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Write a parametric formula for the line that has xy-equation (y + 2) = 1.52 (x - 1). give a vector parallel to this line. give a vector perpendicular to this line. give a point on this line.

Respuesta :

DeanR
We call our real parameter [tex]t[/tex] and try to save some work by setting [tex]t=x-1[/tex] so

[tex]x=t+1[/tex]

and

[tex]y = 1.52 t - 2[/tex]

or

[tex](x,y)=(1, -2) + t(1, 1.52) [/tex]

That's enough to answer the question.  

[tex](1, 1.52) [/tex] is the parallel vector, often called the direction vector.

To get the perpendicular vector, we swap the coordinates and negate one of them; it doesn't matter which.

[tex](1.52, -1) [/tex] is the perpendicular vector.

[tex](1, -2) [/tex] is a point on the line, given by [tex]t=0.[/tex]




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