What is the equation of the line perpendicular to #y=-7x # that passes through # (6,-1) #?

2 Answers
Mar 18, 2017

Answer:

#y=1/7x-13/7#

Explanation:

In general an equation of the form
#color(white)("XXX")y=color(green)mx+color(blue)b#
has a slope of #color(green)(m)#

#y=color(green)(-7)x# is equivalent to #y=color(green)(-7)x+color(blue)0#
and thus has a slope of #color(green)(""(-7))#

If a line has a slope of #color(green)m# then all lines perpendicular to it have a slope of #color(magenta)(""(-1/m))#

Therefore any line perpendicular to #y=color(green)(-7)x#
has a slope of #color(magenta)(1/7)#

If such a perpendicular line passes through the point #(color(red)x,color(brown)y)=(color(red)6,color(brown)(-1))#
we can use the slope-point formula:
#color(white)("XXX")(y-(color(brown)(-1)))/(x-color(red)6) = color(magenta)(1/7)#

Simplifying,
#color(white)("XXX")7y+7=x-6#
or
#color(white)("XXX")y=1/7x-13/7color(white)("XX")#(in slope-intercept form)

Mar 18, 2017

Answer:

# x-7y-13=0.#

Explanation:

Slope of the line # L :y=-7x# is #-7.#

Knowing that, the Product of Slopes of mutually #bot# lines is

#-1#, the slope of the reqd. #bot# line #(-1/-7)=1/7.#

Also, the reqd. line passes thro. the pt. #(6,-1.)#

Therefore, by the Slope-Point Form, the eqn. of reqd. line is,

#y-(-1)=1/7(x-6), i.e., 7y+7=x-6.#

#:. x-7y-13=0.#

Enjoy Maths.!