Question

Given the vertices of ∆ABC are A (2,5), B (4,6) and C (3,1), find the vertices following each of the transformations FROM THE ORIGINAL vertices:

a. Rx-axis

b. Ry = 3


c. T<-2,5>

d. T<3,-6>


e. r(90◦, o)

Answer 1

Answer with explanation:

a)

Reflection over x-axis.

We know that when any point is reflected over the x-axis, then the rule of transformation that is applied is:

(x,y) → (x,-y)

Hence,

A (2,5)→ A'(2,-5)

B (4,6) → B'(4,-6)

and C (3,1) → C'(3,-1)

b)

Reflection about the line y=3

We know that any point after reflection is at a fixed distance from the line as it was before reflection.

Hence,

A (2,5)→ A'(2,1)

B (4,6) → B'(4,0)

and C (3,1) → C'(3,5)

c)

T<-2,5>

The rule that is applied to this translation is:

(x,y) → (x-2,y+5)

Hence,

A (2,5)→ A'(0,10)

B (4,6) → B'(2,11)

and C (3,1) → C'(1,6)

d)

T<3,-6>

The rule that is applied to this translation is:

(x,y) → (x+3,y-6)

Hence,

A (2,5)→ A'(5,-1)

B (4,6) → B'(7,0)

and C (3,1) → C'(6,-5)

e)

r(90◦, o)

It is a rotation of a point 90 degree clockwise about the origin.

Hence, the rule that describes this transformation is:

(x,y) → (y,-x)

Hence,

A (2,5)→ A'(5,-2)

B (4,6) → B'(6,-4)

and C (3,1) → C'(1,-3)

Answer 2

Part a)
R(x-axis) is the reflection of the original triangle ABC on the x-axis. The new coordinates are given as A' (2, -5), B' (4, -6), and C' (3, -1)

Part b)
R(y=3) is the reflection of the original triangle ABC on the line with equation y=3.
The new coordinates would be A' (2, 1), B' (4, 0), and C' (3, 5)

Part c)
T(-2, 5) is the translation of the original triangle ABC two units left and five units up. The new coordinates would be A'(0, 10), B' (2, 11), and C'(1, 6)

Part d)
T(3, -6) is the translation of the original triangle ABC three units right and six units down. The new coordinates would be A'(5, -1), B'(7, 0), and C'(6, -5)

Part e)
r(90°, 0) is the rotation of the original triangle ABC on the origin by 90° clockwise. The new coordinates would be A'(5, -2), B'(6, -4) and C'(1 -3)