Question 1203459
Triangle ABC has vertices A(2, 3), B(1, -8) and C(-3, 2).

The line containing the altitude is perpendicular to the line containing B({{{1}}}, {{{-8}}}) and C({{{-3}}}, {{{2}}})

find a slope of that line:

{{{m=(2-(-8))/(-3-1)=(2+8)/-4=10/-4=-5/2}}}

The line containing the altitude is perpendicular, so slope will be {{{m=-1/(-5/2)=2/5}}}
then, using point slope formula ,  the line containing the altitude will be

{{{y-y1=m(x-x1)}}}

the point A({{{2}}}, {{{3}}}) and {{{m=2/5}}}

{{{y-3=(2/5)(x-2)}}}

{{{y-3=(2/5)x-2(2/5)}}}

{{{y=(2/5)x-4/5+3}}}

{{{y=(2/5)x-4/5+3}}}

{{{y=(2/5)x+11/5}}}

{{{y=(2/5)x+2.2}}}

then

 the point P({{{0}}}, {{{y}}}) will be

{{{y=(2/5)0+11/5=11/5}}}

 the point ({{{0}}}, {{{2.2}}}) 


{{{ drawing( 600, 600, -10, 10, -10, 10,
green(line(2,3,1,-8)),green(line(2,3,-3,2)),green(line(-3,2,1,-8)),
circle(2,3,.12), circle(1,-8,.12),circle(-3,2,.12), circle(0,2.2,.12), 
locate(2,3,A), locate(1,-8,B), locate(-3,2,C), locate(0.1,2.2,P(0,2.2)), 
graph( 600, 600, -10, 10, -10, 10, (2/5)x+2.2), (2/5)x+2.2) }}}