Is the point on the line? Test with vector equation [x,y] = [2,-3] + t[4,7]

[DespicableMe]

Determine if each point P is on the line [x,y] = [2,-3] + t[4,7]

a) P(-2,-10)

b) P(6,5)

c) p(10,14)

d) P(4, 0.5)



Answers: Yes, No, No, Yes



We just learned this today and I'm not sure of the difference between vector r and vector "r not" from r = r0 + tm. I'm not sure where to substitute the given points.

 

[DespicableMe]

Nevermind, I get it now.
We solve for x and y in the original equation using parametric equations, then isolate for t in the x equation. Substitute that into the "t" in the "y" equation and then sub in the points.

 

[Mark44]

Determine if each point P is on the line [x,y] = [2,-3] + t[4,7]

a) P(-2,-10)

b) P(6,5)

c) p(10,14)

d) P(4, 0.5)



Answers: Yes, No, No, Yes



We just learned this today and I'm not sure of the difference between vector r and vector "r not" from r = r0 + tm. I'm not sure where to substitute the given points.

For each of your given points, if you can solve for t so that [x,y] = [2,-3] + t[4,7] is a true statement. For example, to verify that the first point, P(-2, -10) is on the line, solve for t:
<-2, -10> = <2, -3> + t<4, 7>.

Solving, this vector equation, I get t = -1, wbich means that (-2, -10) is on the line.

r₀ is just some specific vector. In your line equation, r₀ corresponds to the point at which t = 0. It's written "r naught" or "r nought".