IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

heavensblaze

a point (4,1) undergoes following tranformations:

a) reflection about y=x

b) tranlation by 2 units in direction of x-axis

c) rotation about origin by an angle $\pi/4$ in anticlockwise sense

Find the coordinates of the point after all these transformations.

joyfrancis

Point after reflection about y=x is A(1,4)

After translating 2 units along the x axis the co-ordinates are : B(3,4)

|OB| = 5 ; and the slope of the line OB = 4/3 ;and if the angle made by OB with the pos x axis=@ then tan@ = 4/3

if OB is rotated pi/4 in the counterclockwise dir. B point would become

5cos(pi/4+@),5sin(pi/4+@)

= (-1/root2 , 7/root2)....(ans)

heavensblaze

your ans is correct could u tell me the solution

allamraju

After the transformations

a)(1,4) b)(3,4)[taking it as along +ve direction of X-axis]

c)The slope of line joining the origin and itself is 4/3.After rotating,it makes an angle Tan^-14/3+Tan^-11=Tan^-1(-7).So,cosk=-1/5rt2 and sink=7/5rt2.As it is at a distance of 5 units from origin,using parametric eqn.s,it is (5cosk,5sink)=(-1/rt2,7/rt2)

mukundmadhav

After a) (1,4) (Interchange y and x)
b) If you mean in direction of positive x axis, then (3,4)
c) Distance from origin is now 5 units and slope is 4/3
New slope will be (4/3 + 1)/(1-4/3) = -7
New coordinates will be (5x(-7)/root50,5x1/root50)

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After the transformations

a)(1,4) b)(3,4)[taking it as along +ve direction of X-axis]

c)The slope of line joining the origin and itself is 4/3.After rotating,it makes an angle Tan-14/3+Tan-11=Tan-1(-7).So,cosk=-1/5rt2 and sink=7/5rt2.As it is at a distance of 5 units from origin,using parametric eqn.s,it is (5cosk,5sink)=(-1/rt2,7/rt2)

c)The slope of line joining the origin and itself is 4/3.After rotating,it makes an angle Tan^-14/3+Tan^-11=Tan^-1(-7).So,cosk=-1/5rt2 and sink=7/5rt2.As it is at a distance of 5 units from origin,using parametric eqn.s,it is (5cosk,5sink)=(-1/rt2,7/rt2)