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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 [#permalink]
A distance of 5 from point (1 , 1) in the coordinate plane will be given by all the points that lie on a circle with radius of 5 and the center at (1 , 1)

Using the equation of a circle, every point that is 5 units from (1 , 1) will lie on the circumference of the following graph:

(P - 1)^2 + (Q - 1)^2 = (5)^2


(1st) we know for sure that if we move 5 units horizontally or vertically from point (1 , 1), we will find the radii of the circle with integer coordinates.

These 4 points are found at:

Count 5 units to the left of (1 , 1) —- at (-4 , 1)

Count 5 units up from (1 , 1) —- at (1 , 6)

Count 5 units to the right of (1 , 1) —- at (6 , 1)

Count 5 units down from (1 , 1) ——- at (1 , -4)


(2nd) the distance formula, used to measure the distance between 2 points, is simply an application of the Pythagorean Theorem.

Given that the distance we are looking for is 5 and we are required to have integer coordinates, if we can draw a right triangle with a hypotenuse of 5 and legs of 3 and 4, we will be able to find integer coordinates that fall on the circumference of this circle.


Quadrant 2:

From point (1, 1):

(1st)
move 4 units to the left of center (1 , 1) to point (-3 , 1)

And

Then Move 3 units up from that point to point (-3 , 4)

The distance from point (-3 , 4) to point (1 , 1) will be 5 units


(2nd)
This time move 3 units to the left

And

From that point move 4 units up to land on point (-2 , 5)

From point (-2 , 5) to point (1 , 1) there will be a distance of 5 units

For Quadrants I, III, and IV —— we can also find 2 points with integer coordinates using the same reasoning.


Thus total points with integer coordinates are:

(4) * (2 more per quadrant) * (4 quadrants) =

12

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Re: In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 [#permalink]
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Re: In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 [#permalink]
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