Question

1. point h is located at (-7, 5). where is h after a 270° clockwise rotation? (5, 7)
2. point r is located at (8, 13). where is r after a translation 8 units left and 6 units down?
3. point l is located at (-3, 6). where is l after a reflection over the y - axis? (3, 6)
4. point b is located at (4, - 16). where is b after a dilation with a scale factor of 3/4? (3, - 12)
5. point w is located at (6, 2). where is w after a 180° counterclockwise rotation? (-6, - 2)
6. point q is located at (-9, 10). where is q after a translation 4 units right and 11 units down?

Explanation:

Step1: Recall translation rules

For a translation \(a\) units left and \(b\) units down on a point \((x,y)\), the new - point \((x',y')=(x - a,y - b)\); for \(a\) units right and \(b\) units down on a point \((x,y)\), the new - point \((x',y')=(x + a,y - b)\); for reflection over the \(y\) - axis, if the point is \((x,y)\), the new point is \((-x,y)\); for a \(180^{\circ}\) counter - clockwise rotation about the origin, if the point is \((x,y)\), the new point is \((-x,-y)\); for a \(270^{\circ}\) clockwise rotation about the origin, if the point is \((x,y)\), the new point is \((y,-x)\); for dilation with a scale factor \(k\) centered at the origin, if the point is \((x,y)\), the new point is \((kx,ky)\).

Step2: Solve for point \(H\)

Point \(H\) is at \((-7,5)\). A \(270^{\circ}\) clockwise rotation about the origin: Using the rule \((x,y)\to(y,-x)\), substituting \(x=-7\) and \(y = 5\), we get \((5,7)\).

Step3: Solve for point \(R\)

Point \(R\) is at \((8,13)\). Translating 8 units left and 6 units down. Using the rule \((x,y)\to(x - 8,y - 6)\), substituting \(x = 8\) and \(y=13\), we have \((8-8,13 - 6)=(0,7)\).

Step4: Solve for point \(L\)

Point \(L\) is at \((-3,6)\). Reflecting over the \(y\) - axis. Using the rule \((x,y)\to(-x,y)\), substituting \(x=-3\) and \(y = 6\), we get \((3,6)\).

Step5: Solve for point \(W\)

Point \(W\) is at \((6,2)\). A \(180^{\circ}\) counter - clockwise rotation about the origin. Using the rule \((x,y)\to(-x,-y)\), substituting \(x = 6\) and \(y=2\), we get \((-6,-2)\).

Step6: Solve for point \(B\)

Point \(B\) is at \((4,-16)\). Dilation with a scale factor of \(\frac{3}{4}\) centered at the origin. Using the rule \((x,y)\to(kx,ky)\) where \(k=\frac{3}{4}\), \(x = 4\) and \(y=-16\), we have \((\frac{3}{4}\times4,\frac{3}{4}\times(-16))=(3,-12)\).

Step7: Solve for point \(Q\)

Point \(Q\) is at \((-9,10)\). Translating 4 units right and 11 units down. Using the rule \((x,y)\to(x + 4,y - 11)\), substituting \(x=-9\) and \(y = 10\), we get \((-9 + 4,10-11)=(-5,-1)\).

Answer:

1. \((5,7)\)
2. \((0,7)\)
3. \((3,6)\)
4. \((-6,-2)\)
5. \((3,-12)\)
6. \((-5,-1)\)