What is the equation of the line that is perpendicular to #y=7/9x+15# and goes through the point #(-1,2)#?

Question

1676 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-13T20:12:22

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)(7/9)x + color(blue)(15)#

Therefore, the slope is: #color(red)(7/9)#

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:

#m_p = -1/(7/9) => -9/7#

Substituting this into the slope-intercept formula gives:

#y = color(red)(-9/7)x + color(blue)(b)#

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

#2 = (color(red)(-9/7) xx -1) + color(blue)(b)#

#2 = 9/7 + color(blue)(b)#

#-color(red)(9/7) + 2 = -color(red)(9/7) + 9/7 + color(blue)(b)#

#-color(red)(9/7) + (7/7 xx 2) = 0 + color(blue)(b)#

#-color(red)(9/7) + 14/7 = color(blue)(b)#

#(-color(red)(9) + 14)/7 = color(blue)(b)#

#5/7 = color(blue)(b)#

We can now substitute this into the formula with the slope to give the equation:

#y = color(red)(-9/7)x + color(blue)(5/7)#