Question 1159250

a line that passes through the point ({{{-1}}},{{{2}}}) and is {{{perpendicular}}} to the line 

{{{y=(1/2)x-3}}}-> slope is {{{m=1/2}}}

perpendicular line will have a slope negative reciprocal to {{{m=1/2}}}

it is {{{m=-1/(1/2)=-2}}}

use point slope formula:

{{{y-y[1]=m(x-x[1])}}} where m is a slope {{{-2}}}, {{{x[1]}}} and {{{y[1]}}} coordinates of given point ({{{-1}}},{{{2}}})


{{{y-2=-2(x-(-1))}}}

{{{y-2=-2(x+1)}}}

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

{{{y=-2x-2+2}}}

{{{y=-2x}}}-> your equation


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-1,2,.12),locate(-1,2.3,p(-1,2)),
graph( 600, 600, -10, 10, -10, 10, -2x, (1/2)x-3)) }}}