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In the xy-plane, the distance between the point P (1,3) and
[#permalink]
13 Oct 2020, 06:29
13 Oct 2020, 06:29
Expert Reply
00:00
Question Stats:
80% (01:08) correct
19% (02:09) wrong based on 41 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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In the xy-plane, the distance between the point P (1,3) and
[#permalink]
22 Aug 2023, 18:48
22 Aug 2023, 18:48
1
Explanation:
Given: P (1,3) and Q (9,y) D=17. y > 0
d=√((x2-x1)²+(y2-y1)²)
Sqauring both sides
d^2 = (x2-x1)²+(y2-y1)²
17^2 = (9-1)^2 + (y-3)^2
289-64 = (y-3)^2
√(225) = y-3
15 = y-3 (as y>0, Taking +ve value)
y=18
Given: P (1,3) and Q (9,y) D=17. y > 0
d=√((x2-x1)²+(y2-y1)²)
Sqauring both sides
d^2 = (x2-x1)²+(y2-y1)²
17^2 = (9-1)^2 + (y-3)^2
289-64 = (y-3)^2
√(225) = y-3
15 = y-3 (as y>0, Taking +ve value)
y=18
gmatclubot
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