coordinate geometry: help finding square vertices

alan82

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Hi, i am from the UK and teaching myself A Level Maths (the equivilent of grades 11 and 12 in the US) and am having trouble with a question on coordinate geometry in my text book. It is as follows:

The equations of two sides of a square are y=3x-1 and x+3y-6=0. If (0,-1) is one vertex of the square find the coordinates of the other vertices.

So far i have solved the two line equations simultaneously to get a second vertex (9/10, 17/10) and then found the length of the line between the two sets coordinates i have, which is 9/10*sqrt(10)

Now i don't know how to use this length to find the other two coordinates or where to go from here.

any help?

thanks
 
You must now decide the relationship between your two points. Are they points that define a single side or are they opposite points? Either way, you can then determine the length of the sides. Let's see what you get.
 
The equations of two sides of a square are y=3x-1 and x+3y-6=0. If (0,-1) is one vertex of the square find the coordinates of the other vertices.

So far i have solved the two line equations simultaneously to get a second vertex (9/10, 17/10) and then found the length of the line between the two sets coordinates i have, which is 9/10*sqrt(10)

Now i don't know how to use this length to find the other two coordinates or where to go from here.
Draw the two lines, y = 3x - 1 and y = -(1/3)x + 2. Draw the two points, (0, -1) on the first line, and (9/10, 17/10) at the intersection of the two lines.

The third line, through point (0, -1), must be parallel to which line? So it must have what slope? (here) So it must have what equation? (here)

Since you know the lengths of the sides, you know that the next corner must be that length along this new line. To which side (right or left) of the point (0, -1) must this next point lie? Given the line's equation and the fact that you need to find the point on that line which is the given distance away, what point is the result? (here)

What must be the slope of the line representing the fourth side? This line passes through the point just found above. What then is its equation? With what of the original lines does this new line intersect? What then is the coordinate of that intersection?

If you get stuck, please reply showing your work and answers to the above questions. Thank you! ;)
 
so the third line through point (0,-1) must be parallel to y=-1/3x+2 with the equation of y=-1/3x-1 as -1 is the y intercept? so the new third line lies to the right of (0,-1).

now im not entirely sure how to convert the line length i have into the third set of coordinates given the information i have deduced
 
so [a] third line through point (0,-1) must be parallel to y=-1/3x+2 with the equation of y=-1/3x-1 as -1 is the y intercept?
Yes.


so the new third line lies to the right of (0,-1).
Well, that's one possibility. And, I would not say that a line lies to the right of (0,-1). The line passes through (0,-1). You're thinking of a side, not a line.

The other possibility is that the side lies to the left of (0,-1).

If you have drawn a sketch, you should be able to form two different squares satisfying the given information.


now [i'm] not entirely sure how to convert the line length i have into the third set of coordinates given the information i have deduced
You can pick symbols for the coordinates of the other point on the third line, and then use them to form a system of two equations.

For example, let (h, k) be the other point on the line y = -1/3 x - 1.

Substitute x=h and y=k into both the equation of the line and the distance formula. Solve the system for (h,k). You will get two solutions because there are two squares. Please show your work, if you need more help. :cool:
 
this is where i get stuck

so let the unknowns x and y equal a and b respectively


substituting them into the equation of the line gives me a=-1/3b-1 or 3a+b+3=0

substituting them into the distance equation makes no sense to me but i tried:

(9/10*sqrt(10))^2 = sqrt((0-a)^2 + (-1-b)^2) --> (9/10*sqrt(10))^2 = sqrt(a^2 + b^2 + 2b + 1)
and if i square root both sides i get 9/10*sqrt(10) = a+b+sqrt(2b)+1

ive no idea if that makes sense, is correct or how i can use these answers together to get the desired sets of coordinates
 
substituting them into the distance [formula] …

(9/10*sqrt(10))^2 = sqrt((0-a)^2 + (-1-b)^2)
That's not correct.

You have squared the left-hand side without also squaring the right-hand side.


… = sqrt(a^2 + b^2 + 2b + 1)

… if i square root both sides … i get … a + b + sqrt(2b) + 1
First Note: If we take the square root of a square root, we get a radical inside another radical (i.e., a root of a root). You want to square, instead of taking the square root.

Second Note: If you misspoke and were instead trying to take the square root of a^2+b^2+2b+1, thinking that you would get a+b+sqrt(2b)+1, then you need to learn that we cannot simplify symbolic square roots like that.

sqrt(a^2 + b^2) does not mean a + b.



Here's the correct set-up, using coordinates (a, b) in the distance formula:

sqrt[(0 - a)^2 + (-1 - b)^2] = 9/10*sqrt(10)

Now, square each side, and simplify.

At this point, you have two equations, each containing the unknowns a and b.


i dont quite understand what you mean by "solve the system"

We have two equations, each containing the unknowns a and b. Using a process (algebraic or otherwise) to find values of a and b which satisfy both equations is called "solving the system".

I had assumed that your class already covered systems of equations. The process I'm thinking of is known as the Substitution Method. That's where we solve one of the equations for one of the unknowns, and then substitute the result for the same unknown in the other equation. Doing this creates a new equation that contains only one unknown, so we can solve for it. Once we know the value of one unknown, we can find the other through substitution, also.

