Question

find the coordinates of the figure after reflecting in the x-axis. e(0, −1), f(8, −2), g(6, −7), h(1, −8) e’: f’: g’: h’:

Explanation:

Step1: Recall reflection over x - axis rule

The rule for reflecting a point \((x,y)\) over the \(x\) - axis is \((x,y)\to(x, - y)\). This is because when we reflect a point across the \(x\) - axis, the \(x\) - coordinate remains the same, and the \(y\) - coordinate changes its sign.

Step2: Find \(E'\)

For point \(E(0,-1)\), using the reflection rule \((x,y)\to(x, - y)\), where \(x = 0\) and \(y=-1\). Then \(-y=-(-1) = 1\). So \(E'\) has coordinates \((0,1)\).

Step3: Find \(F'\)

For point \(F(8,-2)\), with \(x = 8\) and \(y = - 2\). Using the rule \((x,y)\to(x, - y)\), we get \(-y=-(-2)=2\). So \(F'\) has coordinates \((8,2)\).

Step4: Find \(G'\)

For point \(G(6,-7)\), where \(x = 6\) and \(y=-7\). Applying the reflection rule, \(-y=-(-7) = 7\). So \(G'\) has coordinates \((6,7)\).

Step5: Find \(H'\)

For point \(H(1,-8)\), with \(x = 1\) and \(y=-8\). Using the rule \((x,y)\to(x, - y)\), we have \(-y=-(-8)=8\). So \(H'\) has coordinates \((1,8)\).

Answer:

\( E'(0, 1) \), \( F'(8, 2) \), \( G'(6, 7) \), \( H'(1, 8) \)