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Find the equations for the lines through the point (a,
c.that are parallel to and perpendicular to the line y = mx + b where m ≠ 0. use y for the dependent variable and all letters in lower case.

Respuesta :

the slope of the perpendicular line  is -1/m.

it passes through the point (a,c) therefore:-

y - c = -1/m(x - a)

y  = -1/m x + a/m + c  answer


Parallel line:-

c = m(a) + b
b = c - ma

y = mx + c - ma    Answer
MrRoyal

Lines can be parallel, perpendicular to one another.

  • The equation of the parallel line is [tex]y = mx - am + c[/tex].
  • The equation of the perpendicular line is [tex]y = -\frac xm + \frac am + c[/tex]

A linear equation is represented as:

[tex]y = mx + b[/tex]

Where:

[tex]m \to[/tex] slope

(a) Parallel Equation

A line parallel to [tex]y = mx + b[/tex] will have the same slope as [tex]y = mx + b[/tex]

i.e. the slope of the equation is m

The equation is then calculated as:

[tex]y = m(x -x_1) + y_1[/tex]

Where:

[tex](x_1,y_1) = (a,c)[/tex]

Substitute [tex](x_1,y_1) = (a,c)[/tex] in [tex]y = m(x -x_1) + y_1[/tex]

[tex]y = m(x - a) + c[/tex]

Open brackets

[tex]y = mx - am + c[/tex]

Hence, the equation of the line is [tex]y = mx - am + c[/tex]

(b) Perpendicular Equation

The slope (m2) of a line perpendicular to [tex]y = mx + b[/tex] is:

[tex]m_2 = -\frac 1m[/tex]

The equation is then calculated as:

[tex]y = m_2(x -x_1) + y_1[/tex]

Where:

[tex](x_1,y_1) = (a,c)[/tex]

[tex]m_2 = -\frac 1m[/tex]

Substitute [tex](x_1,y_1) = (a,c)[/tex] in [tex]y = m(x -x_1) + y_1[/tex]

[tex]y = -\frac 1m(x - a) + c[/tex]

Open brackets

[tex]y = -\frac xm + \frac am + c[/tex]

Hence, the equation of the line is [tex]y = -\frac xm + \frac am + c[/tex]

Read more about parallel and perpendicular equations of a line at:

https://brainly.com/question/21740769

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