Question 1157269

the equation of a line that goes through the point ({{{4}}}, {{{-3}}}) and is perpendicular to the line {{{x-4y=16}}}

perpendicular lines have slopes negative reciprocal to each other

so, first find a slope of given line:

{{{x-4y=16}}}.......solve for {{{y}}}

{{{x-16=4y}}}

{{{x/4-16/4=y}}}

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

then negative reciprocal is {{{-1/(1/4)=-4}}}

=>perpendicular line have slope {{{m=-4}}}


so far, the equation of a line will be 


{{{y=-4x+b}}}...........use iven point ({{{4}}}, {{{-3}}}) to calculate {{{b}}}


{{{-3=-4*4+b}}}

{{{-3=-16+b}}}

{{{-3+16=b}}}

{{{13=b}}}


and, your equation is:

{{{y=-4x+13}}}


{{{drawing ( 600, 600, -10, 10, -10, 15,
circle(4,-3,.12), locate(4,-3,p(4,-3)),
graph( 600, 600, -10, 10, -10, 15, (1/4)x-4, -4x+13)) }}}