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.

tr

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.

f
Since T is 2 units above the x-axis, T′ is 2 units below the x-axis. T′ has coordinates (-3,-2).

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.

z

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.

z



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.

z

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.

z

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.

z

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.

z

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.

xz