# Finding 2 vertices of rectangle with 2 known vertices

• Aug 13th 2013, 03:23 AM
Rodimus79
Finding 2 vertices of rectangle with 2 known vertices
Rectangle ABCD
A is (-1,4), B is (4,8)
Equation of BC is 4y + 5x= 52
equation of AD is 4y + 5x = 11
equation of AB is 5y= 4x+ 24

Find C and D
Answer is (12,-2) and (7,-6)

I've tried gradient,length,midpoint formulas but to no avail.
Thanks
• Aug 13th 2013, 04:29 AM
MarkFL
Re: Finding 2 vertices of rectangle with 2 known vertices
The problem, as posed, has no unique solution. Is there some other attribute of the rectangle which you have not posted, such as its area?
• Aug 13th 2013, 06:18 AM
Rodimus79
Re: Finding 2 vertices of rectangle with 2 known vertices
Apologies..I left out
Length of BC is twice the length Of AB
• Aug 13th 2013, 07:11 AM
HallsofIvy
Re: Finding 2 vertices of rectangle with 2 known vertices
Quote:

Originally Posted by Rodimus79
Rectangle ABCD
A is (-1,4), B is (4,8)
Equation of BC is 4y + 5x= 52
equation of AD is 4y + 5x = 11
equation of AB is 5y= 4x+ 24

And, you now tell us, "Length of BC is twice the length Of AB"
It is easy to calculate that the length of AB is $\sqrt{(4-(-1)^2+ (8- 4)^2}= \sqrt{25+ 16}= \sqrt{41}$
so the length of BC is $2\sqrt{41}$ and, because this is a rectangle, BC and AD are the same length.

Take D= (x, y)= (x, (11- 5x)/4). Then we must have [tex]\sqrt{(x- (-1))^2+ ((11- 5x)/4- 4)^2}= 2\sqrt{41}[tex]
so that, squaring both sides gives $(x+1)^2+ (5x+ 5)^2/16= 164$. Solve that quadratic equation for x. Do the same with BC.

Quote:

Find C and D
Answer is (12,-2) and (7,-6)

I've tried gradient,length,midpoint formulas but to no avail.
Thanks
• Aug 13th 2013, 10:35 AM
Rodimus79
Re: Finding 2 vertices of rectangle with 2 known vertices
Thanks so much, HallsofIvy!! The key step was to have y = 11-5x/4, to express y in terms of x, since it's a point on the line and satisifies the equation! Thanks once again!!!
• Aug 14th 2013, 08:58 PM
ibdutt
Re: Finding 2 vertices of rectangle with 2 known vertices
• Aug 16th 2013, 06:49 AM
bjhopper
Re: Finding 2 vertices of rectangle with 2 known vertices
given the two points and the fact that the length of the rectangle is twice width(AB) there is a simple method to find the other vertices
slope diagram AB = 4/5
slope diagram AD =-5/4
actual slope diagram AD =-10/8 since AD =2AC
Point D x =8-1 =7 y =4 + absolute value y =10 y=-6
same procedure gives point C
• Aug 17th 2013, 02:41 AM
ibdutt
Re: Finding 2 vertices of rectangle with 2 known vertices
Quote:

Originally Posted by bjhopper
given the two points and the fact that the length of the rectangle is twice width(AB) there is a simple method to find the other vertices
slope diagram AB = 4/5
slope diagram AD =-5/4
actual slope diagram AD =-10/8 since AD =2AC
Point D x =8-1 =7 y =4 + absolute value y =10 y=-6
same procedure gives point C

The slope has nothing to do with the length of line segment? thus the slope of Ad is -5/4
• Aug 17th 2013, 04:45 AM
bjhopper
Re: Finding 2 vertices of rectangle with 2 known vertices
Quote:

Originally Posted by ibdutt
The slope has nothing to do with the length of line segment? thus the slope of Ad is -5/4

Actual slope of AD = -10/8. This slope defines the length of AD which is not used in my solution