Write a vector equation of the line that passes through point P and is parallel to a. Then write parametric equations of the line.

P(-1, 3); a=(-6, -1)
MY ANSWER: (x+1, y-3) = t(-6, -1); x= -1-6t, y=3-t

P(0, 5); a=(2, -9)
MY ANSWER: (x, y-5) = t(2, -9); x=2t, y=5-9t

To write the vector equation of a line passing through a given point P and parallel to a given vector a, you can use the formula:

r = P + ta

Where r is the position vector of a point on the line, P is the position vector of the given point, t is a scalar parameter, and a is the given vector.

For the first example:
P(-1, 3); a = (-6, -1)

The vector equation is:
r = (-1, 3) + t(-6, -1)

Expanding and combining like terms:
r = (-1 - 6t, 3 - t)

So the vector equation of the line is:
x = -1 - 6t
y = 3 - t

Now, to obtain the parametric equations of the line, you simply separate out the x and y components and express them as functions of a parameter t:

For x:
x = -1 - 6t

For y:
y = 3 - t

Therefore, the parametric equations of the line are:
x = -1 - 6t
y = 3 - t

For the second example:
P(0, 5); a = (2, -9)

Following the same process:

The vector equation is:
r = (0, 5) + t(2, -9)

Expanding and combining like terms:
r = (2t, -9t + 5)

So the vector equation of the line is:
x = 2t
y = -9t + 5

And the parametric equations of the line are:
x = 2t
y = -9t + 5