Find the vector equation of the line through the points (1, 2, −1) and (2, 4, 5). is the point (−5, −10, −37) on this line?

(1+t, 2+2t, -1+6t); C is on this line.

Step-by-step explanation:

Let's find the vector equation of the line.

Given information:

Point A is (1, 2, -1)

Point B is (2, 4, 5)

Point C is (-5, -10, -37)

The difference between the two first given points (A, B) is:

B-A=(2, 4, 5)-(1, 2, -1)=(2-1, 4-2, 5-(-1))=(1, 2, 6) which is called the direction vector, so the equation is:

(x, y, z)=(1, 2, -1) + t*(1, 2, 6)=(1+t, 2+2t, -1+6t)

Now let's find if point C is on the line:

(x, y, z)=(1, 2, -1) + t*(1, 2, 6)

(-5, -10, -37)=(1, 2, -1) + t*(1, 2, 6)

(-5, -10, -37)-(1, 2, -1)=t*(1, 2, 6)

(-5-1, -10-2, -37+1)=t*(1, 2, 6)

(-6, -12, -36)=t*(1, 2, 6) from where we can obtain:

1*t=-6 --> t=-6

2*t=-12 --> t=-6

6*t=-36 --> t=-6

In conclusion, because from all the equations we obtained t=-6, then point C is on this line.

answer: b.

step-by-step explanation:

the y coordinate for point b is 0, and answer b is the only one with that y coordinate.