FIND TRANSLATION OF A POINT ALONG THE GIVEN VECTOR

Problem 1 :

Find the image equation when 3x+2y  =  8 is translated under the vector <-1, 3>.

Solution :

Let us find any two points on the straight line 3x+2y  =  8

Converting 3x+2y  =  8 to intercept form, we get

(3x/8)+(2y/8)  =  8/8

x/(8/3) + y/4  =  1

Point on x-intercept is A(8/3, 0).

Point on y-intercept is B(0, 4).

Here (h, k)  ==>  (-1, 3)

x'  =  x-1 and y'  =  y+3

Before translation

A(8/3, 0)

B(0, 4)

After translation

A'(8/3-1, 0+3)  =>  A'(5/3, 3)

B'(0-1, 4+3)  ==>  B'(-1, 7)

Problem 2 :

Find the image equation when 2x-y  =  6 is translated under the vector <-3, 0>.

Solution :

Let us find any two points on the straight line 2x-y  =  6

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

(2x/6)-(y/6)  =  6/6

x/3 - y/6  =  1

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

Point on y-intercept is B(0, -6).

Here (h, k)  ==>  (-3, 0)

x'  =  x-3 and y'  =  y+0

Before translation

A(3, 0)

B(0, -6)

After translation

A'(3-3, 0+0)  =>  A'(0, 0)

B'(0-3, -6+0)  ==>  B'(-3, -6)

Problem 3 :

Find the image equation when y  =  x2 is translated under the vector <0, 3>.

Solution :

Let us find any three points on the parabola y  =  x2

Finding vertex  ==>  V(0, 0)

If x  =  -2, then y  =  4  ==>  A(-2, 4)

If x  =  2, then y  =  4  ==>  B(2, 4)

Here (h, k)  ==>  (0, 3)

Before translation

V(0, 0)

A(-1, 1)

B(1, 1)

After translation

V'(0-0, 0+3)  =>  V'(0, 3)

A'(-1-0, 1+3)  ==>  A'(-1, 4)

A'(1-0, 1+3)  ==>  A'(1, 4)

Problem 4 :

Find the image equation when xy  =  -8 is translated under the vector <3, -2>.

Solution :

Let us find any three points on the rectangular parabola xy  =  -8

Finding 4 point on the rectangular hyperbola,

y  =  -8/x

If x = -2, then y = 4  ==>  A(-2, 4)

If x = -1, then y = 8  ==>  B(-1, 8)

If x = 1, then y = -8  ==>  C(1, -8)

If x = 2, then y = -4  ==>  D(2, -4)

Here (h, k)  ==>  (3, -2)

x'  =  x+3 and y'  =  y-2

x coordinate is +3, so we have to move old x coordinate 3 units right.

y coordinate is -2, so we have to move old y coordinate 2 units down.

Before translation

A(-2, 4)

B(-1, 8)

C(1, -8)

D(2, -4)

After translation

A'(-2+3, 4-2)  ==>  A'(1, 2)

B'(-1+3, 8-2)  ==>  B'(2, 6)

C'(1+3, -8-2)  ==>  C'(4, -10)

D'(2+3, -4-2)  ==>  D'(5, -6)

Problem 5 :

Find the image equation when y  =  2x is translated under the vector <0, -3>.

Solution :

Finding 3 points on the exponential function.

y  =  2x

If x = -1, then y = 1/2  ==>  A(-1, 1/2)

If x = 0, then y = 1  ==>  B(0, 1)

If x = 1, then y = 2  ==>  C(1, 2)

Here (h, k)  ==>  (0, -3)

x'  =  x+0 and y'  =  y-3

No need of horizontal movements, but move the y-coordinate 3 units down.

Before translation

A(-1, 1/2)

B(0, 1)

C(1, 2)

After translation

A'(-1+0, 1/2-3)  => A'(-1, -5/2)

B'(0+0, 1-3)  ==>  B'(0, -2)

C'(1+0, 2-3)  ==>  C'(1, -1)

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