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A triangle in the coordinate plane has vertices P=(c, c), Q=(c+3 t, c
[#permalink]
16 Oct 2026, 10:09
16 Oct 2026, 10:09
Expert Reply
00:00
Question Stats:
50% (04:23) correct
50% (03:55) wrong based on 4 sessions
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A triangle in the coordinate plane has vertices \(P=(c, c), Q=(c+3 t, c+t) \), and \(R= (c+t, c+3 t)\) , where \(c>0\) . Vertex Q lies to the left of the y -axis, and the area of the triangle is 36 .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
2 t+1 |
-3 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Re: A triangle in the coordinate plane has vertices P=(c, c), Q=(c+3 t, c
[#permalink]
20 Oct 2026, 19:30
20 Oct 2026, 19:30
1
With a question like this, a quick sketch goes a long way to seeing the relationship between the variables. Pick simple numbers to start, then adapt them to the specifics of the question.
Since the question only compares t, we can choose any value for c. Choose (1,1) (though really we could choose (0,0) to make things a little simpler). Since P is to the right of the y-axis and Q is to the left of the y-axis, t < 0. Set t=-1, and a quick sketch reveals that the points form an isosceles triangle with the base QR equal to the height (midpoint of QR to P). Points Q(-2,0) and R(0,-2) have a midpoint at (-1,-1), so the base and the height of the triangle (with t=-1) are both \(2\sqrt{2}\), and the area is 4. For the area to be 36, the base and height need to be \(\sqrt{72}\) = \(6\sqrt{2}\), and t = -3. This would put Q at (-8,-2) and R at (-2,-8), and the midpoint at (-5,-5).
Quantity A = 2t+1 = 2(-6)+1 = -11
Quantity B = -3
Quantity B is greater
Since the question only compares t, we can choose any value for c. Choose (1,1) (though really we could choose (0,0) to make things a little simpler). Since P is to the right of the y-axis and Q is to the left of the y-axis, t < 0. Set t=-1, and a quick sketch reveals that the points form an isosceles triangle with the base QR equal to the height (midpoint of QR to P). Points Q(-2,0) and R(0,-2) have a midpoint at (-1,-1), so the base and the height of the triangle (with t=-1) are both \(2\sqrt{2}\), and the area is 4. For the area to be 36, the base and height need to be \(\sqrt{72}\) = \(6\sqrt{2}\), and t = -3. This would put Q at (-8,-2) and R at (-2,-8), and the midpoint at (-5,-5).
Quantity A = 2t+1 = 2(-6)+1 = -11
Quantity B = -3
Quantity B is greater
gmatclubot
Re: A triangle in the coordinate plane has vertices P=(c, c), Q=(c+3 t, c [#permalink]
20 Oct 2026, 19:30
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