Transformations: Question 4

Worked solution

Part (a): reflection in the line $y = 1$

The mirror line $y = 1$ is horizontal, so reflecting in it leaves the $x$-coordinate unchanged and reflects the $y$-coordinate about the value $1$. For a point $(x, y)$:

$(x, y) \longmapsto (x,\ 2(1) - y) = (x,\ 2 - y)$

Apply this rule to each vertex of $L$:

$A(2,3) \to (2,\ 2-3) = (2,-1)$ $B(5,3) \to (5,\ 2-3) = (5,-1)$ $C(2,6) \to (2,\ 2-6) = (2,-4)$

$\boxed{A'(2,-1),\quad B'(5,-1),\quad C'(2,-4)}$

Check: $A(2,3)$ is $3-1=2$ units above the mirror line $y=1$, so its image should be $2$ units below it, at $y=1-2=-1$. This matches $A'(2,-1)$. ✓

Part (b): rotation of $180^\circ$ about $(4, 2)$

For a $180^\circ$ rotation about a centre $(a, b)$, each image point is found using:

$(x, y) \longmapsto (2a - x,\ 2b - y)$

Here the centre is $(a, b) = (4, 2)$, so $2a = 8$ and $2b = 4$:

$(x, y) \longmapsto (8 - x,\ 4 - y)$

Apply this rule to each vertex of $L$:

$A(2,3) \to (8-2,\ 4-3) = (6,1)$ $B(5,3) \to (8-5,\ 4-3) = (3,1)$ $C(2,6) \to (8-2,\ 4-6) = (6,-2)$

$\boxed{A''(6,1),\quad B''(3,1),\quad C''(6,-2)}$

Check (centre is the midpoint of each object–image pair, since a $180^\circ$ rotation is a half-turn): the midpoint of $A(2,3)$ and $A''(6,1)$ is $\left(\dfrac{2+6}{2},\ \dfrac{3+1}{2}\right) = (4,2)$, which is exactly the given centre. ✓

Final answers

• (a) $A'(2,-1)$, $B'(5,-1)$, $C'(2,-4)$
• (b) $A''(6,1)$, $B''(3,1)$, $C''(6,-2)$