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How many points (x, y) lie on the line segment
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08 Jun 2026, 01:12
08 Jun 2026, 01:12
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Question Stats:
75% (02:06) correct
25% (01:54) wrong based on 4 sessions
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How many points (x, y) lie on the line segment between \((22, 12 \frac{2}{3} )\) and \(( 7,17\frac{2 }{ 3 } )\) such that $x$ and $y$ are both integers?
A. 4
B. 5
C. 7
D. 8
E. 9
A. 4
B. 5
C. 7
D. 8
E. 9
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Re: How many points (x, y) lie on the line segment
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11 Jun 2026, 22:20
11 Jun 2026, 22:20
1
Expert Reply
Explanation
The endpoints are
\((22, 12\frac{2}{3})=(22,\frac{38}{3}) and (7,17\frac{2}{3})=(7,\frac{53}{3}).\)
First find the equation of the line.
The slope is
(53/3)-(38/3)]/[7/22]=\(\frac{5}{-15}=-\frac{1}{3}\)
Using point-slope form with \((22, \frac{38}{3}),\)
\(y-\frac{38}{3}=-\frac{1}{3}(x−22).\)
Multiplying by 3,
\(3y-38=-x+22,\)
so
\(x+3y=60.\)
For x and y both to be integers, we need integer solutions of
\(x+3y=60.\)
This implies that x ranges from 7 to 22, so the possible integer values are
\(x=9,12,15,18,21.\)
These give
(9,17), (12,16), (15,15), (18,14), (21,13),
All of which lie on the segment and these are 5 sets of possible values
Answer: B
The endpoints are
\((22, 12\frac{2}{3})=(22,\frac{38}{3}) and (7,17\frac{2}{3})=(7,\frac{53}{3}).\)
First find the equation of the line.
The slope is
(53/3)-(38/3)]/[7/22]=\(\frac{5}{-15}=-\frac{1}{3}\)
Using point-slope form with \((22, \frac{38}{3}),\)
\(y-\frac{38}{3}=-\frac{1}{3}(x−22).\)
Multiplying by 3,
\(3y-38=-x+22,\)
so
\(x+3y=60.\)
For x and y both to be integers, we need integer solutions of
\(x+3y=60.\)
This implies that x ranges from 7 to 22, so the possible integer values are
\(x=9,12,15,18,21.\)
These give
(9,17), (12,16), (15,15), (18,14), (21,13),
All of which lie on the segment and these are 5 sets of possible values
Answer: B
gmatclubot
Re: How many points (x, y) lie on the line segment [#permalink]
11 Jun 2026, 22:20
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