Question 927664
{{{drawing(450,300,-18,27,-3,27,
grid(0),blue(triangle(0,24,-15,0,24,0)),
red(circle(0,24,0.2)),
red(circle(24,0,0.2)),
red(circle(-15,0,0.2)),
locate(0.5,24.5,A(0,24)),
locate(22,3,C(24,0)),
locate(-17,2,B(-15,0)),
green(line(0,0,0,24)),
green(line(-15,0,4.5,19.5)),
green(line(-360/89,1560/89,24,0))
)}}} BC is the horizontal line {{{y=0}}} (the x-axis), and AC is the line {{{x+y=24}}}<--->{{{y=-x+24}}} .
Since BC is the horizontal line {{{y=0}}} (the x-axis), the altitude to BC is the vertical line passing through A, {{{x=0}}} (the y-axis).
Since AC is a line with {{{slope=-1}}} , the altitude to AC is the line with {{{slope=1}}} passing through B(-15,0).
That line is {{{y-0=1(x-(-15))}}}<--->{{{y=x+15}}} .
The three altitudes intersect at one point.
The point where the two altitudes found intersect can be found by solving the system
{{{system(x=0,y=x+15)}}}--->{{{system(x=0,y=15)}}} .
That intersection is the point (0,15).