Answer:
Remember that a vector translation can be written as:
T(a, b)
And if we apply this to a random point, (x, y), the translation gives:
T(a, b)(x, y) = (x + a, y + b)
now, remember that a general line can be written as:
y = m*x + s
Then a point of that line can be written as: (x, m*x + s)
Then if we apply the translation to a point in the line, we get:
T(a, b)(x, m*x + s) = (x + a, m*x + s + b)
Here we have two lines:
2x - 3y - 1 = 0
2x - 3y + 5 = 0
First, let's rewrite both of these in the slope-intercept form:
y = (2/3)*x - 1/3
y = (2/3)*x + 5/3
Now let's assume that we apply a translation to the first line, that has points of the form (x, (2/3)*x - 1/3), such that we want to get points of the form:
(x, (2/3)*x + 5/3).
Then we must have:
T(a, b)(x, (2/3)*x - 1/3) = (x + a, (2/3)*x - 1/3 + b) = (x, (2/3)*x + 5/3).
Then we need to solve:
(x + a, (2/3)*x - 1/3 + b) = (x, (2/3)*x + 5/3).
This means that:
x + a = x
(2/3)*x - 1/3 + b = (2/3)*x + 5/3
From the first equation, we can see that a = 0
Now we can solve the second one to find the value of b.
(2/3)*x - 1/3 + b = (2/3)*x + 5/3
subtracting (2/3)*x in both sides, we get:
-1/3 + b = 5/3
b = 5/3 + 1/3
b = 6/3 = 2
b = 2
Then the vector translation is:
T(0, 2)
So it moves the whole line 2 units upwards.
:D 