Question

write the coordinates of the vertices after a reflection over the y-axis.
r(( ), ( ))
s(( ), ( ))
t(( ), ( ))
u(( ), ( ))

Explanation:

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

The rule for reflecting a point \((x,y)\) over the \(y\) - axis is \((x,y)\to(-x,y)\).

Step2: Find the coordinates of \(R\)

From the graph, the coordinates of \(R\) are \((-10,-4)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(R(-10,-4)\), we have \(x=-10\) and \(y = - 4\). Then \(R'\) has \(x\) - coordinate \(-(-10)=10\) and \(y\) - coordinate \(-4\). So \(R'(10,-4)\).

Step3: Find the coordinates of \(S\)

From the graph, the coordinates of \(S\) are \((-10,2)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(S(-10,2)\), we have \(x=-10\) and \(y = 2\). Then \(S'\) has \(x\) - coordinate \(-(-10)=10\) and \(y\) - coordinate \(2\). So \(S'(10,2)\).

Step4: Find the coordinates of \(T\)

From the graph, the coordinates of \(T\) are \((0,3)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(T(0,3)\), we have \(x = 0\) and \(y=3\). Then \(T'\) has \(x\) - coordinate \(-0 = 0\) and \(y\) - coordinate \(3\). So \(T'(0,3)\).

Step5: Find the coordinates of \(U\)

From the graph, the coordinates of \(U\) are \((0,-3)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(U(0,-3)\), we have \(x = 0\) and \(y=-3\). Then \(U'\) has \(x\) - coordinate \(-0 = 0\) and \(y\) - coordinate \(-3\). So \(U'(0,-3)\).

Answer:

\(R'(10,-4)\), \(S'(10,2)\), \(T'(0,3)\), \(U'(0,-3)\)