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Bunuel
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In the xy-coordinate system, the distance between the point (0,0) and 26 Mar 2015, 04:33
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In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)
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D
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In the xy-coordinate system, the distance between the point (0,0) and 26 Mar 2015, 04:40
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Distance formula =sqrt((x2-x1)^2 + (y2-y1)^2))

Say(x1,y1) is (0,0)
In that way, just plug in values to see, which one gets you to sqrt(40)

Only option D works out.
Answer: D
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In the xy-coordinate system, the distance between the point (0,0) and 27 Mar 2015, 08:38
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Bunuel
In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)

Let the cordinates of P be (x,y)

now distance between (0,0) and (x,y) =
square root of 40 = square root of (x^2+y^2)

Square root from both sides gets cancelled. now we can check with options given and which ever options gives us x^2+y^2 = 40 will be the answer.

Answer D
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In the xy-coordinate system, the distance between the point (0,0) and 27 Mar 2015, 19:24
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Bunuel
In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)


If the hypotnuse of a right angled triangle is \(\sqrt{40}\) then the other two sides can be 6,2 and hence D .

40=6^2 + 2^2
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In the xy-coordinate system, the distance between the point (0,0) and 27 Mar 2015, 20:40
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Bunuel
In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)

Distance of any point from origin is the square root of sum of square of the x and y coordinates of the point i.e. \(\sqrt{x^2 + y^2}\). Only option satisfying the sum of squares to be equal to 40 is D, hence D is the correct answer.
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In the xy-coordinate system, the distance between the point (0,0) and 27 Mar 2015, 20:49
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Distance between 2 points =\(\sqrt{(x1-x2)^2 + (y1-y2)^2}\)

Distance between two points (0,0) and (x,y) is \(\sqrt{40}\)

Substituting, \(\sqrt{x^2 + y^2} = \sqrt{40}\)

==> \(x^2 + y^2\) = 40

So only possible option is D (6,2)
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In the xy-coordinate system, the distance between the point (0,0) and 30 Mar 2015, 04:07
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Bunuel
In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)

MAGOOSH OFFICIAL SOLUTION:
Attachment:
dist40_exp.png
dist40_exp.png [ 42.73 KiB | Viewed 16005 times ]
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In the xy-coordinate system, the distance between the point (0,0) and 31 Mar 2015, 12:17
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Bunuel
In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)

Distance between 2 points =\sqrt{(x1-x2)^2 + (y1-y2)^2}

Since one of the points is (0,0), we only need to square the coordinates of the second point and add them to see if the total is 40.
Only option (D) satisfies this.

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In the xy-coordinate system, the distance between the point (0,0) and 17 Apr 2016, 13:29
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only option D gives the answer as 6^2 + 2^2 = 40 which gives the distance as \(\sqrt{40}\) . so correct answer is option D
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In the xy-coordinate system, the distance between the point (0,0) and 14 Jul 2016, 07:45
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Bunuel
In the xy-coordinate system, the distance between the point (0,0) and point P is \(\sqrt{40}\). Which of the following could be the coordinates of point P?

A. (4,7)
B. (4,10)
C. (5,6)
D. (6,2)
E. (20,20)

Humm.. Intersting, we got no slope , no other point except origin and the distance of the other point from origin.
Well lets work on that
Distance formula says that distance between two points =\(\sqrt{(x2-x1)^2+(y2-y1)^2}\)
\(\sqrt{(x2-0)^2+(y2-0)^2}\) (since pair of x,y is given as 0,0 in the question itself)
\(\sqrt{(x)^2+(y)^2}\)=\(\sqrt{40}\)

Now plugin the values:- only \(6^2 (=36) +2^2 (=4)\) gives us \(\sqrt{40}\)
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In the xy-coordinate system, the distance between the point (0,0) and 16 Feb 2022, 04:46
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Bunuel is second to none. Thank you Bunuel
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