Question

!50 POINTS! (4 SIMPLE GEOMETRY QUESTIONS)

QUESTIONS BELOW:
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Answer 1

For 1st:

`\tt Blank \: 1\:=\boxed{(0,2)}\\\tt Blank \: 1\:=\boxed{(2,-2)}\\\tt Blank \: 1\:=\boxed{(4,4)}`

For 2nd: b. 6

For 3rd: b. 1

For 4th: c. reflection

Explanation:

For 1st Question:

The reflection of a point P (x, y) on the x-axis is a point P' (x, -y).

This means that the x-coordinate of the reflected point is the same as the original point, but the y-coordinate is the opposite sign.

P(x, y) on the x-axis is a point P'(x, -y).

R(0,-2)
`\tt ----------\longrightarrow` R'(0,2)

E(2,2 )
`\tt ----------\longrightarrow` E'(2,-2)

F(4,-4)
`\tt ----------\longrightarrow`F'(4,4)

`\hrulefill`For 2nd Question:

A regular hexagon has 6 lines of symmetry. These lines of symmetry divide the hexagon into 6 congruent parts.

• The lines of symmetry of a regular hexagon are the lines that pass through the center of the hexagon and connect opposite vertices or connect the midpoints of opposite sides.

Therefore, answer is b. 6

For 3rd Question:

If a trapezoid is located entirely in quadrant II, then all of its points will have a positive x-coordinate and a negative y-coordinate. When the trapezoid is reflected across the x-axis, the x-coordinates of all of its points will stay the same, but the y-coordinates will become positive. Therefore, the new trapezoid will be located entirely in quadrant I.

Quadrant II
`\tt ----------\longrightarrow` Quadrant I

(x, -y)
`\tt ----------\longrightarrow` (x, -y)

As you can see, the point (x, y) is located in quadrant II, but its reflection (x, -y) is located in quadrant I.

Therefore, the new trapezoid will be located in quadrant 1.

So the answer is b. 1.

`\hrulefill`

For 4th Question:

If all the coordinates of triangle ABC are from quadrant II to quadrant I, then the triangle must have been reflected across the x-axis.

This is because the x-coordinates of all the points in quadrant II are negative, but the x-coordinates of all the points in quadrant I are positive.

When a point is reflected across the x-axis, its x-coordinate stays the same, but its y-coordinate is negated.

Quadrant II
`\tt ----------\longrightarrow` Quadrant I

(x, y)
`\tt ----------\longrightarrow`(x, -y)

As you can see, the point (x, y) is located in quadrant II, but its reflection (x, -y) is located in quadrant I.

Therefore, the triangle must have been reflected across the x-axis.

So the answer is c. reflection.