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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

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
Limited cohort · enrollment openTarget 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

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

Parallelogram question - need help

Expert replies
by onesome » Sat Jul 05, 2008 3:45 pm
Question => Two opposite pair of vertices of parallelogram are = (1,1) & (4,4)
If one of the remaining vertices is (m,n), what is the fourth vertex of a parallelogram in terms of m & n ?

A) (1-m, 1-n)
B) (2-m, 3-n)
C) (3-m, 4-n)
D) (m-4, n-4)
E) (5-m, 5-n)


Answer is : E
Last edited by onesome on Sat Jul 05, 2008 5:56 pm, edited 2 times in total.
Join the discussion
Source: — Problem Solving |

by ildude02 » Sat Jul 05, 2008 4:16 pm
We need to use the concept that the diagnols bisect each other to solve this problem. So if you draw diagnols for all the points, the mid point of the diagnoal connecting (4, 4) and (1, 1) is (5/2, 5/2). Now you know that (5/2, 5/2) point is th emid point between (m, n) and (x, y). So you can find (x, y) based on this concept,

m+x / 2 = 5/2; n+x / 2 = 5/2. ;
This will give E for x, y values.

Jus wondering if there is an easier approach?
Join the discussion

by onesome » Sat Jul 05, 2008 4:45 pm
ildude02 - thanks :)

I just found this "Mid-point theorem" here and realised this one was simple.

https://www.mathopenref.com/coordmidpoint.html

Thanks again.
Join the discussion