How do you write an equation of a line passing through (5, -3), perpendicular to # y=6x + 9#?

1 Answer
Jul 6, 2017

Answer:

See a solution process below:

Explanation:

The equation in the problem is in slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y = color(red)(6)x + color(blue)(9)#

The slope of this line is: #color(red)(m = 6)#

Let's call the slope of a perpendicular line: #m_p#

The formula for the slope of a perpendicular line is: #m_p = -1/m#

Substituting gives us:

#m_p = -1/6#

We can now use the point-slope formula to find an equation of the line from the problem. The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #(color(red)(x_1, y_1))# is a point the line passes through.

Substituting the slope we calculated and the values from the point in the problem gives:

#(y - color(red)(-3)) = color(blue)(-1/6)(x - color(red)(5))#

#(y + color(red)(3)) = color(blue)(-1/6)(x - color(red)(5))#

If we want the equation in slope-intercept form we can solve for #y#:

#y + color(red)(3) = (color(blue)(-1/6) xx x) - (color(blue)(-1/6) xx color(red)(5))#

#y + color(red)(3) = -1/6x - (-5/6)#

#y + color(red)(3) = -1/6x + 5/6#

#y + color(red)(3) - 3 = -1/6x + 5/6 - 3#

#y + 0 = -1/6x + 5/6 - (6/6 xx 3)#

#y = -1/6x + 5/6 - 18/6#

#y = color(red)(-1/6)x - color(blue)(13/6)#