Solution:

The translation is (x, y) → (x+2, y-2)

The translation tells you to add 2 to the x-value and subtract 2 from the y- value.

So, the image of (3, 2) → (3+2, 2-2) = (5, 0)

Solution:

To translate then the above triangle, we proceed as follows:

(x, y) → (x+9, y-7)

(-6, 6) → (-6 + 9, 6 -7) = (3, 1)

(-6, 1) → (-6 + 9, 1 - 7) = (3, -6)

(-1, 1) → (-1 + 9, 1-7) = (8, -6)

Solution:

The vertices of \(\triangle\) are A(0, 6), B(3, -2) and C(4, 0).

Translation vector (T) =\(\begin{pmatrix} 0\\5 \end{pmatrix}\)

When , T =\(\begin{pmatrix} a\\b \end{pmatrix}\), P(x, y) \(\rightarrow\) P'(x+a, y+a)

\(\therefore\) A(0, 6)\(\rightarrow\) A'(0+0, 6+5) = A'(0, 11)

B(3, -2)\(\rightarrow\) B'(3+0, -2+5) = B'(3, 3)

C(4, 0) \(\rightarrow\) C'(4+0, 0+5) = C'(4, 5)

\(\triangle\)ABC and \(\triangle\)A'B'C' are shown on the graph.

Solution:

According to the question,

Presenting the points A(4, -5) in the graph, under the translation of 3 units right and 4 units up.

A(4, -5) \(\rightarrow\) A'(4+3, -5+4) = A'(7, -1)

Here,

A(4, -5) \(\rightarrow\)A' (7, -1)

Showing in the graph.

Solution:

The translation tells us to add 5 to the x-value and add 4 units to the y-axis.

According to the Question,

\(\triangle\)ABC \(\triangle\)A'B'C'

A(4,5)\(\rightarrow\) A'(9,9)

B(1,3)\(\rightarrow\) B'(6,7)

C(4,3)\(\rightarrow\) C'(9,7)

Presenting in the graph.

Solution:

a) False

b) True

c) True

Solution:

Given points, P(-1, -3) and Q(4,5)

The translation is (x,y) \(\rightarrow\)(x+2,y-3)

The translation tells you to add 2 to the x-value and subtract 3 from the y-value.

P(-1,-3)\(\rightarrow\)P(-1 +2,-3+3) = P'(1,0)

Q (4,5)\(\rightarrow\)Q'(4+2,5+3) = Q'(6,8)

Presenting in the Graph.

Solution:

Given points are, A(1,0), B(4,5) and C(7,-2). The translation tells you to add 3 to the x-value and subtract 5 from the y-value.

So, A(1,0)\(\rightarrow\)A(1+3, 0-5) = A'(4, -5)

B(4,5)\(\rightarrow\)B'(4+3, 5-5) = B'(7,0)

C(7,-2)\(\rightarrow\)C'(7+3, -2-5)=C'(10,-7)

Presenting \(\triangle\)ABC in the graph.

Solution:

Let the given points (4,6), (7,5), (5,1), (2,2) be A, B, C and D

To translate the above triangle, we proceed as follows:

A(4,6)\(\rightarrow\)A'(4-4, 6-5) = A'(0,1)

B(7,5)\(\rightarrow\)B'(7-4, 5-5) = B'(3,0)

C(5,1)\(\rightarrow\)C'(5-4, 1-5)= C'(1-4)

D(2,2)\(\rightarrow\)D'(2-4, 2-5)=D'(-2, -3)

Presenting in graph.

Solution:

Here, points A(4,1) is translated to,

A(4,1)\(\rightarrow\)A' (4+5, 1+4) = A'(9,5)

Again, image A'(9,5) is translated to,

A'(9,5) \(\rightarrow\) A''(9+2, 5-5) = A''(11,0)

Showing in the graph:

####
Under the translation of the points (-3,5) to (4,5), how much units are required. Find in the graph?

Solution:

Let, a unit is added in x-axis and b unit is added to y-axis to translate the points (-3,5) to (4,5).

Therefore,

(-3 +a, 5 +b) = (4,5)

or, -3+a = 4 and 5+b =5.

or, a = 7 and b=0

Here, 7 units are added to the x-axis and 0 units are added to y-axis to translate the points(-3,5) to (4,5).

Presenting in the Graph.

Solution:

Translation on Vector T=(\(\frac{3}{4}\))

When translation on vector T=(\(\frac{a}{b}\)), then,

P(x,y) = P'(x+a, y+b)

P(3,1) = P'(3+3, 1+4) = P'(6,5)

Q(4,3) = Q'(4+3, 3+4) = Q' (7,7)

R(1,4) = R' (1+3, 4+4) = R' (4,8)

Solution:

Here, A(-2, 2), B(-3, -2), C(3, -2) and D(2, 2) is given,

The translation tells us subtract 3 to the x-value and add 4 units to the y-axis.

So, A(-2, 2)\(\rightarrow\) A(-5, 6)

B(-3, -2)\(\rightarrow\) B(-6, 2)

C(3, -2) \(\rightarrow\) C(0, 2)

D(2, 2)\(\rightarrow\) D(-1, 6)

Presenting in the graph.