FIND TRANSLATION OF A POINT ALONG THE GIVEN VECTOR

A translation moves an object from one place to another. Every point on the object moves the same distance in the same direction.

If P(x, y) is translated

h units in the x-direction and

k units in the y-direction

to become P'(x', y'), then

x' = x + h and y' = y + k

x'  =  x + h and y'  =  y + k are called transformation equations.

Example 1 :

Triangle OAB with vertices O(0, 0), A(2, 3) and B(-1, 2) is translated under the vector <3, 2>. Find the image vertices and illustrate the object and the image.

Solution :

To find vertices of image after translation, we use 

x'  =  x+h and y'  =  y+k

Here (h, k)  ==>  (3, 2)

x'  =  x+3 and y'  =  y+2

Example 2 :

Find the image equation when 2x-3y  =  6 is translated under the vector <-1, 2>.

Solution :

Let us find any two point on the straight line 2x-3y  =  6.

Converting 2x-3y  =  6 into intercept form, we get

(2x/6)-(3y/6)  =  6/6

x/3 - y/2  =  1

Point on x-intercept is (3, 0).

Point on y-intercept is (0, -2).

Here (h, k)  ==>  (-1, 2)

x'  =  x-1 and y'  =  y+2

Example 3 :

Find the image of y = 2x² under a translation with vector <3, -2>.

Solution :

Let us find any three points on the parabola.

Vertex of the parabola is (0, 0)

If x  =  -1, then y  =  2

If x  =  1, then y  =  2

So three point on the parabola are V(0, 0), A(-1, 2) and B(1, 2)

The image has to be translated along the vector <3, -2>.

(h, k) ==>  (3,-2)

x'  =  x+3 and y'  =  y-2

To find equation of new cure, let us find value of x and y from x'  =  x+3 and y'  =  y-2

x  =  x'-3 and y  =  y+2

Given curve :

y  =  2x²

Equation of curve after translation :

y+2  =  2(x-3)²

y+2  =  2(x²-6x+9)

y+2  =  2x²-12x+18

y  =  2x²-12x+16

Example 4 :

Find the image of xy  =  5 under a translation with vector <-4, 1>.

Solution :

xy  =  5

Finding 4 point on the rectangular hyperbola,

y  =  5/x

If x  =  1, then y  =  5  ==> A(1, 5)

If x  =  5, then y  =  1  ==> B(5, 1)

If x  =  -1, then y  =  -5  ==> C(-1, -5)

If x  =  -5, then y  =  -1  ==> D(-5, -1)

Here (h, k)  ==>  (-4, 1)

x'  =  x-4 and y'  =  y+1

Equation of curve before translation :

xy  =  5

Equation of curve after translation :

x'  =  x-4 and y'  =  y+1

x  =  x'+4 and y  =  y'-1

By applying x as x'+4 and y as y-1, we get

(x+4)(y-1)  =  5

xy-4x+4y-4-5  =  0

xy-4x+4y-9  =  0

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