Question

triangle xyz is reflected across the line ( y = x ).
the result is ( \triangle xyz ), 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
( x(-2,-4)\to x(square,square) )
( y(8,2)\to y(square,square) )
( z(4,-9)\to z(square,square) )
(b) choose the general rule below that describes the reflection mapping ( \triangle xyz ) to ( \triangle xyz ).
( \bigcirc(x,y)\to(-y,-x))( \bigcirc(x,y)\to(-y,x) )
( \bigcirc(x,y)\to(x,-y))( \bigcirc(x,y)\to(y,-x) )
( \bigcirc(x,y)\to(-x,-y))( \bigcirc(x,y)\to(-x,y) )
( \bigcirc(x,y)\to(y,x) )

Explanation:

Step1: Recall the reflection rule across \(y = x\)

When a point \((x,y)\) is reflected across the line \(y=x\), the transformation rule is \((x,y)\to(y,x)\).

Step2: Apply the rule to point \(X(-2,-4)\)

For \(X(-2,-4)\), swap \(x\) and \(y\) coordinates. So \(X'(-4,-2)\).

Step3: Apply the rule to point \(Y(8,2)\)

For \(Y(8,2)\), swap \(x\) and \(y\) coordinates. So \(Y'(2,8)\).

Step4: Apply the rule to point \(Z(4,-9)\)

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

Step5: Identify the general rule

Since the transformation is \((x,y)\to(y,x)\) (from the reflection across \(y = x\)).

Answer:

(a) \(X(-2, -4)\to X'(-4,-2)\); \(Y(8, 2)\to Y'(2,8)\); \(Z(4, -9)\to Z'(-9,4)\)
(b) \((x,y)\to(y,x)\)