How do you write an equation of a line passing through (4, -2), perpendicular to #y = 4x + 2#?

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Answer 1 · 2017-09-05T12:32:12

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)(4)x + color(blue)(2)#

The slope of this line is: #color(red)(m = 4)#

Let's call the slope of the line perpendicular to the line in the problem: #m_p#

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

Substituting gives the slope of the line we are looking for as:

#m_p = -1/4#

We can substitute this into the slope-intercept formula giving:

#y = color(red)(-1/4)x + color(blue)(b)#

We can now substitute the values from the point in the problem for #x# and #y# in the formula and solve for #color(blue)(b)# giving:

#-2 = (color(red)(-1/4) xx 4) + color(blue)(b)#

#-2 = color(red)(-1) + color(blue)(b)#

#1 - 2 = 1 color(red)(- 1) + color(blue)(b)#

#-1 = 0 + color(blue)(b)#

#-1 = color(blue)(b)#

Substituting this back into the formula gives:

#y = color(red)(-1/4)x + color(blue)(-1)#

#y = color(red)(-1/4)x - color(blue)(1)#