Question

triangle xyz is reflected across the y - axis. the result is \\( \triangle x ^ { prime } y ^ { prime } z ^ { prime } \\), 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 \\( \
ightarrow \\) final coordinates \\( x ( 2,4 ) \
ightarrow x ^ { prime } ( \square, \square ) \\) \\( y ( 6,4 ) \
ightarrow y ^ { prime } ( \square, \square ) \\) \\( z ( 3, - 5 ) \
ightarrow z ^ { prime } ( \square, \square ) \\) (b) choose the general rule below that describes the reflection mapping \\( \triangle x y z \\) to \\( \triangle x ^ { prime } y ^ { prime } z ^ { prime } \\). \\( ( x, y ) \
ightarrow ( - y, - x ) \\) \\( ( x, y ) \
ightarrow ( - x, y ) \\) \\( ( x, y ) \
ightarrow ( - y, x ) \\) \\( ( x, y ) \
ightarrow ( x, - y ) \\) \\( ( x, y ) \
ightarrow ( - x, - y ) \\) \\( ( x, y ) \
ightarrow ( y, - x ) \\) \\( ( x, y ) \
ightarrow ( y, x ) \\)

Explanation:

Step1: Recall the rule for reflection across the y - axis

When a point \((x,y)\) is reflected across the \(y\) - axis, the \(y\) - coordinate remains the same, and the \(x\) - coordinate changes its sign. The rule is \((x,y)\to(-x,y)\).

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

For \(X(2,4)\), using the rule \((x,y)\to(-x,y)\), we substitute \(x = 2\) and \(y = 4\). So, \(X'\) has coordinates \((-2,4)\).

Step3: Apply the rule to point \(Y(6,4)\)

For \(Y(6,4)\), substituting \(x = 6\) and \(y = 4\) into the rule \((x,y)\to(-x,y)\), we get \(Y'(-6,4)\).

Step4: Apply the rule to point \(Z(3,-5)\)

For \(Z(3,-5)\), substituting \(x = 3\) and \(y=-5\) into the rule \((x,y)\to(-x,y)\), we obtain \(Z'(-3,-5)\).

Step5: Confirm the general rule

Since the rule for reflection across the \(y\) - axis is \((x,y)\to(-x,y)\) (by checking the transformation of each point \(X(2,4)\to X'(-2,4)\), \(Y(6,4)\to Y'(-6,4)\) and \(Z(3,-5)\to Z'(-3,-5)\)).

Answer:

(a) \(X(2,4)\to X'(-2,4)\), \(Y(6,4)\to Y'(-6,4)\), \(Z(3,-5)\to Z'(-3,-5)\)
(b) \((x,y)\to(-x,y)\)