Soln:

Construction procedure:

(i) Draw AA' from the point A which is parallel and equal to \(\overrightarrow{P}\).

(ii) Draw BB' from the point B which is parallel and equal to\(\overrightarrow{P}\).

(iii) Also, draw CC' from the point C which is parallel and equal to\(\overrightarrow{P}\).

Then join A', B' and C'. So that the triangle A'B'C' is the image of triangle ABC which is shown in the figure by shaded portion.

Soln:

Construction procedure:

As shown in the above figure, EE' is drawn from the point E which is parallel and equal to\(\overrightarrow{P}\). Similarly, FF' and DD' are drawn from the points F and D which are also equal and parallel to\(\overrightarrow{P}\). Then D', E' and F' are joined so that the triangle D'E'F' is the image of triangle DEF under translation through\(\overrightarrow{p}\).

Soln:

Construction procedure:

As shown in the above figure, GG', HH' and II' are drawn from the points G, H and I respectively. which are equal and parallel to\(\overrightarrow{P}\). Then by joining G', H' and I' we get the image triangle G'H'T' of the triangle GHI under the translation vector\(\overrightarrow{P}\).

Soln:

Construction procedure:

Here, in the figure, AB and DC are parallel and equal to the given vector\(\overrightarrow{P}\). So, images of A and D are A' and D' at B and C respectively. Again, CC' and BB' are drawn from the point C and B which are parallel and equal to the vector\(\overrightarrow{P}\). Then join A', B', C', and D' so that the parallelogram A"B'C'D' is the image of parallelogram under the translation through the vector\(\overrightarrow{P}\).

Soln:

Let, E \(\begin{pmatrix}x\\y\\ \end{pmatrix}\) translates the point (5, 7) into (3,9)

Then we have, \(\begin{pmatrix}x\\y\\ \end{pmatrix}\) + \(\begin{pmatrix}5\\7\\ \end{pmatrix}\) =\(\begin{pmatrix}3\\9\\ \end{pmatrix}\)

or, \(\begin{pmatrix}x\\y\\ \end{pmatrix}\) =\(\begin{pmatrix}3\\9\\ \end{pmatrix}\) -\(\begin{pmatrix}5\\7\\ \end{pmatrix}\) =\(\begin{pmatrix}3-5\\9-7\\ \end{pmatrix}\) =\(\begin{pmatrix}-2\\2\\ \end{pmatrix}\)

∴ Translation component =\(\begin{pmatrix}-2\\2\\ \end{pmatrix}\)

Again, translation of P(8, 4) by T =\(\begin{pmatrix}-2\\2\\ \end{pmatrix}\) is

∴ P'(y', x') = (8 - 2, 4 + 2) = (6, 6). Ans

Soln:

Given, translation component =\(\begin{pmatrix}3\\5\\ \end{pmatrix}\)

Then image of triangle OPQ is given as:

O(0, 0)→ O'(3 + 0, 5 + 0) = O'(3, 5)

P(6, 3)→ P'(3 + 6, 5 + 3) = P'(9, 8)

Q(4, 7)→ Q'(3 + 4, 5+7) = Q'(7, 12)

The graph of triangle OPQ and its image triangle O'P'Q' are as through below: Where the image is the shaded region.

Soln:

Let the component\(\begin{pmatrix}x\\y\\ \end{pmatrix}\) translates A(-3, -1)→ C(5, 1)

So, \(\begin{pmatrix}x\\y\\ \end{pmatrix}\) + \(\begin{pmatrix}-3\\-1\\ \end{pmatrix}\) =\(\begin{pmatrix}5\\1\\ \end{pmatrix}\)

or,\(\begin{pmatrix}x-3\\y-1\\ \end{pmatrix}\) =\(\begin{pmatrix}5\\1\\ \end{pmatrix}\)

or, x - 3 = 5 ∴x = 8

or, y - 1 = 1 ∴y = 2

∴ component of \(\overrightarrow{AC}\)\(\begin{pmatrix}8\\2\\ \end{pmatrix}\). Ans

Alternatively,

translation vector = \(\overrightarrow{AC}\)

= \(\overrightarrow{OC}\)-\(\overrightarrow{OA}\)

=\(\begin{pmatrix}5\\1\\ \end{pmatrix}\)-\(\begin{pmatrix}-3\\-1\\ \end{pmatrix}\)

=\(\begin{pmatrix}8\\2\\ \end{pmatrix}\)

Soln:

We have,\(\overrightarrow{AC}\) =\(\begin{pmatrix}8\\2\\ \end{pmatrix}\).

