IMO E
OA?
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
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
Rectangular Coordinate
Source: Beat The GMAT — Problem Solving |
IMO - B [Edit my post.. Initially typo C]
Since y=x is the perpendicular bisector of AB and A lies in the first quadrant => B lies in the first quadrant.
We are also told that y axis is the perpendicular bisector of BC. Hence C must lie in the second quadrant.
i.e X negative and y positive.
Choose the answer option that satisfies this.
I am not able to solve this mathematically.. Could someone help.
Since y=x is the perpendicular bisector of AB and A lies in the first quadrant => B lies in the first quadrant.
We are also told that y axis is the perpendicular bisector of BC. Hence C must lie in the second quadrant.
i.e X negative and y positive.
Choose the answer option that satisfies this.
I am not able to solve this mathematically.. Could someone help.
Last edited by raghavsarathy on Tue Jul 21, 2009 4:45 pm, edited 2 times in total.
Raghavsarathy, thats right. Even I got the answer. But couldn't solve it mathematically. I need help to solve it mathematically.
Well Mathematically thats what u can do..
1) You knwo line y = x is perpendicular bisector of segment AB and A is (3,2) ===> The segment AB could be of form y = -x + c , since perpendicular bisector is perpendicular to AB ..now get the line eauqntion for AB as 2 = -3 + c which gives c = 5 , hence segment AB is y = 5 - x
2) Now take out the co-ordinate where AB and perpendicular bisector insecxts , which u cna easily get using 2 eqns y = x and y = 5 - x
You get co-rdinates of points of intersection as (2.5,2.5)
Since perpendicular bisector bisects ( or divides exactly in half) the segment AB , so you know easily B is ( 2, 3)
3) Now coming to BC segment where its said y axis is the perpendicular bisector , which means BC is parallel to x axis . The midpoint of BC lies on Y axis which is nothing but (0 , 3) since segemnt BC is parallel to x axis.
Now you can derive the co-ordinates for C similaryly as you derived for B in step 2 . This gives C a (-2,3)
1) You knwo line y = x is perpendicular bisector of segment AB and A is (3,2) ===> The segment AB could be of form y = -x + c , since perpendicular bisector is perpendicular to AB ..now get the line eauqntion for AB as 2 = -3 + c which gives c = 5 , hence segment AB is y = 5 - x
2) Now take out the co-ordinate where AB and perpendicular bisector insecxts , which u cna easily get using 2 eqns y = x and y = 5 - x
You get co-rdinates of points of intersection as (2.5,2.5)
Since perpendicular bisector bisects ( or divides exactly in half) the segment AB , so you know easily B is ( 2, 3)
3) Now coming to BC segment where its said y axis is the perpendicular bisector , which means BC is parallel to x axis . The midpoint of BC lies on Y axis which is nothing but (0 , 3) since segemnt BC is parallel to x axis.
Now you can derive the co-ordinates for C similaryly as you derived for B in step 2 . This gives C a (-2,3)
