State a direction vector for each of the following lines. A line parallel to x=9-3t, y=-4+t
Since x=9-3t, this means that x-9=-3t, which implies t=-(x-9)/3.
Similarly, since y=-4+t, this means that t=y+4.
Therefore, we can express the line as:
x-9=-3(y+4)
Simplifying this equation, we get:
x+3y+3=0
This is the equation of the line in the standard form Ax+By+C=0, where A=1, B=3, and C=3.
So a direction vector for this line is <1, 3>.
To find the direction vector, we need to compare the coefficients of t in the parametric equations of the line.
For the line x=9-3t, y=-4+t, the direction vector will have the coefficients of t as its components. Therefore,
Direction vector = (-3, 1)
This means that any vector in the direction of the line will be a scalar multiple of (-3, 1).