Now, ytranslation of parallelogram ABCD by**\**(\begin{pmatrix}8\\2\\ \end{pmatrix}\), we get

A(-3, -1)→ A'(8-3, 2-1) = A'(5, 1)

B(-1, -1)→ B'(8 - 1, 2 - 1) = B'(7, 1)

C(5, 1)→ C'(8 + 5, 2 + 1) = C'(13, 3)

and D(3, 1) → D'(8 + 3, 2 + 1) = D'(11, 3)

The graph of parallelofram ABCD and its image parallelogram A'B'C'D' are as shown in graph where the image is shaded.

Soln:

Here, the translation compponent \(\begin{pmatrix}x\\y\\ \end{pmatrix}\) translates A(-3, 0) to D(3, -1). So, we have

\(\begin{pmatrix}x\\y\\ \end{pmatrix}\) +\(\begin{pmatrix}-3\\0\\ \end{pmatrix}\) = \(\begin{pmatrix}3\\-1\\ \end{pmatrix}\)

or,\(\begin{pmatrix}x-3\\y+0\\ \end{pmatrix}\) =\(\begin{pmatrix}3\\-1\\ \end{pmatrix}\)

or, x - 3 = 3 ∴ x = 6

and y + 0 = -1 ∴ y = -1

Soln:

Given translation component =\(\begin{pmatrix}4\\7\\ \end{pmatrix}\)

A(-1, 1)→ A'(4, -1), 7+1) = A'(3, 8)

∴ T(A) =(3, 8) [ p(a, b)→ P'(a+x, b+y) where T =\(\begin{pmatrix}x\\y\\ \end{pmatrix}\) ]

B(-1, -3)→B'(4-1, 7-3) = B'(3, 4)

∴T(B) = (3, 4)

C(-5, 7)→ C'(4-5, 7+7) = C'(-1, 14)

∴T(C) = (-1`, 14)

D(6, 0)→ D'(6+4, 0+7) = D'(10, 7)

∴T(D) = (10, 7)

E(0, 7)→ E'(0+4, 7+7) = E'(4, 14)

∴ T(E) = (4, 14)

F(6, -6)→ F'(4+6, -6 +7) = F'(10, 1)

∴ T(F) = (10, 1)

G(8, 8)→ G'(4+8, 8+7) = G'(12, 15)

∴ T(G) = (12, 15)

H(5, 9)→ H'(5+4, 9+7) = H'(9, 16)

∴T(H) =(9, 16)

I(2, 11)→ I'(2+4, 11+7) = I'(6, 18)

∴ T(I) = (6, 18)

j(-3, -3)→ J'(4-3, 7-3 ) = J'(1, 4)

∴ T(J) = (1, 4)

Here, the component vector =(\(\frac{6}{-1}\)) which translate quadrilateral ABCD as follows:

A(-3,0)→A'(6-3,-1+0)=A'(3,-1)

B(-2,2)→B'(6-2,-1+2)=B'(4,1)

C(1,3)→C'(6+1,-1+3)=C'(7,2)

D(3,-1)→D'(6+3,-1-1)=D(9,-2)

∴The graph of quadrilateral ABCD and its image A'B'C'D' are shown below where the shaded region gives the image.

**Construction procedure: **Here from the points Q,R,S,T,U,V,W,X,Y and Z the lines QQ',RR',TT', UU',VV',WW',XX',YY',and ZZ' are drawn equal and parallel to the given vector \(\overrightarrow{P}\).Then the points Q',R',S',T',U',V',W',X',Y', and Z', are joined so that the figure Q'R'S'T'U'V'W'X'Y'Z' is the image of the given figure QRSTUVWXYZ under the translation vector \(\overrightarrow{P}\).

**Construction procedure:**As shown in the above figure from the points E, F,G,H,I,J,K,L,M, and N, lines EE',FF',GG',HH',II',JJ',KK',LL',MM',and NN', are drawn equal and parellel to the vector\(\overrightarrow{P}\).Then the points E',F',G',H',I',J',K',L',M' and N' are joined.Then the figure E'F'G'H'I'J'K'L'M'N' is the image of the figure EFGHIJKLMN under the given vector\(\overrightarrow{P}\).