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 →

difficult coordinate geometry question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 182
Joined: Sun Aug 02, 2009 7:19 pm
Thanked: 18 times
GMAT Score:680

by sanjana » Fri Oct 09, 2009 1:25 am
Since we know c(20,0) and A(-8,0) the distance between the 2 points is 20-(-8) = 28 (If the y coordinate is the same simply subtract the x coordinates to get the distance)

Therefore, AC=28
Given that : AC=BD,Hence BD=28.

Therefore we know that the y coordinate for B has to be 28,we can thus eliminate answers A,C,D

Now,the answer has to be (6,28) OR (14,28)

Now given that AB=DC
Therefore by distance formula,
(-8-x)^2+y^2=(x-20)^2
This reduces to 336-56x=y^2
when y=0,x=6
hence the point B(6,28) IMO.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 26
Joined: Tue Jul 28, 2009 8:12 pm
Thanked: 4 times

by ashis979 » Fri Oct 09, 2009 1:40 am
Started out the way you did, but then I got confused. Just to be clear, so you're saying, B(x,y) and D(x,0), correct?
y=28, that's pretty much a given since AC=BD, and like you said that distance is 28.
But CD is not equal to 20, now is it? Say the origin is O, therefore it is OC that is equal to 20. OA=8 and OC=20, that's the only way AC=28. There's that one piece OD that we need to take into account, yes? Length of OD is essentially x. So the equation should be:
(x+8)^2+28^2=(20-x)^2
...reduce to x= -8, but this can't happen since x(or the length of OD) can't be negative. Where am I wrong?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 55
Joined: Mon Sep 07, 2009 9:11 pm
Location: California
Thanked: 4 times

by nithi_mystics » Sat Oct 10, 2009 9:20 am
AC = 8 + 20 = 28
Since AC = BD, y coordinate of B is 28

We have only 2 options with y co-ordinate 28.

Now lets take CD = x. Therefore, OD = 20-x
O is the origin. And we need to find the x coordinate (thats 20-x )

In triangle ABD,

AB^2 = BD^2 + AD^2
CD^2 = 28^2 + (8+20-x)^2
x^2 = 28^2 + (28-x)^2

Solving this, we get x=14 ====> 20-x = 6

Hence (6,28) is the answer.
Thanks
Nithi
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1