\)

2. Translation \((x,y)\to(x - 3,y)\) for \(\triangle XYZ\) with \(X(2,3)\), \(Y(-3,2)\), \(Z(-4,-3)\)
Step 1: Apply translation to \(X\)

For point \(X(2,3)\), using the rule \((x,y)\to(x - 3,y)\), we subtract 3 from the \(x\) - coordinate and keep the \(y\) - coordinate the same. So \(X(2,3)\to X'(2-3,3)=X'(-1,3)\)

Step 2: Apply translation to \(Y\)

For point \(Y(-3,2)\), using the rule \((x,y)\to(x - 3,y)\), we have \(Y(-3,2)\to Y'(-3-3,2)=Y'(-6,2)\)

Step 3: Apply translation to \(Z\)

For point \(Z(-4,-3)\), using the rule \((x,y)\to(x - 3,y)\), we get \(Z(-4,-3)\to Z'(-4 - 3,-3)=Z'(-7,-3)\)

3. Translation \((x,y)\to(x + 3,y - 1)\) for \(\triangle XYZ\) with \(X(2,3)\), \(Y(-3,2)\), \(Z(-4,-3)\)
Step 1: Apply translation to \(X\)

For point \(X(2,3)\), using the rule \((x,y)\to(x + 3,y - 1)\), we add 3 to the \(x\) - coordinate and subtract 1 from the \(y\) - coordinate. So \(X(2,3)\to X'(2 + 3,3-1)=X'(5,2)\)

Step 2: Apply translation to \(Y\)

For point \(Y(-3,2)\), using the rule \((x,y)\to(x + 3,y - 1)\), we have \(Y(-3,2)\to Y'(-3+3,2 - 1)=Y'(0,1)\)

Step 3: Apply translation to \(Z\)

For point \(Z(-4,-3)\), using the rule \((x,y)\to(x + 3,y - 1)\), we get \(Z(-4,-3)\to Z'(-4+3,-3 - 1)=Z'(-1,-4)\)

4. Translation \((x,y)\to(x + 4,y + 1)\) for \(\triangle XYZ\) with \(X(2,3)\), \(Y(-3,2)\), \(Z(-4,-3)\)
Step 1: Apply translation to \(X\)

For point \(X(2,3)\), using the rule \((x,y)\to(x + 4,y + 1)\), we add 4 to the \(x\) - coordinate and add 1 to the \(y\) - coordinate. So \(X(2,3)\to X'(2 + 4,3+1)=X'(6,4)\)

Step 2: Apply translation to \(Y\)

For point \(Y(-3,2)\), using the rule \((x,y)\to(x + 4,y + 1)\), we have \(Y(-3,2)\to Y'(-3 + 4,2+1)=Y'(1,3)\)

Step 3: Apply translation to \(Z\)

For point \(Z(-4,-3)\), using the rule \((x,y)\to(x + 4,y + 1)\), we get \(Z(-4,-3)\to Z'(-4 + 4,-3+1)=Z'(0,-2)\)

5. Composition of translations for \(\triangle PQR\) with \(P(0,-4)\), \(Q(1,3)\), \(R(2,-5)\) (First \((x,y)\to(x + 1,y + 2)\) then \((x,y)\to(x - 4,y + 1)\))
Step 1: Apply first translation to \(P\)

For point \(P(0,-4)\), using \((x,y)\to(x + 1,y + 2)\), we get \(P_1(0 + 1,-4+2)=P_1(1,-2)\)

Step 2: Apply second translation to \(P_1\)

Using \((x,y)\to(x - 4,y + 1)\) on \(P_1(1,-2)\), we have \(P'(1-4,-2 + 1)=P'(-3,-1)\)

Step 3: Apply first translation to \(Q\)

For point \(Q(1,3)\), using \((x,y)\to(x + 1,y + 2)\), we get \(Q_1(1 + 1,3+2)=Q_1(2,5)\)

Step 4: Apply second translation to \(Q_1\)

Using \((x,y)\to(x - 4,y + 1)\) on \(Q_1(2,5)\), we have \(Q'(2-4,5 + 1)=Q'(-2,6)\)

Step 5: Apply first translation to \(R\)

For point \(R(2,-5)\), using \((x,y)\to(x + 1,y + 2)\), we get \(R_1(2 + 1,-5+2)=R_1(3,-3)\)

Step 6: Apply second translation to \(R_1\)

Using \((x,y)\to(x - 4,y + 1)\) on \(R_1(3,-3)\), we have \(R'(3-4,-3 + 1)=R'(-1,-2)\)

6. Composition of translations fo…

Answer:

s:

  1. \(X'(2,5)\), \(Y'(-3,4)\), \(Z'(-4,-1)\)
  2. \(X'(-1,3)\), \(Y'(-6,2)\), \(Z'(-7,-3)\)
  3. \(X'(5,2)\), \(Y'(0,1)\), \(Z'(-1,-4)\)
  4. \(X'(6,4)\), \(Y'(1,3)\), \(Z'(0,-2)\)
  5. \(P'(-3,-1)\), \(Q'(-2,6)\), \(R'(-1,-2)\)
  6. \(P'(-1,0)\), \(Q'(0,7)\), \(R'(1,-1)\)
  7. (For assumed vertices) \(A'(7,2)\), \(B'(5,4)\), \(C'(3,1)\)
  8. (For assumed vertices) \(E'(1,3)\), \(F'(4,3)\), \(G'(5,6)\), \(H'(2,6)\)
  9. 2