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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Coordinate geometry ...

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 23
Joined: Sat May 15, 2010 3:30 pm
Location: Houston, TX
Thanked: 1 times
by ketandoshi » Fri Aug 13, 2010 11:29 am
Right triangle PQR is to be constructed in the xy plane so that the right angle is at P and PR is parallel to the x-axis. The x-and y-coordinates of P, Q and R ae to be integers that satisfy the inequalities -4<=x<=5 and 6<=y<=16. How many different triangles with these properties could be constructed?



Thanks
Ketan
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sat Jul 11, 2009 5:48 am
Thanked: 1 times
GMAT Score:760

by CompBanker » Fri Aug 13, 2010 1:05 pm
My take on this problem:

I believe the best way to tackle this problem is to determine how many possible places the coordinate "P" can sit on the x-axis and how many possible places the coordinate "Q" can sit on the y-axis. For every possible placement of "P" on the x-axis, we will be able to construct a triangle for every possible placement of "Q" on the y-axis.

Knowing that the x-coordinates of the points on the triangle are bound between -4 and 5, we know that the number of possibilities for P must be:

9 = (5) - (-4) + (1) - (1).

The reason we subtract 1 is because point P and point R can not form a triangle if they sit on the same point, so the last coordinate in the range is reserved for R.

We can use the same methodology for determining all the possible y-coordinates of point "Q."

10 = (16) - (6) + (1) - (1)

So now we know that there are 9 different x-coordinates for point "P" and 10 different y-coordinates for point "Q". Be careful here. Simply multiplying 9 * 10 = 90 will give a wrong answer. You have to take into consideration that point "P" can move up and down the y-axis as well. To compensate for this, we need to multiply the number of x-coordinate possibilities for point "P" by the number of y-coordinate possibilities for point "Q" FOR EVERY y-coordinate possibility for point "P". The formula looks as follows:

9 = Number of x-coordinate possibilities for point "P"

10 = Number of y-coordinate possibilities for point "Q" when point "P" is at y = 6.
9 = Number of y-coordinate possibilities for point "Q" when point "P" is at y = 7.
8 = Number of y-coordinate possibilities for point "Q" when point "P" is at y = 8.
7 = Number of y-coordinate possibilities for point "Q" when point "P" is at y = 9.
etc. etc.

Answer = 9 * (10+9+8+7+6+5+4+3+2+1) = 9*55 = 495
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 345
Joined: Wed Mar 18, 2009 6:53 pm
Location: Sao Paulo-Brazil
Thanked: 12 times
GMAT Score:660

by shibal » Sat Aug 14, 2010 7:08 am
The lowest coordinate point for x can be (-4, 6).

Go horizontally [from -4] to select a point and vertically [from -4] to select another point.

Since you have already chosen -4 as x-coordinate, you have horizontally 9 points [from -3 to + 5 inclusive, i.e., 9 points].
Similarly, since you have chosen 6 as y-coordinate, you have vertically 10 points [from 7 to 16 inclusive, i.e., 10 points].

So, at (-4, 6), you have 10*9 or 90 right triangles.

Now shift the point right to another point, (i.e.-3, 6). Again, you have 9 points; although P now moves by 1 point horizontally, total points do not decrease, because it can now take -4's original position at left also.

Similarly for (-3, 6), it has 10 points vertically. Again, we have 90 triangles.

we can conclude that we have 10 values for x coordinate & 11 values for y-coordinate. Total 10*11=110 points.

No. of total triangles = 110*90 = 9900
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Wed Aug 11, 2010 8:47 pm
Thanked: 3 times

by Prashantbhardwaj » Sat Aug 14, 2010 10:25 am
A right angle triangle PQR with right angle P will be formed in a coordinate system only if coordinates of the three points are:

P (x,y) Q(x,y') R(x',y)

i.e. PR and PQ remain straight lines perpendicular to each other

As such the variations possible are:

P for product of total no. of values for both x and y.

Q 1 for x and 1 less values of y for P.

and R 1 for y and 1 less values of x for P.

So the arrangements of coordinates are:

P= 10*11=110 [-4<=x<=5 givning 10 values]
Q= 1*10=10 [6<=y<=16 giving 11 values]
R= 9*1=9

now forming a triangle is an and condition as all points must exist together therefore total no. of triangles is 110*10*90=9900.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 222
Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members
Thanked: 3 times

by venmic » Sat Aug 14, 2010 7:16 pm
I understood all of it except
and R 1 for y and 1 less values of x for P.

how does it work for R
Prashantbhardwaj wrote:A right angle triangle PQR with right angle P will be formed in a coordinate system only if coordinates of the three points are:

P (x,y) Q(x,y') R(x',y)

i.e. PR and PQ remain straight lines perpendicular to each other

As such the variations possible are:

P for product of total no. of values for both x and y.

Q 1 for x and 1 less values of y for P.

and R 1 for y and 1 less values of x for P.

So the arrangements of coordinates are:

P= 10*11=110 [-4<=x<=5 givning 10 values]
Q= 1*10=10 [6<=y<=16 giving 11 values]
R= 9*1=9

now forming a triangle is an and condition as all points must exist together therefore total no. of triangles is 110*10*90=9900.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Wed Aug 11, 2010 8:47 pm
Thanked: 3 times

by Prashantbhardwaj » Sun Aug 15, 2010 9:39 pm
venmic wrote:I understood all of it except
and R 1 for y and 1 less values of x for P.

how does it work for R
Prashantbhardwaj wrote:A right angle triangle PQR with right angle P will be formed in a coordinate system only if coordinates of the three points are:

P (x,y) Q(x,y') R(x',y)

i.e. PR and PQ remain straight lines perpendicular to each other

As such the variations possible are:

P for product of total no. of values for both x and y.

Q 1 for x and 1 less values of y for P.

and R 1 for y and 1 less values of x for P.

So the arrangements of coordinates are:

P= 10*11=110 [-4<=x<=5 givning 10 values]
Q= 1*10=10 [6<=y<=16 giving 11 values]
R= 9*1=9

now forming a triangle is an and condition as all points must exist together therefore total no. of triangles is 110*10*90=9900.

Q 1 for x and 1 less values of y for P.

and R 1 for y and 1 less values of x for P.

Let me clarify it. I skipped a detail.

What I meant to say here is that the value of y for R will be the same as the value of y for P and value of x for Q will also be the same as the value of x for P. As such the variations in x for Q and y for R have been counted for in P.

Thus there is 1 value of x for Q and there is one value of y in R for a given value of P(x,y).

I have attached an image if it helps image.
Attachments
PQR Right Angle.png
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Wed Aug 11, 2010 8:47 pm
Thanked: 3 times

by Prashantbhardwaj » Sun Aug 15, 2010 9:52 pm
And from above since one value of the coordinates of R is already the same as that of P's x so if the other i.e. the value of y is also taken to be same as that of P's y then there will be no triangle as P and R become the same point.

So the variations for R are 10-1 = 9 where the one value less is the value of x for P which R can never have.

And same goes for Q for value of y.
we get 10 variable values for Q
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1