Question

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

Answer 1

`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)

Answer 2

`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.!