Solve Isosceles Triangle in Coordinate System

[thereddevils]

Homework Statement

I will to explain this without a diagram. Consider a coordinate system(1st quadrant) where the x and y-axis both stop at 7 units (i mean the boundary), there are two points P(3,3) and Q(4,4). How many points can R be positioned such that PQR is an isosceles triangle.

Homework Equations

The Attempt at a Solution

I got 6.

 

[Mentallic]

Well I had quite a few and started counting them... Eventually I counted up to infinite different possible points

There are actually 3 paths the point R can trace, a line and 2 circles. See if you can find them.

 

[HallsofIvy]

Homework Statement

I will to explain this without a diagram. Consider a coordinate system(1st quadrant) where the x and y-axis both stop at 7 units (i mean the boundary), there are two points P(3,3) and Q(4,4). How many points can R be positioned such that PQR is an isosceles triangle.

Homework Equations

The Attempt at a Solution

I got 6.

Then you haven't given all the information. Are you talking about a "mesh" where the points must have integer components? As you give it, any point on the perpendicular bisector of PQ works and there are an infinite number of them.

 

[thereddevils]

Then you haven't given all the information. Are you talking about a "mesh" where the points must have integer components? As you give it, any point on the perpendicular bisector of PQ works and there are an infinite number of them.

yeah sorry, it must be of integer components where you can precisely represent that particular point on the plane.

My 6 points are

(5,2) , (2,5) , (6,1) , (1,6) , (7,0) , (0,7)

 

[Mentallic]

must be of integer components where you can precisely represent that particular point on the plane.

Yet the distance between P and Q is not an integer.
Anyway, this means we have the two equations

$$\sqrt{(x-4)^2+(y-4)^2}=n$$

$$\sqrt{(x-3)^2+(y-3)^2}=n$$

where n is an integer, to solve simultaneously.
After simplifying, we find the perpendicular bisector of PQ to be the line x+y-7=0
Now just use all possible integer values of x and y such that x+y=7.

I don't know if I should scrap the idea of the point R can trace that are circles rather than this line, because that means we have to use a non-integer value for another length.

 

[thereddevils]

Yet the distance between P and Q is not an integer.
Anyway, this means we have the two equations

$$\sqrt{(x-4)^2+(y-4)^2}=n$$

$$\sqrt{(x-3)^2+(y-3)^2}=n$$

where n is an integer, to solve simultaneously.
After simplifying, we find the perpendicular bisector of PQ to be the line x+y-7=0
Now just use all possible integer values of x and y such that x+y=7.

I don't know if I should scrap the idea of the point R can trace that are circles rather than this line, because that means we have to use a non-integer value for another length.

sigh.. 8 points then including (4,3) , (3,4)