Find symmetric equations for the line. (Use the parameter t.)

The line through
(−8, 4, 7)
and parallel to the line
x/2= y/3= z+1

well, the given line goes through (0,0,-1) right?

The new line must have the same direction as x/2 = y/3 = z+1, which means that the new line must have direction (2,3,1) and must pass through the point (−8, 4, 7).

parametric equations of new line:
x = -8 + 2t
y = 4 + 3t
z = 7 + 1

The fact that (0,0,-1) in on the given line, has no impact on the new line

no, it does not. But the given symmetric equations can be used as a model for the desired equations. (not parametric, btw)

which can also be derived from yours, by eliminating t.

missed the "symmetric" part and assumed parametric, by bad

so we could just go ...

(x+8)/2 = (y-4)/3 = z -7

To find the symmetric equations for the line passing through the point (-8, 4, 7) and parallel to the line given by the equation x/2=y/3=z+1, we need to use the concept of direction vectors.

Step 1: Find the direction vector of the given line.
The direction vector of the line x/2=y/3=z+1 can be found by looking at the coefficients of x, y, and z. From the equation, we can see that the direction vector is (2, 3, 1).

Step 2: Use the point (-8, 4, 7) and the direction vector to write the symmetric equations.
The symmetric equations of a line can be written as:
x = x0 + at
y = y0 + bt
z = z0 + ct

where (x0, y0, z0) is a point on the line and (a, b, c) is the direction vector.

Substituting (-8, 4, 7) for (x0, y0, z0), and (2, 3, 1) for (a, b, c), we have:
x = -8 + 2t
y = 4 + 3t
z = 7 + t

So, the symmetric equations for the line passing through (-8, 4, 7) and parallel to x/2=y/3=z+1 are:
x = -8 + 2t
y = 4 + 3t
z = 7 + t

These equations represent all the points on the line as t varies.