ABCD is a rectangle and A and B are the points (4,2) and (2,8) respectively. Given that the equation of AC is y=x-2, find a) Equation of BC b) coordinates of B and C

A(4,2), B(2,8)

AC: y=x-2

Point point form for a line through (a,b) and (c,d) is (c-a)(y-b)=(d-b)(x-a)

AB is  (2 - 4)(y - 2) = (8-2)(x-4)

-2(y - 2)=6(x-4)

y - 2 = -3(x-4)

y = -3x + 14

BC is perpendicular through B, so slope 1/3 and we calculate the constant as y-(1/3)x:

y = (1/3) x  + (8 - (1/3)(2) ) = (1/3) x + 22/3

a)  BC is  y =  (1/3) x + 22/3

C is the meet of AC and BC,

x - 2 = (1/3) x + 22/3

3x - 6 = x + 22

2x = 28

x = 14

y = x-2 = 12

Check: y = (1/3)x + 22/3 = (14+22)/3 = 36/3 = 12  good

C(14,12)

The remaining corner is D.   We have A-B=D-C or

D = A+C-B+C = (4,2)+(14,12)-(2,8) = (16, 6)

b) B(2,8) [given, but asked for]  C(14,12), D(16,6)