Question 125064
 the answer is: {{{y = -(3/2)x + 7)}}} 

since given line {{{y=(2/3)x-5}}}which contains point ({{{2}}},{{{4}}}), we will find a line perpendicular to {{{y=(2/3)x-5}}} using what we know: 

 we know 
 
the slopes of the perpendicular lines are negative reciprocal; so  the slope {{{2/3}}} from the line{{{y = (2/3)x - 5)}}} will have negative reciprocal equal to {{{-3/2}}}

so, the slope of the unknown line must be {{{-3/2}}};
 
 
Slope {{{m[p] = -(1/m)}}}

{{{m[p] = -(1/(2/3))= -(3/2)}}}

we also know that line goes through ({{{2}}},{{{4}}}), and we can find the equation by plugging in this info into the {{{point-slope}}} formula

Point-Slope Formula:

{{{y-y1=m(x-x1)}}} where m is the slope and (x,x1) is the given point

Plug in, {{{x1=2}}} and {{{y1=4}}}, 

{{{y-4=-(3/2 )(x-2)}}}

{{{y-4= -(3/2 )x -(3/2 )( 2)}}}

{{{y-4= -(3/2 )x +3}}}

{{{y= -(3/2 )x + 3 + 4}}}

{{{y= -(3/2 )x + 7}}}