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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

PS - Rectangle Coord.

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1665
Joined: Thu Nov 03, 2011 7:04 pmFollowed by: 70 members
Thanked: 165 times
by karthikpandian19 » Wed Jul 11, 2012 7:27 pm
Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a - 3, a), respectively. If Point R lies at (11, 2), what is the value of a?


(A) 1

(B) 4

(C) 5

(D) 7

(E) 8
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" :) :)---
---Never stop until cracking GMAT---
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 520
Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members
Thanked: 339 times
GMAT Score:770

by eagleeye » Wed Jul 11, 2012 8:42 pm
karthikpandian19 wrote:Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a - 3, a), respectively. If Point R lies at (11, 2), what is the value of a?
Draw a line joining the diagonal PR. Since PQRS is a rectangle, we can use pythagoras theorem in triangle PQR (with 90 at Q).
We have
PQ^2 + QR^2 = PR^2
Now P=(a+2, 2a), Q=(a-3,a), and R=(11,2)
Then PQ^2 + QR^2 = PR^2 becomes:

=> (a+2-a+3)^2+(2a-a)^2 + (a-3-11)^2 + (a-2)^2 = (a+2-11)^2 + (2a-2)^2
=> 5^2 + a^2 + (a-14)^2 + (a-2)^2 = (a-9)^2 + 4(a-1)^2
=> 25+ a^2 + a^2+196-28a + a^2 +4 -4a = a^2 + 81 -18a +4a^2 +4 -8a
=> a^2*(5-3) + a*(32-26) + (81 - 196 - 25) = 0
=> 2a^2 + 6a - 140 = 0
=> a^2 + 3a = 70

Check from the options, for middle value 5, 5^2 + 5*3 = 40, smaller than 70, so try next one (7):
7^2+3*7 = 49+21 = 70. Awesome.
Hence D is the correct answer.

Let me know if this helps :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1665
Joined: Thu Nov 03, 2011 7:04 pmFollowed by: 70 members
Thanked: 165 times

by karthikpandian19 » Wed Jul 11, 2012 11:34 pm
I solved it like this:

Since PQRS is a rectangle and PR,QS are the diagonals
Therefore PQ is to be perpendicular to QR

Find Slope of PQ:
P(a + 2, 2a) and Q(a - 3, a)
= a-2a/(a-3-(a+2))
= a/5 is the slope

Find Slope of QR:
Q(a - 3, a) and R(11,2)
=2-a/(11-(a-3))
=2-a/(14-a) is the slope

Since both are perpendicular:
2-a/(14-a) = -5/a
2a-a^2=-70+5a
0=a^2+3a-70
(a+10)(a-7)=0
So "a" as per the available answer should be 7
Answer is D


eagleeye wrote:
karthikpandian19 wrote:Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a - 3, a), respectively. If Point R lies at (11, 2), what is the value of a?
Draw a line joining the diagonal PR. Since PQRS is a rectangle, we can use pythagoras theorem in triangle PQR (with 90 at Q).
We have
PQ^2 + QR^2 = PR^2
Now P=(a+2, 2a), Q=(a-3,a), and R=(11,2)
Then PQ^2 + QR^2 = PR^2 becomes:

=> (a+2-a+3)^2+(2a-a)^2 + (a-3-11)^2 + (a-2)^2 = (a+2-11)^2 + (2a-2)^2
=> 5^2 + a^2 + (a-14)^2 + (a-2)^2 = (a-9)^2 + 4(a-1)^2
=> 25+ a^2 + a^2+196-28a + a^2 +4 -4a = a^2 + 81 -18a +4a^2 +4 -8a
=> a^2*(5-3) + a*(32-26) + (81 - 196 - 25) = 0
=> 2a^2 + 6a - 140 = 0
=> a^2 + 3a = 70

Check from the options, for middle value 5, 5^2 + 5*3 = 40, smaller than 70, so try next one (7):
7^2+3*7 = 49+21 = 70. Awesome.
Hence D is the correct answer.

Let me know if this helps :)
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" :) :)---
---Never stop until cracking GMAT---
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 520
Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members
Thanked: 339 times
GMAT Score:770

by eagleeye » Thu Jul 12, 2012 2:42 am
karthikpandian19 wrote:I solved it like this:

Since PQRS is a rectangle and PR,QS are the diagonals
Therefore PQ is to be perpendicular to QR

Find Slope of PQ:
P(a + 2, 2a) and Q(a - 3, a)
= a-2a/(a-3-(a+2))
= a/5 is the slope

Find Slope of QR:
Q(a - 3, a) and R(11,2)
=2-a/(11-(a-3))
=2-a/(14-a) is the slope

Since both are perpendicular:
2-a/(14-a) = -5/a
2a-a^2=-70+5a
0=a^2+3a-70
(a+10)(a-7)=0
So "a" as per the available answer should be 7
Answer is D


eagleeye wrote:
karthikpandian19 wrote:Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a - 3, a), respectively. If Point R lies at (11, 2), what is the value of a?
Draw a line joining the diagonal PR. Since PQRS is a rectangle, we can use pythagoras theorem in triangle PQR (with 90 at Q).
We have
PQ^2 + QR^2 = PR^2
Now P=(a+2, 2a), Q=(a-3,a), and R=(11,2)
Then PQ^2 + QR^2 = PR^2 becomes:

=> (a+2-a+3)^2+(2a-a)^2 + (a-3-11)^2 + (a-2)^2 = (a+2-11)^2 + (2a-2)^2
=> 5^2 + a^2 + (a-14)^2 + (a-2)^2 = (a-9)^2 + 4(a-1)^2
=> 25+ a^2 + a^2+196-28a + a^2 +4 -4a = a^2 + 81 -18a +4a^2 +4 -8a
=> a^2*(5-3) + a*(32-26) + (81 - 196 - 25) = 0
=> 2a^2 + 6a - 140 = 0
=> a^2 + 3a = 70

Check from the options, for middle value 5, 5^2 + 5*3 = 40, smaller than 70, so try next one (7):
7^2+3*7 = 49+21 = 70. Awesome.
Hence D is the correct answer.

Let me know if this helps :)
Your way is definitely less error-prone and faster. Awesome job :)!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1