Coordinate Geometry????

The coordinates of ABC are A(1,5), B(4,4) and C(8,6). Given that P is the foot of the perpendicular from A to BC find the coordinates of P and equation of AP

i tried to use the midpoint formula for P but got it wrong the book says the answer is (2,3)

how would i solve this

first we need to find the equation of the line BC
the gradient of BC is: (6-4) / (8-4) = 1/2

now we substitute the information into the equation:
y = mx + c
4 = 1/2 (4) +c
make c the subject
c = 4-2 = 2
therefore
y = 1/2 x +2

if line AP is perpendicular to line BC, the gradient is going to be the negative reciprocal of the gradient BC: 1/2 --> -2

we substitute the information into the equation:
y = mx + c
5 = -2 (1) + c
solve for c
c = 5+2 = 7
therefore
y = -2x +7

now we solve the simultaneous equation to find the intersect of the two lines also know as point P
equation 1: y = 1/2 x +2
equation 2: y = -2x +7

1/2 x +2 = -2x +7
make the x to one side and the numbers to the other:
5/2 x = 5
5x = 10
x=2
now substitute x back to any of the equation to find y
y = -2x +7
y = -2(2) +7 = -4 +7 =3

hence x=2 and y=3
point P = (2, 3)

1) Slope of BC is: (6-4)/(8-4) = 2/4 = 1/2

2) Since AP is perpendicular to BC, its slope is negative reciprocal of BC. So slope of AP = -2.
Hence equation of AP: Using slope-point form: [Slope = -2 and point (1,5)]

y - 5 = -2(x - 1)

==> y - 5 =-2x + 2

==> 2x + y = 7

3) Equation of BC: y - 4 = (1/2)(x-4)

==> 2y - 8 = x - 4;

==> x - 2y = -4

4) Solving the equations of BC and AP, we will get the point P;

Substituting x = 2y-4 from BC on AP, 4y-8+y = 7; ==> y = 3; so x = 2

Thus P is (2,3)

[Note: The foot of the perpendicular from one vertex to the opposite base of a triangle, is not the mid point of the base. Only in equilateral and isosceles triangle it is mid point].

slope of BC=(6-4)/(8-4)=1/2 slope of AP(perpendicular to BC)=-2 eqn of APis y-5=-2(x-1), 2x+y=7-(1)

eqn of BC is x-4y+4=0 ---(2) solve eqns (1) and (2) you get the point P