Vector Equation of a Line in 2-space (full lesson) | MCV4U

[peterlohnes]

Another way to do this is to realize the 4t from the y equation and the 2t from the x equation is actually rise over run: 4/2 = 2. You have a point on this line given by the 3 in the x equation (3, y), and the -5 in the y equation (3, -5). You can use this point to find b by substituting : y=mx + b (put in your m that you found from rise/run)---> y=2x + b (put in your point to solve for b) -5=6 + b, b=-11. Rewrite your line equation using m and b: y=2x -11. Or in standard form: 2x-y-11=0. Saves you from the fractions. Mind you if your rise/run ends up being a fraction, you're back to solving fractions.

[Cawil.Maxamad]

Do I need to watch every unit 6 & 7 lesson go understand 8? Our teacher split vectors into 3 units

[peterlohnes]

Again a faster method at 22:10. Anything is parallel if they have the same m...m is right in the vector equation. In line 2 the rise is 12, the run is 3, so m=4. In line 1, the rise is 4, the run is 2, so m=2. Not parallel. You should learn the scalar trick for 3 dimensional vectors, so thats why its taught this way.