Find the parametric and symmetric equations for the line passing through the points p1=(2,2,3) and p2=(1,3,−1) . note: this problem does not have a unique answer unless we specify how to choose the direction vector v (see pages 610 and 611). in order to obtain the required solution for this problem use the vector v=p1−p2 (see page 585). further, you need to use the given point p1 in the formula. parametric: x(t)= y(t)= z(t)= symmetric: (quotient involving x ) = (quotient involving y) = (quotient involving z ) =

parametric equation: q = (2-t)x + (2+t)y + (3-4t)z

symm equation:

Step-by-step explanation:

Parametric equation is just like the point slope equation:

y= b + mx

but in this case we have multiple variable so it'll be

q= p₀ + dt

p₀ will be any of your point. d will be the vector containing points shared by both your vector. It can be found by subtracting your two points together.

t is just an arbitrary variable.

Solve for d=p₂ - p₁ , you can also do  d=p₁ - p₂

Now plug it into your q= p₀ + dt equation.

You can plug p₂ or p₁ in for p₀. I used p₁.

The result should give you the parametric equation.

Now solve for all of them in term of t. This step is just basic algebra.

since they're all t=..., then you can set them all equal to each other to get the symmetric equation.

the answer is c.

for continuous functions, they continue in the same pattern and you have to find that pattern kinda like slope

step-by-step explanation: