Question

map the coordinates of the reflection of trapezoid qrst with vertices -1), r(6, -1), s(8, -6) and t(2, -6): in the line y = x

Explanation:

Step1: Recall reflection over \( y = x \)

The rule for reflecting a point \((x, y)\) over the line \( y = x \) is to swap the \( x \)-coordinate and the \( y \)-coordinate, so the image is \((y, x)\).

Step2: Reflect point \( R(6, -1) \)

Using the rule, for \( R(6, -1) \), swap \( x = 6 \) and \( y = -1 \). So the reflected point \( R' \) is \((-1, 6)\).

Step3: Reflect point \( S(8, -6) \)

For \( S(8, -6) \), swap \( x = 8 \) and \( y = -6 \). The reflected point \( S' \) is \((-6, 8)\).

Step4: Reflect point \( T(2, -6) \)

For \( T(2, -6) \), swap \( x = 2 \) and \( y = -6 \). The reflected point \( T' \) is \((-6, 2)\).

(Note: The vertex \( Q \) is partially given as \( Q(\_, -1) \), assuming \( Q \) has the same \( y \)-coordinate as \( R \) (since it's a trapezoid, likely \( Q \) is \((2, -1)\) to make \( QR \) and \( ST \) the bases). Reflecting \( Q(2, -1) \) would give \( Q'(-1, 2) \).)

Answer:

• If \( Q \) is \( (2, -1) \), reflected vertices are:

\( Q'(-1, 2) \), \( R'(-1, 6) \), \( S'(-6, 8) \), \( T'(-6, 2) \)

• For given vertices \( R(6, -1) \), \( S(8, -6) \), \( T(2, -6) \), their reflections are \( R'(-1, 6) \), \( S'(-6, 8) \), \( T'(-6, 2) \) respectively.