The equation of line QR is y = - 1 /2 x + 1. How do you write an equation of a line perpendicular to line QR in slope-intercept form that contains point (5, 6)?

1 Answer
smendyka
Aug 25, 2017

See a solution process below:

Explanation:

First, we need to find the slope of the for the two points in the problem. The line QR 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)(-1/2)x + color(blue)(1)#

Therefore the slope of QR is: #color(red)(m = -1/2)#

Next, let's call the slope for the line perpendicular to this #m_p#

The rule of perpendicular slopes is: #m_p = -1/m#

Substituting the slope we calculated gives:

#m_p = (-1)/(-1/2) = 2#

We can now use the slope-intercept formula. Again, 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.

Substituting the slope we calculated gives:

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

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

#6 = (color(red)(2) xx 5) + color(blue)(b)#

#6 = 10 + color(blue)(b)#

#-color(red)(10) + 6 = -color(red)(10) + 10 + color(blue)(b)#

#-4 = 0 + color(blue)(b)#

#-4 = color(blue)(b)#

Substituting this into the formula with the slope gives:

#y = color(red)(2)x + color(blue)(-4)#

#y = color(red)(2)x - color(blue)(4)#

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