Here's a simple example:

x^2 + y^2 = 1

2x + y = 1

Solve the 2nd equation for y:

y = 1 - 2x

Substitute this result for y in the other equation:

x^2 + (1 - 2x)^2 = 1

Solve this new equation for x:

x = 0 or x = 4/5

Now that we know one unknown, we find the other through substitution, also:

y = 1 - 2(0) = 1

or

y = 1 - 2(4/5) = -3/5

Therefore, the solutions to the given system of equations are

(0, 1) or (4/5, -3/5)

Now, try your exercise again. Let us know, if you get stuck. :cool:
 
hi again,

ive not taken any classes, im teaching myself but i have covered solving simultaneous equations by substitution just never seen it called 'solving a system'.

i eventually got the solutions although it was an incredibly lengthy process (hence why i dont want to show my working ;) ) many thanks for guiding me through this.
finally i just wanted to know exactly why i plug in the (x, y) coordinates i have found into the equation of the line they are on (that i deduce) and the distance formula, then solve the system. i can see it is correct but i dont know exactly why i am doing it. is it because those coordinates are satisfied by both the length of the line and the equation of the line?

anyhow i ended up with 8 sets of coordinates in total, 2 of which i discounted by comparing with graph (is that correct?) or is there a logical way to work out which coordinates are correct from the double result you get?

so i have the final answer and it tallies with what is given in the book. the coordinates of the other vertices are (9/10, 17/10) and (36/10, 8/10), (27/10, -19/10) or (-18/10, 26/10), (-27/10, -1/10)
 
let the unknowns x and y equal a and b, respectively

substituting them into the equation of the line gives me a=-1/3b-1 or 3a+b+3=0
Oops, I did not notice this mistake, before.

You substituted y=a and x=b, instead of x=a and y=b.

The equation should be:

a + 3b + 3 = 0
 
… it was an incredibly lengthy process …
I'm glad that you figured it out. I will post my work, tomorrow. Perhaps, you'll see something that helps you to shorten the process.


… finally i just wanted to know exactly why i plug in the (x, y) coordinates i have found into the equation of the line they are on … and the distance formula … is it because those coordinates are satisfied by both the length of the line and the equation of the line?
Yes.


anyhow i ended up with 8 sets of coordinates in total, 2 of which i discounted by comparing with graph (is that correct?) or is there a logical way to work out which coordinates are correct from the double result you get?
Comparing with the graph is good; without seeing what you did, I'm not sure why you obtained extras.


so i have the final answer and it tallies with what is given in the book. the coordinates of the other vertices are (9/10, 17/10) and (36/10, 8/10), (27/10, -19/10) or (-18/10, 26/10), (-27/10, -1/10)
The book showed (36/10, 8/10) and (-18/10, 26/10)? I would reduce fractions to lowest terms.

Vertices of Square #1 (clockwise):

(0, -1)

(9/10, 17/10)

(18/5, 4/5)

(27/10, -19/10)

Vertices of Square #2 (counterclockwise)

(0, -1)

(9/10, 17/10)

(-9/5, 13/5)

(-27/10, -1/10)
 
yea i caught the mistake when i started from scratch :)
 
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Comparing with the graph is good; without seeing what you did, I'm not sure why you obtained extras.

when solving the system of the fourth and fifth (previously unknown) equations of the line and the distance equation, it yields 2 distinct pairs of x and y values, as did the third line, which had values either side of (0, -1)... if that makes sense. i may post my work for clarity later

edit: i see now that i didnt need to use the distance equation repeatedly. i could have just used the equation of the lines i found and solved those by simultaneous elimination
 
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edit: i see now that i didnt need to use the distance equation repeatedly. i could have just used the equation of the lines i found and solved those by simultaneous elimination
Right. I suspected that you had used the Distance Formula more than once; that would take extra time. So, instead of showing all my work, I'll just comment on a couple things.

When using the Distance Formula for something other than finding a distance, it's often easier to work with the square of the distance, instead of the distance itself. That's what I did, when trying to find the points on line y=-1/3x-1 that are 9/10*sqrt(10) units from the vertex (0,-1).

In other words, I solved this system of equations:

b = -1/3*a - 1

(a - 0)^2 + (b + 1)^2 = 81/10

Substituting -1/3*a - 1 for b in the second equation gives:

(a - 0)^2 + (-1/3*a - 1 + 1)^2 = 81/10

a^2 + (-1/3*a)^2 = 81/10

a^2 + 1/9*a^2 = 81/10

10/9*a^2 = 81/10

a^2 = (9/10)*(81/10) = 729/100

|a| = sqrt(729/100) = 27/10

a = 27/10 or a = -27/10


Then, I used the Point-Slope form, to get equations for the lines passing through (27/10, -19/10) and (-27/10, -1/10) parallel to the given line with slope 3.

y + 19/10 = 3(x - 27/10) gives y = 3x - 10

y + 1/10 = 3(x + 27/10) gives y = 3x + 8

And then, as you said, the remaining vertices are found by finding the intersection points of the known lines.

Solving -1/3*x + 2 = 3x - 10 gives x = 18/5

Solving -1/3*x + 2 = 3x + 8 gives x = -9/5

Good work! :cool:
 
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