Question

period: 2nd a/b day: a for exercises 1 & 2, the vertices of δxyz are x(1, -4), y(-2, -1) and z(3,1). for each translation, give the vertices of δxyz 1. ( t_{(-4,-3)}(delta xyz) = ) 2. ( t_{(5,-3)}(delta xyz) = ) 3. what is the rule for the translation shown? image of a quadrilateral with coordinates for exercises 4 & 5, write the composition of translations as one translation 4. ( t_{(7,8)} circ t_{(-3,-4)} ) 5. ( t_{(0,3)} circ t_{(4,6)} )

Explanation:

Problem 1:

Step1: Translate X(1, -4)

To translate a point \((x, y)\) by \(T_{(h, k)}\), we use the rule \((x + h, y + k)\). For \(X(1, -4)\) and \(T_{(-4, -3)}\), we calculate \(x' = 1 + (-4)= -3\) and \(y' = -4 + (-3)= -7\), so \(X'(-3, -7)\).

Step2: Translate Y(-2, -1)

For \(Y(-2, -1)\), \(x' = -2 + (-4)= -6\) and \(y' = -1 + (-3)= -4\), so \(Y'(-6, -4)\).

Step3: Translate Z(3, 1)

For \(Z(3, 1)\), \(x' = 3 + (-4)= -1\) and \(y' = 1 + (-3)= -2\), so \(Z'(-1, -2)\).

Step1: Translate X(1, -4)

Using the translation rule \((x + h, y + k)\) for \(T_{(5, -3)}\) and \(X(1, -4)\), \(x' = 1 + 5 = 6\) and \(y' = -4 + (-3)= -7\), so \(X'(6, -7)\).

Step2: Translate Y(-2, -1)

For \(Y(-2, -1)\), \(x' = -2 + 5 = 3\) and \(y' = -1 + (-3)= -4\), so \(Y'(3, -4)\).

Step3: Translate Z(3, 1)

For \(Z(3, 1)\), \(x' = 3 + 5 = 8\) and \(y' = 1 + (-3)= -2\), so \(Z'(8, -2)\).

Let's take a point, say \(A(-5, -1)\) (from the graph) and its image \(A'(1, -3)\). The horizontal change (x - translation) is \(1 - (-5)=6\), and the vertical change (y - translation) is \(-3 - (-1)= -2\). So the translation rule is \(T_{(6, -2)}\), which means \((x, y)\to(x + 6, y - 2)\).

Answer:

\(\Delta X'Y'Z'\) with vertices \(X'(-3, -7)\), \(Y'(-6, -4)\), \(Z'(-1, -2)\)

Problem 2: