How do you find the parametric equations for the line that has the points (3, -5) and (-2, 2)?

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Answer 1 · 2018-02-26T16:34:41

#x(t)=t#

#y(t)=-7/5t-4/5#

Gradient of line is:

#(y_2-y_1)/(x_2-x_1)#

#(2-(-5))/(-2-3)=7/-5=-7/5#

Using point slope form of a line:

#(y_2-y_1)=m(x_2-x_1)#

Where #m# is the gradient:

#y-(-5)=-7/5(x-3)#

#y=-7/5x+21/5-5#

#y=-7/5x-4/5#

Let #x=t#

Then:

#y=-7/5t-4/5#

#x(t)=t#

#y(t)=-7/5t-4/5#