On a coordinate plane, a line goes through (negative 3, 2) and (2, negative 1). A point is at (3, 0). What is the equation of the line that is perpendicular to the given line and passes through the point (3, 0)

Respuesta :

Answer:            [tex]y=\frac{5}{3}x-5[/tex]

Step-by-step explanation:

Alright, lets get started.

Two points are given as (-3,2) and (2,-1)

Slope will be : [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

Slope will be : [tex]\frac{-1-2}{2-(-3)}[/tex]

Slope will be : [tex]\frac{-3}{5}[/tex]

Our line is perpendicular to this slope, then,

[tex]m \times (\frac{-3}{5}) =-1[/tex]

[tex]m=\frac{5}{3}[/tex]

Line passing through the point (3,0) and having slope [tex]\frac{5}{3}[/tex], is given by :

[tex]y-y_{1}=m(x-x_{1} )[/tex]

[tex]y-0=\frac{5}{3}(x-3)[/tex]

[tex]y=\frac{5}{3}x-5[/tex]    :  Answer

Hope it will help :)

Answer:

y=\frac{5}{3}x-5

Step-by-step explanation:

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