Soln:

Consttruction procedure:

To enlarge the triangle ABC with centre at O and scale factor 2 i. e. join O and A and produce it upto A' where OA' = 2OA. Also join O and B and O and C. Then produce OB and OC upto B' and C' where OB' = 2OB and OC' = 2OC. Then, if we join A', B' and C' then the triangle A'B'C' is the image of triangle ABC under the enlargemant with centre O and scale factor 2.

Soln:

Construction procedure:

Here, at first B, C and D are joined with O and then OB, OC and OD are produced upto B', C' and D' such that OB' = 3OB, OC' = 3OC and OD' = 3OD. Now, B', C' and D' are joined so that the triangle A'B'C'D' is the image of triangle BCD under the enlargement with centre at O and scale factor 3.

Soln:

Construction procedure:

Here, at first CO, DO and Eo are produced upto C', D' and E' respectively such that OC' =$$\frac{1}{2}$$OC, OD' =$$\frac{1}{2}$$OD and OE' =$$\frac{1}{2}$$OE.

Then C", D', E' and O are joined so that the parallelogram OC'D'E' is the image of parallelogram OECD under the enlargement with centre at O and scale factor $$-\frac{1}{2}$$

Soln:

Construction procedure:

Here, OF, OG, OH, OI and OJ are produced upto F', G', H', I' and J' as in the figure so that OF' = $$1\frac{1}{2}$$OF, OG' =$$1\frac{1}{2}$$OG, OH' =$$1\frac{1}{2}$$OH, OI' =$$1\frac{1}{2}$$ and OJ' =$$1\frac{1}{2}$$OJ. Then join O, F, G, I and J so that the hexagon OF'G'H'I'J' is the image of hexagon OFGHIJ under the enlargement with the centre at O and scale factor $$1\frac{1}{2}$$.

Soln:

Construction procedure:

Here, as in the figure, QO, RO and PO are produced upto Q', R' and P' respectively so that OQ' = 2OQ, OR' = 2OR and OP = 2OP'. Now O, Q', R' and P' are joined so that the figure OP'Q'R' is the image of the figure OPQR under the enlargement with centre at O and scale factor -2.

Soln:

Construction procedure:

Here, at first points B, C, D, E and F are joined with O. Then the points A', B', C', E', F' and G' are marked on the line OA, OB, OC, OD, OE, OF and OG respectively where OA' =$$\frac{3}{4}$$OA, OB' =$$\frac{3}{4}$$OB, OC' =$$\frac{3}{4}$$OC, OD' =$$\frac{3}{4}$$OD, OE' =$$\frac{3}{4}$$OE, OF' =$$\frac{3}{4}$$OF and OG' =$$\frac{3}{4}$$OG. Then the points A', B', C' D', E', F' and G' are joined so that the shaded figure A'B'C'E'F'G'F'G' is the image of the given figure under [0,$$\frac{3}{4}$$].

Soln:

Construction procedure:

Here, as in figure, points A, B, C, D and E are joined with O. Then points A', B', C', D' and E' are marked on the line OA, OB, OC, OD and OE respectively where OA' =$$\frac{1}{2}$$OA, OB' =$$\frac{1}{2}$$OB, OC' =$$\frac{1}{2}$$OC, OD' =$$\frac{1}{2}$$OD and OE' =$$\frac{1}{2}$$OE. Now, the points A', B', C', D' and E' are joined so that the pentagon A'B'C'D'E' is the image of the pentagon ABCDE under the enlargement with centre at O and scale factor$$\frac{1}{2}$$.

Soln:

Procedure: Here lines joining AA', BB; and CC; intersect at the point O. So, by the actual measurement,

OA = 4.2 cm

OA' = 7.2 cm

So, $$\frac{OA'}{OA}$$ = $$\frac{7.2}{4.2}$$ =$$\frac{12}{7}$$ = 1.7

∴ Scale factor = 1. 7

In symbol [0,$$\frac{12}{7}$$

Soln:

Procudure: Here, the lines AA', BB', CC' and DD' intersect at the point O. So that O is the centre of enlargement. Also from actual measurement,

OA' = 4 cm

OA = 1.5 cm

$$\frac{OA'}{OA}$$ =$$\frac{4}{1.5}$$ =$$\frac{8}{3}$$

∴ Scale factor =$$\frac{8}{3}$$

Soln: Here given, image of point A(1,p) is A' (q,8) and centre of enlargement (o,o) and scale factor k=2

We know that enlargement of P(x,y) under E[o,k] is P'(kx,ky)

Sp, A (1,p)→A'(2.1)2p)=A'(2,2p)

but, image of A is A'(q,8)

So, (2,2P)=(q,8)

or, 2=q amd 2p=8

∴ p=4 and q=2. Ans.

