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m21 #33 - Coordinate Geometry [#permalink]
 08 Jun 2010, 09:40
Question Stats:
44% (03:02) correct
55% (01:40) wrong based on 0 sessions
What is the area of the triangle formed by lines y = 5-x, 2y = 3x, and y = 0? a. 7.5 b. 8.0 c. 9.0 d 10.5 e. 15.0 The vertices of the triangle formed are (0,0), (5,0) and (2,3) The first two vertices are fine, but how do we arrive at the third vertice of (2,3)? Since the equation for the third line is y = 0x + 0 / y = 0 it should be a horizontal line. Look forward to any explanations.
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Re: m21 #33 - Coordinate Geometry [#permalink]
08 Jun 2010, 09:53
If you solve for the first 2 linear equations -> y = 5-x and 2y = 3x, you will have (2,3) as the solution.
A - 7.5
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Re: m21 #33 - Coordinate Geometry [#permalink]
08 Jun 2010, 10:09
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abhi758 wrote: What is the area of the triangle formed by lines y = 5-x, 2y = 3x, and y = 0? a. 7.5 b. 8.0 c. 9.0 d 10.5 e. 15.0 The vertices of the triangle formed are (0,0), (5,0) and (2,3) The first two vertices are fine, but how do we arrive at the third vertice of (2,3)? Since the equation for the third line is y = 0x + 0 / y = 0 it should be a horizontal line. Look forward to any explanations. We have three equations of lines: y_1=0 (x-axis), y_2=\frac{3x}{2}, y_3= 5-x. Equate the functions to get the x-coordinate of the vertex (x-coordinate of the intersection point) and then substitute this value in either of functions to get y-coordinate of the vertex. First vertex: y_1=0=y_2=\frac{3x}{2} --> x=0, y=0 --> vertex_1=(0,0); Second vertex: y_1=0=y_3=5-x --> x=5, y=0 --> vertex_2=(5,0); Third vertex: y_2=\frac{3x}{2}=y_3=5-x --> x=2, y=3 --> vertex_3=(2,3). Base=5, Height=3 --> Area=\frac{1}{2}*Base*Height=7.5. Answer: A.
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Re: m21 #33 - Coordinate Geometry [#permalink]
08 Jun 2010, 10:25
Thanks Bunnel for the detailed explanation! Very helpful..
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Re: m21 #33 - Coordinate Geometry [#permalink]
04 Nov 2010, 16:17
I got the three co-ordinates as (5,5) (0,0) and (2,3)... Anyone Please tel me where am i going wrong?
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Re: m21 #33 - Coordinate Geometry [#permalink]
04 Nov 2010, 18:16
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Re: m21 #33 - Coordinate Geometry [#permalink]
06 Nov 2010, 10:07
+1 A 
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Re: m21 #33 - Coordinate Geometry [#permalink]
06 Nov 2010, 10:33
Bunuel, what do you use for making the graphs?
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Re: m21 #33 - Coordinate Geometry [#permalink]
06 Nov 2010, 10:38
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Re: m21 #33 - Coordinate Geometry
[#permalink]
06 Nov 2010, 10:38
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