Question

1. triangle abc has vertices at a(-8, -6), b(-3, -6), and c(-2, -3).

(a) draw the image of δabc after a translation by 9 units to the right and 2 units up. label its image δabc. list the coordinates of its vertices below.
(b) draw the image of δabc after a counterclockwise rotation of 90° about the origin. label its image δabc. give the coordinates of its vertices below.
(c) draw the image of δabc after a reflection in the y - axis. label its image δabc. give the coordinates of its vertices below.

Explanation:

Step1: Translation formula

For a translation of \(h\) units to the right and \(k\) units up, the formula is \((x,y)\to(x + h,y + k)\). Here \(h = 9\) and \(k=2\).

• For \(A(-8,-6)\): \(x=-8+9 = 1\), \(y=-6 + 2=-4\), so \(A'(1,-4)\)
• For \(B(-3,-6)\): \(x=-3+9 = 6\), \(y=-6 + 2=-4\), so \(B'(6,-4)\)
• For \(C(-2,-3)\): \(x=-2+9 = 7\), \(y=-3 + 2=-1\), so \(C'(7,-1)\)

Step2: Rotation formula

For a counter - clockwise rotation of \(90^{\circ}\) about the origin, the formula is \((x,y)\to(-y,x)\)

• For \(A'(1,-4)\): \(x=-(-4)=4\), \(y = 1\), so \(A''(4,1)\)
• For \(B'(6,-4)\): \(x=-(-4)=4\), \(y = 6\), so \(B''(4,6)\)
• For \(C'(7,-1)\): \(x=-(-1)=1\), \(y = 7\), so \(C''(1,7)\)

Step3: Reflection formula

For a reflection in the \(y\) - axis, the formula is \((x,y)\to(-x,y)\)

• For \(A''(4,1)\): \(x=-4\), \(y = 1\), so \(A'''(-4,1)\)
• For \(B''(4,6)\): \(x=-4\), \(y = 6\), so \(B'''(-4,6)\)
• For \(C''(1,7)\): \(x=-1\), \(y = 7\), so \(C'''(-1,7)\)

Answer:

(a) \(A'(1,-4)\), \(B'(6,-4)\), \(C'(7,-1)\)
(b) \(A''(4,1)\), \(B''(4,6)\), \(C''(1,7)\)
(c) \(A'''(-4,1)\), \(B'''(-4,6)\), \(C'''(-1,7)\)