Soln: Here, the enlargement of rectangle ABCD under centre at origin and scale factor2 that is [O,2] by using P(x,y)→P'(kx,ky) is as follow:

A(-2,1)→A'(-2×2,1×2)=A'(-4,2)

B(-2,4)→B'(-2×2,4×2)=B'(-4,8)

C(4,4)→C'(4×2,4×2)-C'(8,8)

D(4,1)→D'(4×2,1×2)=D'(8,2)

Here, rectangle A'B'C'D' is the image of the rectangle under the enlargement [0,2], which is shown in the following graph paper.

Soln: Here, let the centre be P (2,1) and scale factor k = 2 and A (2,0) be a point. so,

∴ $$\overrightarrow{PA}$$=$$\begin{pmatrix} 2&-2 \\ o&-1 \\ \end{pmatrix}$$= $$\begin{pmatrix} 0 \\ -1 \\ \end{pmatrix}$$

∴ A (2,0)→ A' (2×0+2,2×-1+1)=A.(2,-1)

Also, Relation of P (2,1) and B (3,5)

$$\overrightarrow{PB}$$= $$\begin{pmatrix} 3&-2 \\ 5&-1 \\ \end{pmatrix}$$=$$\begin{pmatrix} 1 \\ 4 \\ \end{pmatrix}$$

∴ B (3,5)→ B' (2×1+2,2×4+1) = B'(4,9)

Relation of P (2,1) and C (4,-7) is

$$\overrightarrow{PC}$$=$$\begin{pmatrix} 4&-2 \\ -7&-1 \\ \end{pmatrix}$$=$$\begin{pmatrix} 2 \\ -8 \\ \end{pmatrix}$$

∴ C= (4,-7)→C'(2×2+2,2×-8+1)=C'(6,-15)

Relation of P (2,1) and D (-3,8) is

$$\overrightarrow{PD}$$=$$\begin{pmatrix} -3&-2 \\ 8&-1 \\ \end{pmatrix}$$=$$\begin{pmatrix} -5 \\ 7 \\ \end{pmatrix}$$

∴ D(-3,8)→D'(2×-5+2,2×7+1)=D'(-8,15)

Relation of P (2,1) and F (-5,-7) is

$$\overrightarrow{PF}$$= $$\begin{pmatrix} -5 &-2 \\ -7&-8 \\ \end{pmatrix}$$=$$\begin{pmatrix} -7 \\ -8 \\ \end{pmatrix}$$

∴ F(-5,-7)→F'(2×-7+2,2×-8+1)=F'(-12,-15)

Relation of P (2,1) and G (0,-11) is

$$\overrightarrow{PG}$$= $$\begin{pmatrix} 0&-2 \\ -11&-1 \\ \end{pmatrix}$$=$$\begin{pmatrix}-2 \\ -1 \\ \end{pmatrix}$$

∴ G(0,-11)→G'(2×(-2)+2,2×(-12)+1)=G'(-2,-23)

The graph of the given points and their images under the enlargement with centre at P(2,1) and scale factor 2 is given below:

Soln:

Construction procudure: Here as shown in the figure, R is joined with O, Here scale factor is 3, So the lines OP,OQ and OR are produced up to P', Q' and R' respectively so that OP'=3OP, OQ'=3OQ and OR' =3OR.Now, P',Q' and R', are joined so that the $$\triangle$$ P',Q',R' of the images of $$\triangle$$PQR is prepared with centre at O and scale factor 3.

Soln:

Enlargement of $$\triangle$$ ABC under E[0,2]:

A(2,4)→A'(2×2,4×2)=A'(4,8) ∴p(x,y)→P'(kx,ky)

B(-3,5)→B'(-3×2,5×2)=B'(-6,10)

C(-2,-3)→C'(-2×2,-3×2)=C'(-4,-6)

$$\triangle$$A'B'C' is the image of $$\triangle$$ABC under E[0,2].Which is shown in the following graph.

Soln:

Enlargement of $$\triangle$$ ABC under E[0,-2]:

A(2,4)→A'(2×-2,4×-2)=A'(-4,-8) ∴p(x,y)→P'(kx,ky)

B(-3,5)→B'(-3×-2,5×-2)=B'(6,-10)

C(-2,-3)→C'(-2×-2,-3×-2)=C'(4,6)

$$\triangle$$A'B'C' is the image of $$\triangle$$ABC under E[0,-2].Which is shown in the following graph.