Solution:

P(6, -4) $$\rightarrow$$ P' = (x, y)
x = 6, y = -4
or, P'(7, -2) = (x + a, y + b)
or, P'(7, -2) = (6 + a, -4 + b)
Equating the corresponding value
or, 6 + a = 7 or, -4 + b = -2
or, a = 7 - 6 or, b = -2 + 4
$$\therefore$$ a = 1 $$\therefore$$ b = 2
Here, The vector is $$\begin{pmatrix} 1 \\ 2 \end{pmatrix}$$.
Now,
or, A(-4, 6) = A'(x + a, y + b)
or, A(-4, 6) = A'(-4 + 1, 6 + 2)
or, A(-4, 6) = A'(-3, 8)
$$\therefore$$ The tarnslation of given point is A'(-3, 8).

Solution:

Here,
A(6, 4) = A'(x, y)
or, x = 6, y = 4
or, A'(3, 2) = (x + a, y + b)
or, A'(3, 2) = (6 + a, 4 + b)
Equalating the corresponding value
or, 6 + a = 3 or, 4 + b = 2
or, a = 3 - 6 or, b = 2 - 4
$$\therefore$$ a = -3 $$\therefore$$ b = -2
$$\therefore$$ The translation vector is $$\begin{pmatrix} -3 \\ -2 \end{pmatrix}$$
Now,
or, Q(2, -4) = Q'(x + a, y + b)
or, Q(2, -4) = Q'(2 - 3, -3 - 2)
or, (2, -4) = Q'(-1, -6)
$$\therefore$$ The image of Q' is (-1, -6)

Reflect the point over the line.

Since G is 5 units to the right of the y-axis, G′ is 5 units to the left of the y-axis. G′ has coordinates (-5,-5).

Reflect the point over the line.

Solution:

Here,
P(x, y)P' (x + a, y + b)
P(6, 4)P' (4 + (-2), -7 - 4) = P'(2, -11)
P(4, -7)P' (4 + 3, -7 - 1) = P' (7, -8)

Solution:

Here P(4, 5) = P(x, y)

$$\therefore$$ x = 5, y = 5

P'(2, 1) = P'(x+a, y+b)

or, P'(2,1) = P'(4+a, 5+b)

$$\therefore$$ 4+a = 2 and 5+b = 1, equating the corresponding elements.

or, a = 2-4 or, b = 1-5

or, a = -2 or, b = -4

$$\therefore$$ Translation vector =$$\begin{pmatrix} a\\ b\end{pmatrix}$$ =$$\begin{pmatrix} -2 \\ -4 \end{pmatrix}$$

Now, Q(3, -5)$$\rightarrow$$ Q'(x+a, y+b) = Q'(3-2, -5-4)

= Q'(1, -9)

$$\therefore$$The image of Q is Q'(1, -9).

Solution:

The vertices of $$\triangle$$ are A(2, 5), B(6, -4) and C(8, 0).

Translation vector (T) =$$\begin{pmatrix} 0\\3 \end{pmatrix}$$

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

$$\therefore$$ A(2, 5) $$\rightarrow$$ A'(2+0, 5+3) = A'(2, 8)

B(6, -4)$$\rightarrow$$ B'(6+0, -4+3) = B'(6, -1)

C(8, 0) $$\rightarrow$$ C'(8+0, 0+3) = C'(8, 3)

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

Solution:

P(3, -4) $$\rightarrow$$ P' = (x, y)
x = 3, y = -4
or, P'(4, -1) = (x + a, y + b)
or, P'(4, -1) = (3 + a, -4 + b)
Equating the corresponding value
or, 3 + a = 4 or, -4 + b = -1
or, a = 4 - 3 or, b = 4 - 1
$$\therefore$$ a = 1 $$\therefore$$ b = 3
Here, The vector is $$\begin{pmatrix} 1 \\ 3\end{pmatrix}$$.
Now,
or, A(-2, 3) = A'(x + a, y + b)
or, A(-2, 3) = A'(-2 + 1, 3 + 3)
or, A(-2, 3) = A'(-1, 6)
$$\therefore$$ The tarnslation of given point is A'(-1, 6).

