Question 1191010


the equation of the line 

{{{y=mx+b}}}


find the line passing through the point ({{{6}}},{{{3}}}) 

that is perpendicular to the line {{{4x−5y=−10}}}

perpendicular lines have slopes negative reciprocal to each other

so, find a slope of {{{4x−5y=−10}}} 

{{{4x+10=5y}}}


{{{y=(4/5)x+2}}}-> slope is {{{4/5}}}


perpendicular line will have a slope {{{-1/(4/5)=-5/4}}}


{{{y=-(5/4)x+b}}}.....use given point to find {{{b}}}


{{{3=-(5/4)6+b}}}

{{{3+30/4=b}}}

{{{b=21/2}}}


your equation is:

{{{y=-(5/4)x+21/2}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(6,3,.12),locate(6,3,p(6,3)),
graph( 600, 600, -10, 10, -10, 10, (4/5)x+2, -(5/4)x+21/2)) }}}