Question

The diagram shows the rectangle ABCD, where A is (3,2) and B is (1,6).

(i) find the equation of BC.

Given that the coordinates of AC is y=x-1, find

(ii) the coordinates of C
(iii) the perimeter of the rectangle ABCD.

Answer 1

i) The equation of BC is y=2x-4.

ii) The coordinates of C are (3,2).

iii) The perimeter of the rectangle ABCD is (-2+4+2+0)=4.

The diagram shows the rectangle ABCD, where A is (3,2) and B is (1,6).

(i) The equation of BC is of the form y=mx+C.

By substituting (3,2) and (1,6), we get 2=3m+C and 6=m+C.

By subtracting these equations, we get 4=2m, so m=2.

Substituting m=2 into 2=3m+C, we get C=-4.

Therefore, the equation of BC is y=2x-4.

(ii) The coordinates of C can be found by substituting the equation of AC (y=x-1) into the equation of BC (y=2x-4).

2x-4=x-1

x=3

Substituting x=3 into y=x-1, we get y=3-1=2.

Therefore, the coordinates of C are (3,2).

(iii) The perimeter of the rectangle ABCD is the sum of the lengths of its four sides.

The length of AB is the difference of the x-coordinates of B and A: 1-3=-2.

The length of BC is the difference of the y-coordinates of B and C: 6-2=4.

The length of CD is the difference of the x-coordinates of C and D: 3-1=2.

The length of DA is the difference of the y-coordinates of D and A: 2-2=0.

Therefore, the perimeter of the rectangle ABCD is (-2+4+2+0)=4.

Answer 2

(i)

`A(3,2),B(1,6) \\ \\m_(AB)= (6-2)/(1-3)=-2 \\ \\m_(BC)m_(AB)=-1 \\ \\m_(BC)= (-1)/(m_(AB)) = (-1)/(-2)= (1)/(2) \\ \\B(1,6)\Rightarrow x_1=1,y_1=6 \\ \\BC:y-y_1=m_(BC)(x-x_1) \\ \\y-6= (1)/(2)(x-1) \\ \\2y-12=x-1 \\x-2y+11=0`

(ii)

`AC:y=x-1 \\BC:x-2y+11=0 \\ \\C=AC\cap BC \\ \\x-2(x-1)+11=0 \\x-2x+2+11=0 \\x=13 \\y=x-1=13-1=12 \\C(13,12)`

(iii)

`A(3,2)B(1,6),C(13,12) \\P=2(d(A,B)+d(B,C)) \\ \\d(A,B)= √((1-3)^2+(6-2)^2)= √(20) =2 √(5) \\ \\d(B,C)= √((13-1)^2+(12-6)^2)= √(180)= 6√(5) \\ \\P=2(2 √(5)+6 √(5))=2*8 √(5) =16 √(5)`