Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).

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

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
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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

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).

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

Show Tags

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)

-------------------------------- Kudos if the answer helped

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

Show Tags

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.

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

Show Tags

14 Apr 2015, 08:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

16 Sep 2016, 11:03

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

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.

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

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.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...