Alliance Form 3 P1 Q21 — Transformations on a Grid (Enlargement, Reflection, Rotation)

KCSE (Form 3) Form 3 Geometry

Published

The Question

“A small square ABCD has vertices A(6, -2), B(7, -2), C(7, -1) and D(6, -1). (a) The square is enlarged by scale factor 3 with centre (9, -4) to give image A'B'C'D'. (b) A'B'C'D' is reflected in the line x = 0 to give A''B''C''D''. (c) A''B''C''D'' is rotated 90 degrees anticlockwise about the origin to give A'''B'''C'''D'''. State the coordinates of each image, then describe fully the single transformation that maps A'B'C'D' directly onto A'''B'''C'''D'''.”

1

Enlarge ABCD by scale factor 3 about (9, -4)

For an enlargement each vertex moves so that its distance from the centre is multiplied by the scale factor. The quick rule is to take the centre and add three times the vector from the centre to the original point. Doing this for all four corners turns the unit square ABCD into a 3 by 3 square A'B'C'D'.

A′=(9,−4)+3((6,−2)−(9,−4))=(0,2)A'=(9,-4)+3\big((6,-2)-(9,-4)\big)=(0,2)
B′=(9,−4)+3((7,−2)−(9,−4))=(3,2)B'=(9,-4)+3\big((7,-2)-(9,-4)\big)=(3,2)
C′=(9,−4)+3((7,−1)−(9,−4))=(3,5)C'=(9,-4)+3\big((7,-1)-(9,-4)\big)=(3,5)
D′=(9,−4)+3((6,−1)−(9,−4))=(0,5)D'=(9,-4)+3\big((6,-1)-(9,-4)\big)=(0,5)

The side length grows from 1 unit to 3 units, matching the scale factor of 3.

2

Reflect A'B'C'D' in the y-axis (x = 0)

The line x = 0 is the y-axis. Reflecting in the y-axis leaves each y-coordinate unchanged but reverses the sign of each x-coordinate, so the square flips across to the left-hand side of the grid.

(x,y)  →  (−x, y)(x,y)\;\rightarrow\;(-x,\,y)
A′′=(0,2),B′′=(−3,2),C′′=(−3,5),D′′=(0,5)A''=(0,2),\quad B''=(-3,2),\quad C''=(-3,5),\quad D''=(0,5)
3

Rotate A''B''C''D'' 90 degrees anticlockwise about the origin

A quarter-turn anticlockwise about the origin sends any point (x, y) to (-y, x). Applying this rule to each vertex drops the square down into the lower-left region of the grid to give A'''B'''C'''D'''.

(x,y)  →  (−y, x)(x,y)\;\rightarrow\;(-y,\,x)
A′′′=(−2,0),B′′′=(−2,−3),C′′′=(−5,−3),D′′′=(−5,0)A'''=(-2,0),\quad B'''=(-2,-3),\quad C'''=(-5,-3),\quad D'''=(-5,0)
4

Compare A'B'C'D' with A'''B'''C'''D'''

To find the single transformation, look at what happened to each point overall. Comparing the enlarged square with the final square, every point (x, y) has ended up at (-y, -x). This coordinate swap-and-negate rule is the signature of one specific reflection.

A′(0,2)  →  A′′′(−2,0)A'(0,2)\;\rightarrow\;A'''(-2,0)
B′(3,2)  →  B′′′(−2,−3)B'(3,2)\;\rightarrow\;B'''(-2,-3)
(x,y)  →  (−y, −x)(x,y)\;\rightarrow\;(-y,\,-x)

Final Result

The images are A'B'C'D' = (0,2),(3,2),(3,5),(0,5); A''B''C''D'' = (0,2),(-3,2),(-3,5),(0,5); and A'''B'''C'''D''' = (-2,0),(-2,-3),(-5,-3),(-5,0). The single transformation that maps A'B'C'D' directly onto A'''B'''C'''D''' is a reflection in the line y = -x.

Why this method works

A reflection in the line y = -x always sends a point (x, y) to (-y, -x), because that line acts as a mirror that both swaps the roles of the two coordinates and reverses their signs. Reflecting in the y-axis (which negates x) and then rotating 90 degrees anticlockwise (which turns (x, y) into (-y, x)) combine to give exactly this same overall rule. Whenever two separate transformations produce a single consistent (x, y) to (-y, -x) mapping for every point, they are equivalent to that one reflection, which is why a single reflection can replace the pair.

✓

Test D': (0,5) under y = -x gives (-5, 0), which is exactly D''' — confirming the single transformation.