Solution:

The vertices of $$\triangle$$ are A(0, 4), B(3, -2) and C(6, 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, 4)$$\rightarrow$$ A'(0+0, 4+5) = A'(0, 9)

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

C(6, 0) $$\rightarrow$$ C'(6+0, 0+5) = C'(6, 5)

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

Solution:

The vertices of $$\triangle$$ are A(2, 3), B(4, 5) and C(6, 4).

Translation vector (T) = $$\begin{pmatrix} 2\\3 \end{pmatrix}$$

When, T = $$\begin{pmatrix} a \\ b \end{pmatrix}$$, then

P(x, y) $$\rightarrow$$ P'(x+a, y+a)

$$\therefore$$ A(2,3) $$\rightarrow$$ A'(2+2, 3+3) = A'(4, 6)

or, B(4, 5) $$\rightarrow$$ B'(4+2, 5+3) = B'(6, 8)

or, C(6, 4) $$\rightarrow$$ C'(6+2, 4+3) = C'(8, 7)

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

Solution:

The vertices of $$\triangle$$ are A(2, 5), B(1, -4) and C(3, 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(2, 5) $$\rightarrow$$ A'(2+2, 5+3) = A'(4, 8)

or, B(1, -4) $$\rightarrow$$ B'(1+2, -4+3) = B'(3, -1)

or, C(3, 0) $$\rightarrow$$ C'(3+2, 0+3) = C'(5, 3)

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

Solution:

The vertices of $$\triangle$$ are X(3, 1), Y(4, 3) and Z(1, 4).

Translation vector (T) = $$\begin{pmatrix} 3\\4 \end{pmatrix}$$

When, T = $$\begin{pmatrix} a\\b \end{pmatrix}$$, then

P(x, y) $$\rightarrow$$ P'(x+a, y+a)

$$\therefore$$ X(3, 1) $$\rightarrow$$ X'(3+3, 1+4) = X'(6, 5)

or, Y(4, 3) $$\rightarrow$$ Y'(4+3, 3+4) = Y'(7, 7)

or, Z(1, 4) $$\rightarrow$$ Z'(1+3, 4+4) = Z'(4, 8)

$$\triangle$$XYZ and $$\triangle$$X'Y'Z' are shown on the graph.

Solution:

The vertices of $$\triangle$$ are A(-2, 3), B(3, 4) and C(2, 1)

Translation vector (T) = $$\begin{pmatrix} 2\\3 \end{pmatrix}$$

When, T = $$\begin{pmatrix} a\\b \end{pmatrix}$$, then

P(x, y) $$\rightarrow$$ P'(x+a, y+a)

$$\therefore$$ A(-2,3) $$\rightarrow$$ A'(-2+2, 3+3) = A'(0, 6)

or, B(3, 4) $$\rightarrow$$ B'(3+2, 4+3) = B'(5, 7)

or, C(2, 1) $$\rightarrow$$ C'(2 +2, 1+3) = C'(4, 4)

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

Solution:

The vertices of $$\triangle$$ are P(2, 5), Q(3, -4) and R(4, 1).

Translation vector (T) = $$\begin{pmatrix} 1\\3 \end{pmatrix}$$

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

$$\therefore$$ P(2, 5) $$\rightarrow$$ P'(2+1, 5+3) = P'(3, 8)

or, Q(3, -4) $$\rightarrow$$ Q'(3+1, -4+3) = Q'(4, -1)

or, R(4, 1) $$\rightarrow$$ R'(4+1, 1+3) = R'(5, 4)

$$\triangle$$PQR and $$\triangle$$P'Q'R' are shown on the graph.