Respuesta :
No, the point (3,-2) does not lie on the on the equation r=(2,1) + t (1,-3).
What is a straight line?
It is the minimum distance shown between two extremes points with no curves. It is of infinite length when the extremes points are at infinite distance.
Given extremes points:
r(2,1) and t(1,-3)
The equation of the line becomes:
[tex](y - y_{2} ) = \frac{y_{2} - y_{1} }{x_{2} - x_{1} } (x - x_{2} )[/tex]
[tex](y - (-3) ) = \frac{-3 - 1 }{1 - 2 } (x - 1 )[/tex]
[tex](y + 3 ) = \frac{ -4 }{-1 } (x - 1 )[/tex]
[tex](y + 3 ) = 4 (x - 1 )\\y = 4x - 4 -3\\y = 4x -7[/tex]
[tex]y - 4x + 7 = 0[/tex]
Now put the given point (3, -2) and check whether it satisfices the line equation or not:
[tex]-2 - 4(3) + 7 = -7[/tex] ≠ 0
Hence the line equation is not satisfices, RHS and LHS are different, therefore the point (3, -2) not lie on the given equation.
To learn more about straight line, refer to the link:
https://brainly.com/question/27560536
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