Question

The diagram shows a rectangle ABCD. The point A is (2, 14), B is (−2, 8) and C lies on the x-axis. Find

(i) the equation of BC,
(ii) find coordinates of C and D

Answer 1

Explanation:

since AB and BC are perpendicular so AB.BC=0, (scalar product)

AB=(-4, -6), and BC=(xc+2, yc-8)

AB.BC= -4x-8-6y+48 =0

so the equation of BC is 2x+3y-20=0

C is on the line BC, so 2xc+3yc-20=0, but yc=0, so xc=20/2=10

AB and CD are parallel, so det(AB, CD)=0

AB and AD are perpendicular so AB.AD=0

AB=(-4, -6) and CD =(x-10, y), AD= (x-2, y-14)

det(AB, CD)=0

x-10-y = 0 equivalent to x-y = -10

AB.AD=0

-4x+8-6y+84=0 equivalent to 2x+3y=44

let's solve

x-y = -10

2x+3y=44

-5y = -64, y = 64/5, x= -10+64/5=14/5

finally D(14/5, 64/5)