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In the xy-plane, the distance between the point P (1,3) and
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13 Oct 2020, 06:29
13 Oct 2020, 06:29
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Question Stats:
82% (01:18) correct
17% (02:09) wrong based on 47 sessions
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In the xy-plane, the distance between the point P (1,3) and Q (9,y) is 17. If y > 0, then what is the value of y?
A. 15
B. 16
C. 17
D. 18
E. 19
Kudos for the right answer and explanation
A. 15
B. 16
C. 17
D. 18
E. 19
Kudos for the right answer and explanation
Re: In the xy-plane, the distance between the point P (1,3) and
[#permalink]
13 Oct 2020, 10:22
13 Oct 2020, 10:22
1
I plug the answer back to the question (starting from C)
Assume y =17
(1,3) (9,17)
Then the distance equal (17-3)^2 +(9-1)^2
225+64=289
Distance equal √289=17
Ans equal C
Posted from my mobile device
Assume y =17
(1,3) (9,17)
Then the distance equal (17-3)^2 +(9-1)^2
225+64=289
Distance equal √289=17
Ans equal C
Posted from my mobile device
In the xy-plane, the distance between the point P (1,3) and
[#permalink]
14 Oct 2022, 02:58
14 Oct 2022, 02:58
2
\(x^2 + 8^2 = 17^2\)
Then
\(x^2 = 289-64\)
So
\(x = \sqrt{225}=15\)
Since it started at y equals to 3 so 15+3 = 18
D
Then
\(x^2 = 289-64\)
So
\(x = \sqrt{225}=15\)
Since it started at y equals to 3 so 15+3 = 18
D
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