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

Question

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

Eva Abramson · Accepted Answer

See a solution process below:

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)#

Hazel Anglin · Answer

undefined To write the equation of a line passing through (5, -3) and perpendicular to $y = 6x + 9$, you use the point-slope form. The equation of the line is $y = -\frac{1}{6}x - \frac{33}{6}$.