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# What is the area of the triangle formed by lines y = 5-x, 2y

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What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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08 Jun 2010, 08:40
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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

M21-33

[Reveal] Spoiler:
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.
[Reveal] Spoiler: OA
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Re: m21 #33 - Coordinate Geometry [#permalink]

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08 Jun 2010, 08:53
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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]

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08 Jun 2010, 09:09
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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

[Reveal] Spoiler:
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: $$y_2=\frac{3x}{2}=y_3=5-x$$ --> $$x=2$$, $$y=3$$ --> $$vertex_3=(2,3)$$.

$$Base=5$$, $$Height=3$$ --> $$Area=\frac{1}{2}*Base*Height=7.5$$.

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Re: m21 #33 - Coordinate Geometry [#permalink]

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08 Jun 2010, 09:25
Thanks Bunnel for the detailed explanation! Very helpful..
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Re: m21 #33 - Coordinate Geometry [#permalink]

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04 Nov 2010, 15:17
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]

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04 Nov 2010, 17:16
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vigneshpandi wrote:
I got the three co-ordinates as (5,5) (0,0) and (2,3)...
Anyone Please tel me where am i going wrong?

Second vertex: intersection of $$y=0$$ and $$y=5-x$$ --> $$y_1=0=y_3=5-x$$ --> $$x=5$$, $$y=0$$ --> $$vertex_2=(5,0)$$;

Or refer to the graphs below:
Attachment:

graph.php.png [ 19.97 KiB | Viewed 4807 times ]
Hope it's clear.
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Re: m21 #33 - Coordinate Geometry [#permalink]

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06 Nov 2010, 09:33
Bunuel, what do you use for making the graphs?
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Re: m21 #33 - Coordinate Geometry [#permalink]

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06 Nov 2010, 09:38
medanova wrote:
Bunuel, what do you use for making the graphs?

There are many online graph plotters, try this one: http://graph-plotter.cours-de-math.eu/ or: http://www.wolframalpha.com/
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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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10 Mar 2014, 04:04
Bunuel.. I got this question in gmat club tests..

What is the area of a triangle enclosed by line 2x+3y=6, line y−x=2 and X-axis on the coordinate plane?

My question is.. why approach is different ?

If 2x+3y=6 and x=0 then y=2. And when y=0 then x=3.. (0,2) (3,0)

then y cant we do same with y=5-x? (5,0) (0,5)

and 2x=3y.. x=0 and y=0

where m wrong?
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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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10 Mar 2014, 05:30
sanjoo wrote:
Bunuel.. I got this question in gmat club tests..

What is the area of a triangle enclosed by line 2x+3y=6, line y−x=2 and X-axis on the coordinate plane?

My question is.. why approach is different ?

If 2x+3y=6 and x=0 then y=2. And when y=0 then x=3.. (0,2) (3,0)

then y cant we do same with y=5-x? (5,0) (0,5)

and 2x=3y.. x=0 and y=0

where m wrong?

Below is that question. Please point out what differences are you talking about.

What is the area of a triangle enclosed by line 2x+3y=6, line y-x=2 and X-axis on the coordinate plane?

A. 3
B. 4
C. 5
D. 8
E. 10

Look at the diagram below:
Attachment:

Area.png [ 7.92 KiB | Viewed 3763 times ]

Lines $$y=-\frac{2}{3}x+2$$ and $$y=x+2$$ intersect at point (0, 2). So the height of enclosed triangle is 2. Next, X-intercept of line $$y=-\frac{2}{3}x+2$$ is (3, 0) and X-intercept of line $$y=x+2$$ is (-2, 0), so the base of enclosed triangle is 3-(-2)=5. The area is $$\frac{1}{2}*base*height=\frac{1}{2}*5*2=5$$.

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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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11 Mar 2014, 00:36
Bunuel wrote:
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

[Reveal] Spoiler:
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: $$y_2=\frac{3x}{2}=y_3=5-x$$ --> $$x=2$$, $$y=3$$ --> $$vertex_3=(2,3)$$.

$$Base=5$$, $$Height=3$$ --> $$Area=\frac{1}{2}*Base*Height=7.5$$.

actually this highlighted thing is my question.. y have we taken another equation y=3/2x to get the points of equation y=5-x?

What is the area of a triangle enclosed by line 2x+3y=6, line y-x=2 and X-axis on the coordinate plane?

and i tuk this question (in bold text) to show u that , we didnt do same thing in this question. we tuk different vertices by individully solving an equation.

i tried my best to explain my question

thank u
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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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11 Mar 2014, 01:11
sanjoo wrote:
Bunuel wrote:
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

[Reveal] Spoiler:
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: $$y_2=\frac{3x}{2}=y_3=5-x$$ --> $$x=2$$, $$y=3$$ --> $$vertex_3=(2,3)$$.

$$Base=5$$, $$Height=3$$ --> $$Area=\frac{1}{2}*Base*Height=7.5$$.

actually this highlighted thing is my question.. y have we taken another equation y=3/2x to get the points of equation y=5-x?

