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MathRevolution
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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 10 Aug 2018, 02:36
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[Math Revolution GMAT math practice question]

In the x-y coordinate plane, the distance between \((p,q)\) and \((1,1)\) is \(5\). If \(p\) and \(q\) are integers, how many possibilities are there for the point \((p,q)\)?

\(A. 2\)
\(B. 4\)
\(C. 8\)
\(D. 12\)
\(E. 16\)
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D
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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 10 Aug 2018, 03:25
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MathRevolution
[Math Revolution GMAT math practice question]

In the x-y coordinate plane, the distance between \((p,q)\) and \((1,1)\) is \(5\). If \(p\) and \(q\) are integers, how many possibilities are there for the point \((p,q)\)?

\(A. 2\)
\(B. 4\)
\(C. 8\)
\(D. 12\)
\(E. 16\)
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The distance between \((p,q)\) and \((1,1)\) is \(5\)
Or, \((p-1)^2+(q-1)^2=5^2\)

1) \((p-1)^2+(q-1)^2=5^2+0^2\); so\((p-1)^2=0\) and \((q-1)^2=5^2\). Possible pairs of (p,q):-(0,5), (0,-5),(5,0), (-5,0) ----(4 pairs)
2)\((p-1)^2+(q-1)^2=5^2=4^2+3^2\);so \((p-1)^2=4^2\) and \((q-1)^2=3^2\). Possible pairs of (p,q):-(5,4), (5,-2),(-3,4), (-3,-2)---(4 pairs)
3)\((p-1)^2+(q-1)^2=5^2=3^2+4^2\);so \((p-1)^2=3^2 and (q-1)^2=4^2\). Possible pairs of (p,q):-(4,5), (-2,5),(4,-3), (-2,-3)---(4 pairs)
N,B,:- If \((x-a)^2=k\) , then \(\sqrt{(x-a)^2}=|x-a|=+k\) or \(-k\)
Total number of (p,q) pairs=4+4+4=12
Ans. (D)
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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 13 Aug 2018, 06:29
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=>

(p-1)^2 + (q-1)^2 = 5^2
If p – 1 = ±3, and q - 1 = ±4, then p = 1 ± 3, and q = 1 ± 4. There are four possible points: ( p, q ) = ( 4, 5 ), ( 4, -3 ), ( -2, 5 ), ( -2, -3 ).
If p – 1 = ±4, and q - 1 = ±3, then p = 1 ±4, and q = 1 ± 3. There are four possible points: ( p, q ) = ( 5, 4 ), ( 5, -2 ), ( -3, 4 ), ( -3, -2 ).
If p – 1 = 0, and q - 1 = ±5, then p = 1, and q = 1 ±5. There are two possible points: ( p, q ) = ( 1, 6 ), ( 1, -4 ).
If p – 1 = ±5, and q - 1 = 0, then p – 1 = ±5, and q = 1. There are two possible points: ( p, q ) = ( 6, 1 ), ( -4, 1 ).

There are a total of 4 + 4 + 2 + 2 = 12 possibilities for the point (p,q).
Therefore, the answer is D.

Answer: D
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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 05 Aug 2021, 00:03
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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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In the x-y coordinate plane, the distance between (p,q) and (1,1) is 5 14 Jan 2025, 03:02
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Distance between two points \((p,q)\) and \((1,1)\) is given by
\(\sqrt{(p - 1)^2 + (q - 1)^2} = 5\)

=> \((p - 1)^2 + (q - 1)^2 = 5^2 = 25\)
=> Sum of square of two integers = 25

Now, Sum of square of two integers = 25, is possible for following set of numbers
  1. First Number = ±3, Second Number = ±4 => 4 cases (-3,-4), (-3,4), (3,-4), (3,4) as the corresponding values for (p-1, q-1)
    => Values for (p,q) will be (-3+1,-4+1), (-3+1,4+1), (3+1,-4+1), (3+1,4+1)
    => Values for (p,q) will be (-2,-3), (-2,5), (4,-3), (4,5)
  2. First Number = ±4, Second Number = ±3 => 4 cases
    => Values for (p,q) will be (-3+1,-4+1), (-3+1,4+1), (3+1,-4+1), (3+1,4+1)=> Values for (p,q) will be (-2,-3), (-2,5), (4,-3), (4,5)
  3. First Number = 0, Second Number = ±5 => 2 cases
  4. First Number = ±5 , Second Number = 0 => 2 cases

So, Total cases = 4 + 4 + 2 + 2 = 12

So, Answer will be D
Hope it helps!

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