How do you write the equation in point slope form given (-2,-2) perpendicular to y= -1/3x + 9?

1 Answer
Feb 14, 2017

Answer:

#y = color(red)(3)x + color(blue)(4)#

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)(-1/3)x + color(blue)(9)# so we know the slope of this line is #color(red)(m = -1/3)#

The slope of a perpendicular line (let's call it #m_p#) is the negative inverse of the slope of this line or #m_p = -1/m#

Therefore the slope of a perpendicular line is #m_p = 3#

We can now use the point-slope formula to find an equation for the line. 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 point from the problem and the slope we determined gives:

#(y - color(red)(-2)) = color(blue)(3)(x - color(red)(-2))#

#(y + color(red)(2)) = color(blue)(3)(x + color(red)(2))#

We can now solve this for #y# to put it into slope-intercept form:

#y + color(red)(2) = (color(blue)(3) xx x) + (color(blue)(3) xx color(red)(2))#

#y + color(red)(2) = 3x + 6#

#y + color(red)(2) - 2 = 3x + 6 - 2#

#y + 0 = 3x + 4#

#y = color(red)(3)x + color(blue)(4)#