A(0,−5),B(4,3),C(−4,3),P(5,0)
Equation of AB is
y+5=3+54−0(x−0)y+5=2x2x−y=5.......(i)
Image of P in (i) is
x−52=y−0−1=−2(2(5)−1(0)−5)4+1x−52=y−0−1=−2⇒x=1,y=2⇒P′(1,2)
Feet of perpendicular is the mid point of image of the point in given line and the point itself
So the feet of perpendicular from P on AB is
(5+12,0+22)⇒(3,1)
Now equation of AC is
y+5=3+5−4−0(x−0)y+2x+5=0.......(ii)
Image of P in (ii) is
x−52=y−01=−2(5(2)+1(0)+54+1)x−52=y−01=−6⇒x=−7,y=−6⇒P′′(−7,−6)
Feet of perpendicular on AC is the mid point of P and P′′
(−7+52,−6+02)⇒(−1,−3)
Using simple geometric constrution of perpendicular from P on BC as shown in the figure
Equation of BC is
y=4
So feet of perpendicular from P on BC is (5,3)