What is the area of a triangle enclosed by line 2x+3y=6, line y-x=2 and X-axis on the coordinate plane?

and i tuk this question (in bold text) to show u that , we didnt do same thing in this question. we tuk different vertices by individully solving an equation.

i tried my best to explain my question

thank u

The approaches are identical for both questions:

The base in both, is on x-axis. Find x-intercepts of other two lines and find the difference between them to get the length of the base.
Next, find at what point do these two lines intersect. y-coordinate of that point would be the height.

To get intersection point of two lines y=3x/2 and y=5-x, we equate 3x/2 and 5-x to get x-coordinate of intersection point --> 3x/2=5-x --> x=2. Then we substitute it into either of the equations to get y=coordinate of intersection point --> y=5-2=3. So, y=3x/2 and y=5-x intersect at (2,3).

Hope it's clear.
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Triangle problem - help me understand how to find vertices [#permalink]

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12 Apr 2014, 09:23
Q: What is the area of the triangle formed by lines y=5−x, 2y=3x, and y=0?

Having trouble understanding the best approach to calculating vertices of the triangle which are given as follows:

(0,0)
(5,0)
(2,3)

Can anyone explain the best approach for calculating these? I'm fine with calculating the area once I have the vertices.

Thanks.
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Re: Triangle problem - help me understand how to find vertices [#permalink]

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12 Apr 2014, 12:03
rxn wrote:
Q: What is the area of the triangle formed by lines y=5−x, 2y=3x, and y=0?

Having trouble understanding the best approach to calculating vertices of the triangle which are given as follows:

(0,0)
(5,0)
(2,3)

Can anyone explain the best approach for calculating these? I'm fine with calculating the area once I have the vertices.

Thanks.

The vertices are the points where these lines intersect with each other.
so, solve two two equation together:

point where y=0 and 2y = 3x intersect: put y=0 in 2y = 2x => x = 0, so coordinates (x,y) = (0,0)
point where y = 0 and y = 5 - x intersect: put y=0 in y = 5-x => x = 5, so coordinates (x,y) = (5,0)
point where 2y = 3x and y = 5 - x intersect: put y=3x/2 in y = 5-x => x = 2; y = 3, so coordinates (x,y) = (2,3)

--------------------------------
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Re: Triangle problem - help me understand how to find vertices [#permalink]

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13 Apr 2014, 04:54
Alternatively, you could draw a quick graph on your scratch paper using the equations. For 2y= 3x , use the equation y= 3x/2 to draw the line.
Then you just have to read the coordinates to have the coordinates of the vertices.
I still think ind23's method is faster, but it is important to visualize geometry problems as it makes it easier to answer.
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Re: Triangle problem - help me understand how to find vertices [#permalink]

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13 Apr 2014, 05:35
rxn wrote:
Q: What is the area of the triangle formed by lines y=5−x, 2y=3x, and y=0?

Having trouble understanding the best approach to calculating vertices of the triangle which are given as follows:

(0,0)
(5,0)
(2,3)

Can anyone explain the best approach for calculating these? I'm fine with calculating the area once I have the vertices.

Thanks.

Merging similar topics. Please refer to the discussion above.

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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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14 Apr 2015, 08:05
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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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16 Sep 2016, 11:03
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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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15 Oct 2016, 12:14
Bunuel wrote:
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

[Reveal] Spoiler:
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: $$y_2=\frac{3x}{2}=y_3=5-x$$ --> $$x=2$$, $$y=3$$ --> $$vertex_3=(2,3)$$.

$$Base=5$$, $$Height=3$$ --> $$Area=\frac{1}{2}*Base*Height=7.5$$.

Hi Bunuel,

Please help me solidify the understanding for this - the reason you equated the line equations is because intersecting lines have the same slope right? Thank you.
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Re: What is the area of the triangle formed by lines y = 5-x, 2y [#permalink]

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16 Oct 2016, 01:51
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TheLordCommander wrote:
Bunuel wrote:
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

[Reveal] Spoiler:
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: $$y_2=\frac{3x}{2}=y_3=5-x$$ --> $$x=2$$, $$y=3$$ --> $$vertex_3=(2,3)$$.

$$Base=5$$, $$Height=3$$ --> $$Area=\frac{1}{2}*Base*Height=7.5$$.

Hi Bunuel,

Please help me solidify the understanding for this - the reason you equated the line equations is because intersecting lines have the same slope right? Thank you.

No, we are equating to get the x-intercept of the point of intersection.
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Re: What is the area of the triangle formed by lines y = 5-x, 2y   [#permalink] 16 Oct 2016, 01:51
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