1. ## slope and line

A rectangle ABCD, where A(3,2) and B(1,6).
(i) Find the equation of BC.
(ii)Given that the equation of AC is y=x-1, Find the coordinates of C.

2. Hello

It might be helpful to start by plotting the points and the line you are given.

Can you find the equation of a line from two points? Do you know how to find the gradient of a line perpendicular to another line?

The point C must lie on the line which is perpendicualr to AB and passes through point B (1,6).

Using the points A and B you can find the equation of AB and then BC, see above.

AC is a diagonal of the rectangle and you can find the point C by finding the intersection of the lines AC and BC.

If you plot the co-ordinates of A, B and C you might be able to see how to find the perimeter.

Can you proceed from here. Please ask if you need more help.

3. Originally Posted by kaley015
A rectangle ABCD, where A(3,2) and B(1,6).
(i) Find the equation of BC.
(ii)Given that the equation of AC is y=x-1, Find the coordinates of C.
(i)
The equation of BC can be found by the point-slope form of the equation of the line.
(y -y1) = m(x -x1)

You have the point (x1,y1) = B(1,6)
So find the m.
m is the negative reciprocal of the slope of side AB.

You should get the equation of BC as.
(y -6) = (1/2)(x -1)

----------------------
(ii)Given that the equation of AC is y=x-1, Find the coordinates of C.

Get the intersection of
y = (1/2)x +11/2 ----------(1)
and
y = x -1 ------------------(2)
to find the coordinates of point C

---------------------------------------------
(iii) the perimeter of the rectangle ABCD.

The perimeter of ABCD is twice the sum of AB +BC,
P = 2(AB +CD)

Distance, d, between two points is
d = sqrt[(x2 -x1)^2 +(y2 -y1)^2]

So for AB,
AB = sqrt[(1 -3)^2 +(6 -2)^2] = sqrt(20) = 2sqrt(5)

You should get perimeter of ABCD = 16sqrt(5) ------answer.