Question

Reflection across x = 4
F(3,-5), C(3,-4), P(5,-4)

Answer 1

F' = (5, -5)

C' = (5, -4)

P' = (3, -4)

Explanation:

When an object is reflected over a vertical line, the x-value changes but the y-value does not change.

To reflect points across x = 4, determine the distance of the x-value of each of the points to the line of reflection by subtracting 4 from the x-values:

• F: 3 - 4 = -1
• C: 3 - 4 = -1
• P: 5 - 4 = 1

Point F is located 1 unit to the left of the line of reflection. Therefore, the reflection of this point is 1 unit to the right of the line of reflection. So the reflection of point F is 2 units to the right of point F:

• F' = (3+2, -5) = (5, -5)

As point C has the same x-value as point F, the x-value of its reflection in the line of reflection will be the same as point F:

• C' = (3+2, -4) = (5, -4)

Finally, point P is located 1 unit to the right of the line of reflection. Therefore, the reflection of this point is 1 unit to the left of the line of reflection. So the reflection of point P is 2 units to the left of point P:

• P' = (5-2, -4) = (3, -4)

Answer 2

Given:

• Points F(3,-5), C(3,-4), P(5,-4)

The triangle FCP is reflected across the line x = 4.

See the attached with both FCP and F'C'P' triangles shown.

The x-coordinates are 1 unit apart from the x = 4 line, hence the new coordinates are 2 units less or greater than the original one and the y-coordinates remain as is.

Reflected points:

• F(3, -5) → F'(5, - 5)
• C(3, - 4) → C'(5, - 4)
• P(5, - 4) → P'(3, - 4)