Translation refers to moving an object without changing it in any other ways. In the other hand, it is a term used in geometry to describe a function that moves an object a certain distance in fixed direction and gives the congruent image. It is not rotated, reflected or re-sized.
Translation using Co-ordinates
The coordinator of the point which is translated forward and translated upward is the translation by using co-ordinates.
It can be written as,
P(x, y) → P'(x+a, y+b)
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).
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gaurab baral
if it goes left then we should do - to which makes (x-a,y b) isnt it
Mar 14, 2017
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