Question 1189440


 the point-slope formula:

{{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} is y-intercept
 
given:

desired line is perpendicular to {{{7y=x-4}}} 

and passes through the point ({{{-2}}},{{{2}}})

find a slope of the given line:

 {{{7y=x-4}}} 
{{{y=(1/7)x-4/7}}} => slope is {{{(1/7)}}}

perpendicular line will have a slope negative reciprocal to {{{(1/7)}}}

and it is {{{-1/(1/7)=-7}}}

so, the slope of perpendicular line is {{{m=-7}}}

{{{y=-7x+b}}} .........use given point to find {{{b}}}
({{{-2}}},{{{2}}})

{{{2=-7(-2)+b}}}
{{{2=14+b}}}
{{{b=2-14}}}
{{{b=-12}}}

so, your equation is

{{{y=-7x-12}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-2,2,.12),locate(-2,2,p(-2,2)),
graph( 600, 600, -10, 10, -10, 10, (1/7)x-4/7, -7x-12)) }}}