Question

triangle def is reflected across the line ( y = x ).
the result is ( \triangle def ), as shown below.

(a) the arrows below show that the coordinates on the
left are mapped to the coordinates on the right. fill
in the blanks to give the coordinates after the
reflection.

original coordinates ( \to ) final coordinates

( d(-2,-4)\to d(square,square) )
( e(8,2)\to e(square,square) )
( f(4,-9)\to f(square,square) )

(b) choose the general rule below that describes
the reflection mapping ( \triangle def ) to ( \triangle def ).

( (x,y)\to(-y,-x) )
( (x,y)\to(-x,y) )
( (x,y)\to(x,-y) )
( (x,y)\to(-y,x) )
( (x,y)\to(-x,-y) )
( (x,y)\to(y,-x) )
( (x,y)\to(y,x) )

Explanation:

(a)

Step1: Apply reflection rule across \(y = x\)

When reflecting a point \((x,y)\) across the line \(y=x\), the rule is \((x,y)\to(y,x)\).
For point \(D(-2,-4)\), swap \(x\) and \(y\) coordinates. So \(D(-2,-4)\to D'(-4, - 2)\).

Step2: Apply reflection rule across \(y = x\)

For point \(E(8,2)\), swap \(x\) and \(y\) coordinates. So \(E(8,2)\to E'(2,8)\).

Step3: Apply reflection rule across \(y = x\)

For point \(F(4,-9)\), swap \(x\) and \(y\) coordinates. So \(F(4,-9)\to F'(-9,4)\).

(b)

Step1: Identify the general rule

Since the reflection across \(y = x\) transforms \((x,y)\) to \((y,x)\) (as seen from part (a) where for each point \((x,y)\) in \(\triangle DEF\), the corresponding point in \(\triangle D'E'F'\) has coordinates \((y,x)\)).

Answer:

(a)

\(D(-2,-4)\to D'(-4, - 2)\)
\(E(8,2)\to E'(2,8)\)
\(F(4,-9)\to F'(-9,4)\)

(b)

\((x,y)\to(y,x)\) (the option \((x,y)\to(y,x)\) in the multiple - choice list)