contain the third vertex

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

contain the third vertex

by sanju09 » Wed Mar 03, 2010 5:02 am

The area of a triangle is 21. If two of its vertices lie on points (5, 3) and (-4, -3), then which of the following represents the equation of line that would always contain the third vertex of triangle?
(A) 2 x - 3 y = 15
(B) 2 x + 3 y = 15
(C) 3 x + 2 y = 15
(D) 3 x - 2 y = 15
(E) both (B) and (D)

I agree with rohan_vus, so the necessary editing has been carried out in the wordings, this indeed is made for A to be the OA.

Last edited by sanju09 on Thu Mar 04, 2010 1:11 am, edited 1 time in total.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

Quote

Source: Beat The GMAT — Problem Solving | Wed Mar 03, 2010 5:02 am

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

by harsh.champ » Wed Mar 03, 2010 5:11 am

sanju09 wrote:The area of a triangle is 21. If two of its vertices lie on points (5, 3) and (-4, -3), then which of the following represents the equation of line that could contain the third vertex of triangle?
(A) 2 x - 3 y = 15
(B) 2 x + 3 y = 15
(C) 3 x + 2 y = 15
(D) 3 x - 2 y = 15
(E) both (B) and (D)

Even if the line from dat pt. is the altitude we will get the point.
(1,0) will be the mid-pt.
Now,area = 21
My method is a long one:-Calculating from Hero's formula.
Any short method??

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

Quote

rohan_vus: Master | Next Rank: 500 Posts; Posts: 189; Joined: Thu Apr 03, 2008 2:03 pm; Location: USA; Thanked: 21 times

by rohan_vus » Wed Mar 03, 2010 5:26 am

You can get the eqn of line which join vertices points (5, 3) and (-4, -3)
-->y-3/x-5 = 6/9 ==> 2x - 3y - 1 = 0

Now the third verex got to be in some line perpendicular to the line joining points (5, 3) and (-4, -3)

When you calculate area you take base *height/2 Here base you can take as line joining points (5, 3) and (-4, -3) and height a perpendicular dropped from the third vertex. But in this particular case you dont really need to worry about actual area size , etc.

Eqn of line that can be dropped from third vertex perpendicular to base is of form 3x + 2y + c = 0,

Now looking at the choices one choice is of this form , i,e choice C .

Last edited by rohan_vus on Wed Mar 03, 2010 7:14 am, edited 1 time in total.

Quote

sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

by sanju09 » Wed Mar 03, 2010 5:30 am

harsh.champ wrote:
sanju09 wrote:The area of a triangle is 21. If two of its vertices lie on points (5, 3) and (-4, -3), then which of the following represents the equation of line that could contain the third vertex of triangle?
(A) 2 x - 3 y = 15
(B) 2 x + 3 y = 15
(C) 3 x + 2 y = 15
(D) 3 x - 2 y = 15
(E) both (B) and (D)
Even if the line from dat pt. is the altitude we will get the point.
(1,0) will be the mid-pt.
Now,area = 21
My method is a long one:-Calculating from Hero's formula.
Any short method??

A line parallel to the line segment joining the points (5, 3) and (-4, -3), and at a distance of the calculable altitude, would contain all points those are needful here. We really don't need to do that hard work, though.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

Quote

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

by harsh.champ » Wed Mar 03, 2010 5:31 am

rohan_vus wrote:You can get the eqn of line which join vertices points (5, 3) and (-4, -3)
-->y-3/x-5 = 6/9 ==> 2x - 3y - 1 = 0

Now the third verex got to be in some line perpendicular to the line joining points (5, 3) and (-4, -3)

When you calculate area you take base *height/2 Here base you can take as line joining points (5, 3) and (-4, -3) and height a perpendicular dropped from the third vertex. But in this particular case you dont really need to worry about actual area size , etc.

Eqn of line that can be dropped from third vertex perpendicular to base is of form 3x + 2y + c = 0,

Now looking at the choices only one choice is of this form , i,e choice C

Why is it compulsory that the line will be perpendicular ??
Its not given that AD will be necessarily a altitude.

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

Quote

yeahdisk: Senior | Next Rank: 100 Posts; Posts: 45; Joined: Tue Mar 02, 2010 4:27 am; Thanked: 2 times

by yeahdisk » Wed Mar 03, 2010 7:01 am

Surely, any line that doesn't go through both of the vertices already given could create a triangle with any area?

Quote

rohan_vus: Master | Next Rank: 500 Posts; Posts: 189; Joined: Thu Apr 03, 2008 2:03 pm; Location: USA; Thanked: 21 times

by rohan_vus » Wed Mar 03, 2010 7:08 am

harsh.champ wrote:
rohan_vus wrote:You can get the eqn of line which join vertices points (5, 3) and (-4, -3)
-->y-3/x-5 = 6/9 ==> 2x - 3y - 1 = 0

Now the third verex got to be in some line perpendicular to the line joining points (5, 3) and (-4, -3)

When you calculate area you take base *height/2 Here base you can take as line joining points (5, 3) and (-4, -3) and height a perpendicular dropped from the third vertex. But in this particular case you dont really need to worry about actual area size , etc.

Eqn of line that can be dropped from third vertex perpendicular to base is of form 3x + 2y + c = 0,

Now looking at the choices only one choice is of this form , i,e choice C
Why is it compulsory that the line will be perpendicular ??
Its not given that AD will be necessarily a altitude.

Its not about being compulsory . Its about possibility . Infact any line could be part of the third vertex provided it gives you calculabe area.
Ok , lets , go step further

The lenght of base = sqrt((5+4)^2 +(3+3)^2) = sqrt(117) = 3*sqrt(13) , so length of altitude for the triangle could be 14/sqrt(13)

Let say vertex is (a,b) ..The height of the triangle w.r.t its base would be the perpendicular distance from (a,b) to the base.. perpendicular distance to a line 2x-3y-1 = |2a-3b-1|/sqrt(13) , this gives |2a-3b-1| = 14

This eqn is satisfied by choice A (2x-3y-15= 0) . A vertex can be anywhere on this line , it doesnt matter .
But as far as eqn of third vetex goes and question says ( what equation of line that could contain the third vertex ) then even choice C satisfies that

Choice C satisfies eqn ( 3 x + 2 y = 15) for value of X = 75/13 and y = -15/13 , this co-ordinate also gives a perprndicular distance of 14/sqrrt(13) from the base

So there are more than one correct answer here. Question , if would have stated , "eqn of third vertex must always be part of which eqn", then A would have been the answer .