Question

quadrilateral ( abcd ) underwent a sequence of transformations to give quadrilateral ( abcd ). which transformations could have taken place?
a. a reflection across the ( x )-axis followed by a reflection across the ( y )-axis
b. a translation 10 units down followed by a translation 8 units to the right
c. a rotation ( 90^{circ} ) counterclockwise about the origin followed by a reflection across the ( y )-axis
d. a reflection across the line ( y = x ) followed by a reflection across the line ( y = -x )

Explanation:

Step1: Analyze Option A

A reflection across the x - axis changes the sign of the y - coordinate of a point \((x,y)\) to \((x, - y)\). A reflection across the y - axis changes the sign of the x - coordinate of a point \((x,y)\) to \((-x,y)\). Let's take a point from ABCD, say \(A(-6,4)\). After reflection across x - axis: \((-6,-4)\), then reflection across y - axis: \((6,-4)\), which is not the coordinates of \(A'\) (which is \((2,-6)\) approximately? Wait, no, let's check the coordinates properly. Let's find coordinates of ABCD: \(A(-6,4)\), \(B(-6,6)\), \(D(-4,4)\), \(C(-2,6)\). Coordinates of \(A'(2,-6)\), \(B'(2,-4)\), \(D'(4,-6)\), \(C'(6,-4)\). Wait, maybe my initial coordinate reading was wrong. Let's re - read the graph. The x - axis: from - 8 to 8, y - axis from - 8 to 6. For ABCD: \(A(-6,4)\), \(B(-6,6)\), \(D(-4,4)\), \(C(-2,6)\). For \(A'B'C'D'\): \(A'(2,-6)\)? No, looking at the lower part, \(A'\) is at (2, - 6)? Wait, no, the lower quadrilateral: \(A'\) is at (2, - 6)? Wait, the y - coordinate for \(A'\) is - 6? Wait, the grid lines: the upper quadrilateral is in the second quadrant (x negative, y positive), the lower one is in the fourth quadrant (x positive, y negative). Let's check the vertical and horizontal distances. The vertical distance from \(A(-6,4)\) to \(A'\) (let's say \(A'\) is (2, - 6)): no, wait, the vertical change: from y = 4 to y=-6, that's a change of - 10 (down 10 units). Horizontal change: from x=-6 to x = 2, that's a change of + 8 (right 8 units). So for point \(A(-6,4)\): translation down 10 units: \((-6,4 - 10)=(-6,-6)\), then translation right 8 units: \((-6 + 8,-6)=(2,-6)\), which matches \(A'\) (assuming \(A'\) is (2,-6)). Let's check \(B(-6,6)\): translation down 10: \((-6,6 - 10)=(-6,-4)\), then right 8: \((-6 + 8,-4)=(2,-4)\), which matches \(B'\) (2,-4). \(D(-4,4)\): translation down 10: \((-4,4 - 10)=(-4,-6)\), then right 8: \((-4 + 8,-6)=(4,-6)\), which matches \(D'\) (4,-6). \(C(-2,6)\): translation down 10: \((-2,6 - 10)=(-2,-4)\), then right 8: \((-2+8,-4)=(6,-4)\), which matches \(C'\) (6,-4). Now check Option A: reflection across x - axis: \(A(-6,4)\) becomes \((-6,-4)\), then reflection across y - axis: \((6,-4)\), which is not \(A'\) (2,-6). Option C: rotation 90 counterclockwise about origin: \((x,y)\to(-y,x)\). For \(A(-6,4)\): \((-4,-6)\), then reflection across y - axis: \((4,-6)\), which is not \(A'\) (2,-6). Option D: reflection across \(y = x\): \((x,y)\to(y,x)\), so \(A(-6,4)\to(4,-6)\), then reflection across \(y=-x\): \((x,y)\to(-y,-x)\), so \((4,-6)\to(6,6)\), which is not \(A'\). Option B: translation down 10 (y - 10) and right 8 (x + 8) works for all points.

Step2: Confirm with all points

For \(B(-6,6)\): down 10: \((-6,6 - 10)=(-6,-4)\), right 8: \((-6 + 8,-4)=(2,-4)\), which matches \(B'\). For \(D(-4,4)\): down 10: \((-4,4 - 10)=(-4,-6)\), right 8: \((-4 + 8,-6)=(4,-6)\), which matches \(D'\). For \(C(-2,6)\): down 10: \((-2,6 - 10)=(-2,-4)\), right 8: \((-2+8,-4)=(6,-4)\), which matches \(C'\). So Option B works.

Answer:

B. a translation 10 units down followed by a translation 8 units to the right