A perpendicular line is drawn from point A, B and C to the axis of reflection.The line from A be AP, from B be BR and from C be CQ. E longate lines AP,BR and CQ so that AP =PA', BR=RB' and CQ be QC'. Now join the points A', B' and C' and the formed \(\triangle\)A'BC' is the image of\(\triangle\) ABC in the axis m.

A perpendicular line is drawn from point A, B, C and D to the axis of reflection. The line from A be AP, from B be BQ and from C be CR and D be DS. E longate lines AP, BQ, CR and DS so that AP =PA', BQ=QB', CR=RC' and DS =SD'. Now join the points A', B',C' and D' and the formed \(\triangle\)A'BC' is the image of\(\triangle\) A'B'C'D' in the axis m.

A perpendicular line is drawn from point A, B, C,D and E to the axis of reflection. The line from A be AP, from B be BQ and from C be CR, D be DS and E be ET. E longate lines AP,BQ,CR,DS and ET so that AP =PA', BQ=QB', CR=RC', DS =SD' and ET=TE'. Now join the points A', B',C' D' and E' and the formed \(\triangle\)A'BC'D'E' is the image of\(\triangle\) in the axis m.

Solution:

Plot the points A (-2, 5); B (-2, -1); C (-5, -4); D (-5, 2) on the graph paper. Now join AB, BC, CD and DA to get a parallelogram.

When reflected in y-axis, we get A' (2, 5); B' (2, -1); C' (5, -4); D' (5, 2). Now join A'B', B'C', C'D' and D'A'.

Thus, we get the parallelogram A'B'C'D as the image of the parallelogram ABCD in y-axis.

Solution:

Image of the following points when reflected in y-axis.

(i) The image of (-3 , 3) is (3 , 3).

(ii) The image of (2, 4) is (-2, 4).

(iii) The image of (-2 , -6) is (2, -6).

(iv) The image of (5, -7) is (-5, -7).

Solution:

(i) The image of A(-7, 9) is A' (7, 9).

(ii) The image of B (-3, -6) is B' (3, -6).

(iii) The image of C(4, 8) is C' (-4, 8).

(iv) The image of D (5, -7) is D' (-5, -7).

Solution:

(i) The image of A(1, 4) is A’ (-1, -4).

(ii) The image of b (-3, -7) is B’ (3, 7).

(iii) The image of C (-5, 8) is C’ (5, -8).

(iv) The image of D (6, -2) is D’ (-6, 2).

Solution:

Plot the points X (1, -5); Y (6, -1); Z(-4, -3) on the graph paper. Now join XY, YZ and ZX; to get a triangle XYZ.

When reflected in x-axis, we get X' (1, 5); Y' (6, 1); Z' (-4, 3). Now join X'Y', Y'Z' and Z'X'.

Thus, we get a triangle X'Y'Z' as the image of the triangle XYZ in x-axis.

Solution:

(i)The image of (-5 , 2) is (-5 , -2).

(ii) The image of (2, -7) is (2, 7).

(iii) The image of (3, 4) is (3, -4).

(iv) The image of (-5, -4) is (-5, 4).

Solution:

The image of P (-6, -8) is P' (-6, 8).

The image of Q (4, 6) is Q' (4, -6) .

The image of R (-3, 2) is R' (-3, -2).

The image of S (3, -2) is S' (3, 2).

Solution:

When rotated through 90° about the origin in clockwise direction, the new position of the above points are;

(i) The new position of point P (3, 7) will become P' (7, -3)

(ii) The new position of point Q (-4, -7) will become Q' (-7, 4)

(iii) The new position of point R (-7, 5) will become R' (5, 7)

(iv) The new position of point S (2, -4) will become S' (-4, -2)

Solution:

P(x,y)=P'(x,-y)

A'(1,3)\(\rightarrow\)A'(1,-3)

B(4,5)\(\rightarrow\)B'(4,-5)

C(6,2)\(\rightarrow\)C'(6,-2)

Solution:

P(x,y)=P'(-x,y)

P(2,3)=P'(-2,3)

Q(4,5)=Q'(-4,5)

R(6,2)=R'(-6,2)

Solution:

P(x,y)\(\rightarrow\)P'(x,-y)

A(3,2)\(\rightarrow\)A'(3,-2)

B(5,6)\(\rightarrow\)B'(5,-6)

C(8,1)\(\rightarrow\)C'(8,-1)

Solution:

P(x,y)\(\rightarrow\)P'(-x,y)

A(-1,5)\(\rightarrow\)A'(1,5)

B(4,1)\(\rightarrow\)B'(-4,1)

C(-4,-3)\(\rightarrow\)C(4,-3)

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A (1,2), B(4,5) and C(5,1) are vertices of \(\triangle\)ABC.Rotate \(\triangle\) ABC through 90^{o}.

^{o}.

Solution:

P(x,y)\(\rightarrow\)P'(y,-x)

A(1,2)\(\rightarrow\)A'(2,-1)

B(4,5)\(\rightarrow\)B'(5,-4)

C(5,1)\(\rightarrow\)C'(1,-5)

Solution:

The vertices of \(\triangle\)ABC are A(1,4), B(4,6) and C(5,0).

After reflection in x-axis,

P(x,y)\(\rightarrow\)P'(x,-y)

A(1,4)\(\rightarrow\)A'(1,-4)

B(4,6)\(\rightarrow\)B'(4,-6)

C(5,0)\(\rightarrow\)C'(5,0)

Now,

we draw \(\triangle\)ABC and \(\triangle\)A'B'C' in the same graph paper as below.

Solution:

The vertices of \(\triangle\)ABC are A(2,1),B(5,1), and C(4,-2)

A(2,1)\(\rightarrow\)A'(1,2)

B(5,1)\(\rightarrow\)B'(1,5) and

C(4,-2)\(\rightarrow\)C'(-2,4)

Now,

we draw \(\triangle\) ABC and \(\triangle\)A'B'C' in the same graph paper as below.

Solution:

The vertices of \(\triangle\)ABC are A(5,6),B(7,4), and C(-3,2)

After reflection in y-axis,

P(x,y)\(\rightarrow\)P'(-x,y)

A(5,6)\(\rightarrow\)A'(-5,6)

B(7,4)\(\rightarrow\)B'(-7,4)

C(-3,2)\(\rightarrow\)C'(3,2)

Now, we draw \(\triangle\) ABC and \(\triangle\)A'B'C' in the same graph paper as below.

Solution:

Using graph, the image of the given co.ordinates reflecting them in x-axis are as follows:

1. A(1,2)\(\rightarrow\)A'(1,-2)

2. B(-2,3)\(\rightarrow\)B'(-2,-3)

3. C(4,-5)\(\rightarrow\)C'(4,5)

4. D(-6,6)\(\rightarrow\)D'(-6,-6)

5. E(-5,-4)\(\rightarrow\)E'(-5,4)

6. F(-2,5)\(\rightarrow\)F'(-2,-5)

7. G(9,-8)\(\rightarrow\)G'(9,8)

8. H(-3,-9)\(\rightarrow\)H'(-3,9)

9. I(-10,12)\(\rightarrow\)I'(-10,-12)

10. J(7,8)\(\rightarrow\)J'(7,-8)

Solution;

Under reflection about x-axis,

P(x, y) \(\rightarrow\) P'(x, -y)

\(\therefore\) A(1, 3) \(\rightarrow\) A'(1, -3)

B(4, 5) \(\rightarrow\) B'(4, -5) and

C(6, 2)\(\rightarrow\) C'(6, -2)

\(\triangle\) ABC and its image\(\triangle\) A'B'C' are shown on the graph.