Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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

Coordinate Plane Strategy

Expert replies
by phoenix9801 » Sun Jul 22, 2012 11:40 pm
Hello, I have a quick question. Please refer to the Image below. Hence please explain step by step instruction for clarifications. Greatly appreciate it.

Question 1: how did they get 8 as a y-coordinate and 6 x-coordinate as an answer.
Slop = y^2 – y^1 / x^2 – x^1 =
3 – (-5) = 8
1 – 7 = -6
Hence, I don’t understand why is it 6 when it supposed to be negative -6 right?


Question 2: How to you allays know which side is a, b, or c? is the Y-coordinate is always the height and x-coordinate is always the base?
You do not have the required permissions to view the files attached to this post.
Join the discussion
Source: — Problem Solving |

by das.ashmita » Mon Jul 23, 2012 1:04 am
Hi phoenix9801

In the given solution, they are not calculating the slope.
They are calculating the length of the base and height of the right angled triangle.

Hence the sign won't matter.

base = |x2-x1|= |3-(-5)| =8
height = |y2-y1| = |1-7| = 6

For the 2nd question, it doesn't matter which side of the triangle you take as base. But it's advisable to chose |x2-x1| as the base for better understanding.
Join the discussion

by niketdoshi123 » Mon Jul 23, 2012 1:22 am
phoenix9801 wrote: Question 2: How to you allays know which side is a, b, or c? is the Y-coordinate is always the height and x-coordinate is always the base?
Here points a,b & c are just the notations, you can use them to represent any of the sides of the triangle.
Here the co-ordinate plane represents a 2-dimensional plane system, where the coordinates of a point are a pair of numbers that define its exact location on a 2-dimensional plane.
The co-ordinates of a point are written as an ordered pair. for example a(4,5), b(6,2) etc
Let a(x,y)
"x" specifies how far along the x (horizontal) axis the point is.
"y" specifies how far up or down the y (vertical) axis to go.
all are in respect to the origin where O(x,y) = O(0,0).

Slop = y^2 - y^1 / x^2 - x^1 =
Here the notation is wrong
Slop = y2 - y1/ x2 - x1

The question is asking you to find the distance between the two points not the slope of the line, they(the distance b/w two points and the slope of the line) are not same.

The distance can be calculated using the pythagoras' theorem.
Join the discussion