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What is the area of the triangle formed by lines y = 5-x, 2y 08 Jun 2010, 09:40
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35% (medium)

Question Stats:

75% (02:41) correct 25% (02:40) wrong based on 903 sessions

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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

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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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What is the area of the triangle formed by lines y = 5-x, 2y 08 Jun 2010, 10:09
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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

Show 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.
Show more

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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What is the area of the triangle formed by lines y = 5-x, 2y 08 Jun 2010, 10:25
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Thanks Bunnel for the detailed explanation! Very helpful..
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What is the area of the triangle formed by lines y = 5-x, 2y 04 Nov 2010, 16:17
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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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What is the area of the triangle formed by lines y = 5-x, 2y 04 Nov 2010, 18:16
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vigneshpandi
I got the three co-ordinates as (5,5) (0,0) and (2,3)...
Anyone Please tel me where am i going wrong?
Show more

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
graph.php.png [ 19.97 KiB | Viewed 41842 times ]
Hope it's clear.
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What is the area of the triangle formed by lines y = 5-x, 2y 06 Nov 2010, 10:33
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Bunuel, what do you use for making the graphs?
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What is the area of the triangle formed by lines y = 5-x, 2y 06 Nov 2010, 10:38
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Bunuel, what do you use for making the graphs?
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There are many online graph plotters, try this one: https://graph-plotter.cours-de-math.eu/ or: https://www.wolframalpha.com/
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What is the area of the triangle formed by lines y = 5-x, 2y 10 Mar 2014, 05:04
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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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What is the area of the triangle formed by lines y = 5-x, 2y 10 Mar 2014, 06:30
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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?
Show more

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
Area.png [ 7.92 KiB | Viewed 40291 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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).

Answer: C.
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What is the area of the triangle formed by lines y = 5-x, 2y 11 Mar 2014, 01:36
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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

Show 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: \(y_2=\frac{3x}{2}=y_3=5-x\) --> \(x=2\), \(y=3\) --> \(vertex_3=(2,3)\).

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

Answer: A.
Show more

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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What is the area of the triangle formed by lines y = 5-x, 2y 11 Mar 2014, 02:11
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sanjoo
Bunuel
abhi758
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

Show 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: \(y_2=\frac{3x}{2}=y_3=5-x\) --> \(x=2\), \(y=3\) --> \(vertex_3=(2,3)\).

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

Answer: A.

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
Show more

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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What is the area of the triangle formed by lines y = 5-x, 2y 12 Apr 2014, 10:23
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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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What is the area of the triangle formed by lines y = 5-x, 2y 12 Apr 2014, 13:03
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rxn
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.
Show more


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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What is the area of the triangle formed by lines y = 5-x, 2y 13 Apr 2014, 05:54
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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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What is the area of the triangle formed by lines y = 5-x, 2y 13 Apr 2014, 06:35
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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.
Show more

Merging similar topics. Please refer to the discussion above.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1, 2, 3, 7 and 8. Thank you.
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What is the area of the triangle formed by lines y = 5-x, 2y 15 Oct 2016, 13:14
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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

Show 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: \(y_2=\frac{3x}{2}=y_3=5-x\) --> \(x=2\), \(y=3\) --> \(vertex_3=(2,3)\).

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

Answer: A.
Show more

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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What is the area of the triangle formed by lines y = 5-x, 2y 16 Oct 2016, 02:51
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abhi758
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

Show 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: \(y_2=\frac{3x}{2}=y_3=5-x\) --> \(x=2\), \(y=3\) --> \(vertex_3=(2,3)\).

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

Answer: A.

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.
Show more

No, we are equating to get the x-intercept of the point of intersection.
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What is the area of the triangle formed by lines y = 5-x, 2y Updated on: 20 May 2018, 06:58
Originally posted by dave13 on 20 May 2018, 06:37.
Last edited by dave13 on 20 May 2018, 06:58, edited 1 time in total.
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abhi758
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

Show 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: \(y_2=\frac{3x}{2}=y_3=5-x\) --> \(x=2\), \(y=3\) --> \(vertex_3=(2,3)\).

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

Answer: A.
Show more


hi Bunuel

i really dont get your solution so we have this y = 5-x, 2y = 3x, and y = 0 and we have these coordinaty geometry map with x and y axis

y = 5-x, so i plugged in this 2y = 3x --> x = 2y/3 into this y = 5-x and got y = 3 so i have y= 3 and x =2


hence y is height and x is basis ---> y = 3 and x = 2 i am sure you can visualize my wring solution :)

area (3*2)/2 = 3 :?

pushpitkc - help pls:)

why are solving in a totally different way:? are quant gmatclub tests harder than the quant on gmat exam ?
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What is the area of the triangle formed by lines y = 5-x, 2y 20 May 2018, 06:47
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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

For line 2
2y = 3x ==> y=3/2x

If x = 0, y=3/2(0) ==> 0
y=0 , x=0

Where Am'I going wrong? chetan2u, VeritasPrepKarishma , Bunuel , mikemcgarry
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What is the area of the triangle formed by lines y = 5-x, 2y 20 May 2018, 20:15
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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

For line 2
2y = 3x ==> y=3/2x

If x = 0, y=3/2(0) ==> 0
y=0 , x=0

Where Am'I going wrong? chetan2u, VeritasPrepKarishma , Bunuel , mikemcgarry
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I am not sure why you think you are going wrong.

Given 2y = 3x, we get y = (3/2)x + 0 (to put it in the y = mx + c form)
This just means that it is a line with slope 1.5 and y intercept 0. So it must pass through the origin (which you got as - if x = 0 then y = 0)

y = 0 is the equation of the x axis. So we know that these two lines intersect at the origin. So it is one of the three vertices of the triangle.

Similarly, now we find where the third line intersects these two lines to get the other two vertices of the triangle.
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