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Solve for tin one of the equations, Substitute into the second equation ,Simplify
y= 4/5 t + 7/5
i mean not t y= 4/5 x + 7/5
how do i solve that for t?
\[x = rcos \theta, y = rsin \theta\]
you converted to polar form math8g, not parametric
lol, but its a nice polar ;)
misread ...darn sorry
y(t)= 4/5 x(t) + 7/5
the answer the book got was x=5t-2 and y=6-4t
it asks me to find the parametric representation of a line that runs through (-2,6) and (3,2)
well that makes it a lot easier
first find a vector that point from one point to the next what is that vector?
i have no idea. i got the equation of a line in rectangular form though
would it be (5,-4)?
that get's us farther away from where we want to be yes the vector is \(\vec r=<5,-4>\) the formula for a vector equation of a line is....
I should have called that \(\vec v=<5,-4>\) the formula for the line is\[\vec r=\vec r_0+t\vec v\]where \(\vec r_0\) is one of the given points the line passes through
so we have\[\vec r=<-2,6>+t<5,-4>=<-2+5t,6-4t>\]
looking at each component individually we get your books answer
you should really read this to get a good handle on what we just did http://tutorial.math.lamar.edu/Classes/CalcII/EqnsOfLines.aspx
okay thank you
it is possible to get the same rectangular equation, even though the parameter is different and the parametric equations are different depending on the choice on x1 , y1
youll get the same direction vector; but you can use any point on the line to anchor it too; and any scalar of the vector itself