The point (1, 3) undergoes the following transformations successively
i) Reflection with respect to the line y=X
ii) translation through 3 units along the positive direction of the X-axis
iii) Rotation through an angle 6
about the origin in the clockwise direction. The final position of the point P is
O a.
-5
O b.
6+73 -673
2
2
2
(ha
(+
(0/3.1, 6173
(3+1, 43.9)
O c. 63–1 6+3
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Given Point A(4,1)
(i) reflection about y=x
Hence putting coordinates of point A in given eq y=4,x=1
The point becomes A(1,4)
(ii) translation through distance of 2 units in positive X axis
A(1+2,4)⇒A(3,4)
(iii) rotation of point through and angle
4
π
in anticlockwise about origin O
After rotation Point will be A(rcos(α+
4
pi
),rsin(α+
4
pi
))
Converting into polar form
r=
3
2
+4
2
=5
tanα=
3
4
Hence cosα=
5
3
sinα=
5
4
cos(α+
4
π
)=cosαcos
4
π
−sinαsin
4
π
cos(α+
4
π
)=
5
3
2
1
−
5
4
2
1
cos(α+
4
π
)=
5
2
−1
sin(α+
4
π
)=sinαcos
4
π
+cosαsin
4
π
sin(α+
4
π
)=
5
4
2
1
+
5
3
2
1
sin(α+
4
π
)=
5
2
7
A(5×
5
2
−1
,5×
5
2
7
)
A(−
2
1
,
2
7
)
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