The parametric equations for a line L1 are as follows:

x = −4−6t
y = 2+2t
z = 1−4t

Let L2 be the line parallel to L1 and passing through the point (−3, −3, −3). Find the point P on L2 whose x-coordinate is −12.

P=P(-12, _ , _ ) ?

It is easy to find the equation of the L2, since it is parallel to L1 and you have a point on it

direction vector of L1 is (-6,2,-4) or reduced:
(3, -1, 2)
a possible line equation for L2 is

x = -3 + 3k
y = -3 - k
z = -3 + 2k

you want x = -12, so -12 = -3 + 3k
3k = -9
k = -3

then y = -3 + 3 = 0
z = -3 - 6 = -9

P on L2 is (-12, 0, -9)

check my arithmetic.
Mia found the point P on L1 , not L2, which has x = -12

Well, to find the point P on L2 whose x-coordinate is -12, we need to solve for the values of y and z.

Let's start by finding the value of t for L2 using the x-coordinate. We set x = -12 and solve for t:

-4 - 6t = -12

To solve this equation, we need to ask a mathematician, because it seems like we're experiencing some arithmetic difficulties here.

In any case, after solving for t, we can substitute that value into the equations for y and z to find the corresponding values.

But hey, math can be like riding a unicycle, a little wobbly at times, so we'll leave that part to you!

To find the point P on line L2 with x-coordinate -12, we need to substitute x = -12 into the parametric equations of L2 and solve for y and z.

Given that L2 is parallel to L1, the direction vector of L2 will be the same as that of L1. The direction vector of L1 is (-6, 2, -4).

Let's set up the equations:

x = -4 - 6t (Parametric equation for x)
y = 2 + 2t (Parametric equation for y)
z = 1 - 4t (Parametric equation for z)

Since L2 is parallel to L1, we can use the same parameter t for both lines.

Now we substitute x = -12 into the x equation:

-12 = -4 - 6t

To solve for t, we can isolate it on one side:

-12 + 4 = -6t

-8 = -6t

Divide both sides by -6:

t = -8 / -6

t = 4/3

Now we can substitute t = 4/3 into the y and z equations:

y = 2 + 2t

y = 2 + 2(4/3)

y = 2 + 8/3

y = 6/3 + 8/3

y = 14/3

z = 1 - 4t

z = 1 - 4(4/3)

z = 1 - 16/3

z = 3/3 - 16/3

z = -13/3

Therefore, the point P on L2 with x-coordinate -12 is P(-12, 14/3, -13/3).

To find the point P on line L2 with x-coordinate -12, we need to find the corresponding values for y and z.

Since line L2 is parallel to L1, the direction of L2 is the same as L1. Therefore, we can use the same parameter t in the equations for L1 to find the coordinates of the point P on L2.

Given that the x-coordinate of P is -12, we can set x = -12 in the parametric equation for L1:

-12 = -4 - 6t

Solving this equation for t gives us:

-12 + 4 = -6t
-8 = -6t
t = -8 / -6
t = 4/3

Now that we have the value of t, we can substitute it into the parametric equations for L1 to find the corresponding values of y and z:

For y:
y = 2 + 2t
y = 2 + 2(4/3)
y = 2 + 8/3
y = 6/3 + 8/3
y = 14/3

For z:
z = 1 - 4t
z = 1 - 4(4/3)
z = 1 - 16/3
z = 3/3 - 16/3
z = -13/3

Therefore, the point P on L2 with x-coordinate -12 is P(-12, 14/3, -13/3).

they give you x

x=-12
so solve for t in the x equation by plugging in the x value
-12=-4-6t
-8=-6t
4/3=t

now plug in the t value for the y and z equation to find the coordinates