Question 965404


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

given:
the graph of f passes through ({{{-6}}},{{{ 4}}})
 and 
is perpendicular to the line that has an x-intercept of {{{2}}} and a y–intercept of {{{-4}}} 

{{{y=mx+b}}} if ({{{-6}}},{{{ 4}}}) on a line, then we have

{{{4=-6m+b}}}........eq.1

if line is perpendicular to the line that has an x-intercept of {{{2}}} and a y–intercept of {{{-4}}} , then our line has a slope negative reciprocal to the slope of  the line that has an x-intercept of {{{2}}} and a y–intercept of {{{-4}}}

first find the equation of  the line that has an x-intercept of {{{2}}} and a y–intercept of {{{-4}}}

if the line that has an x-intercept of {{{2}}}, then the point ({{{2}}},{{{0}}}) is on a line

so, we have

{{{0=m[p]*2+b}}} 

{{{0=2m[p]+b}}}........... eq.2 



if the line that has an y-intercept of {{{-4}}}, then the point ({{{0}}},{{{-4}}}) is on a line

so, we have

{{{-4=m[p]*0+b}}} 

{{{-4=0+b}}}

{{{b=-4}}}.......substitute in eq.2 and find {{{m[p]}}}


{{{0=2m[p]-4}}}........... eq.2 

{{{2m[p]=4}}}

{{{m[p]=2}}}


so, the equation of the line is perpendicular to the line that has an x-intercept of {{{2}}} and a y–intercept of {{{-4}}} is

{{{y=2x-4}}} 



since the slope is {{{m[p]=2}}}, our line will have the slope {{{m=-1/2}}} which is  is the negative reciprocal of the {{{m[p]=2}}}

so far we have {{{y=-(1/2)x+b}}} and we need to find {{{b}}}
{{{y=2x-4}}} 

use {{{4=-6m+b}}}........eq.1, substitute {{{m=-1/2}}}

{{{4=-6(-(1/2))+b}}}

{{{4=cross(6)(1/cross(2))+b}}}

{{{4=3+b}}}

{{{b=4-3}}}

{{{b=1}}}

so, the equation of a line we are looking for is:  {{{y=-(1/2)x+1}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-6,4,.12), locate(-6,4,p(-6,4)),
 graph( 600, 600, -10, 10, -10, 10, -(1/2)x+1, 2x-4)) }}}