Last visit was: 28 Apr 2024, 12:53 It is currently 28 Apr 2024, 12:53

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Director
Director
Joined: 17 Mar 2014
Posts: 756
Own Kudos [?]: 609 [44]
Given Kudos: 1348
Send PM
Most Helpful Reply
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 2873
Own Kudos [?]: 5206 [16]
Given Kudos: 47
Location: India
GPA: 3.12
Send PM
Senior SC Moderator
Joined: 22 May 2016
Posts: 5330
Own Kudos [?]: 35501 [10]
Given Kudos: 9464
Send PM
General Discussion
Manager
Manager
Joined: 01 Feb 2017
Posts: 246
Own Kudos [?]: 346 [4]
Given Kudos: 148
Send PM
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
3
Kudos
1
Bookmarks
In Triangle ABC, let's consider side BC as base.
Length of BC= √(6^2+3^2) = √45

Now, Let's assume AD as perpendicular to side BC.
Slope of line AD= negative reciprocal of Slope BC= -1/2

Now,
Equation of BC: y=2x-2 &
Equation of AD: y=-x/2+1/2

Therefore, Coordinates of point D [Calculated by equating equations of line AD to BC]= (+1, 0).

So, Length AD= √5

Hence, Area of Triangle ABC= 1/2xBCxAD = 1/2x√45x√5= 15/2

Ans B.
Manager
Manager
Joined: 26 Sep 2017
Status:Enjoying the Journey
Affiliations: ND
Posts: 100
Own Kudos [?]: 246 [0]
Given Kudos: 655
Schools: Rotman '21
WE:Marketing (Consulting)
Send PM
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
pushpitkc wrote:
Given data
x1=-1,y1=1
x2=0,y2=-2
x3=3,y3=4

Given the vertices of the triangle
Area = \(\frac{1}{2} *| (x1y2 - y1x2) + (x2y3 - y2x3) + (x3y1 - y3x1)|\)

Substiuting these values,
Area = \(\frac{1}{2} * |2 + 6 + 7|\) = \(\frac{1}{2} * |15| = \frac{15}{2}\)(Option B)


Hi pushpitkc,

Can you elaborate more on your answer, I understood what you did, but is this a rule or something, can you highlight the rule please. Your answer is very efficient and I would like to know how to use it.

Thanks,
NDND
Director
Director
Joined: 02 Sep 2016
Posts: 528
Own Kudos [?]: 194 [0]
Given Kudos: 275
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
I don't know if this is the right approach, but I solved it like this.

(Using the figure genxer123's post)

I did not extend the rectangle this way. Using the distance between points formula, I found YX and YB i.e. root 10 and 5 respectively. As opposite sides of a rectangle are equal, I took the opposite sides as root 10 and 5.

Area of rectangle= root 10*5

Area of the other triangle in the rectangle= 1/2 *root 10*5

So the area of the triangle that we had to find= 2.5* root 10= approx 8..... So I chose B.
Senior Manager
Senior Manager
Joined: 22 Feb 2018
Posts: 355
Own Kudos [?]: 801 [4]
Given Kudos: 34
Send PM
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
4
Bookmarks
NDND wrote:
pushpitkc wrote:
Given data
x1=-1,y1=1
x2=0,y2=-2
x3=3,y3=4

Given the vertices of the triangle
Area = \(\frac{1}{2} *| (x1y2 - y1x2) + (x2y3 - y2x3) + (x3y1 - y3x1)|\)

Substiuting these values,
Area = \(\frac{1}{2} * |2 + 6 + 7|\) = \(\frac{1}{2} * |15| = \frac{15}{2}\)(Option B)


Hi pushpitkc,

Can you elaborate more on your answer, I understood what you did, but is this a rule or something, can you highlight the rule please. Your answer is very efficient and I would like to know how to use it.

Thanks,
NDND

NDND
The area of triangle with vertices(A,B,C) having following coordinates
Attachment:
triangle.png
triangle.png [ 3.16 KiB | Viewed 37604 times ]

A : \(x_{1},y_{1}\)
B : \(x_{2},y_{2}\)
C : \(x_{3},y_{3}\)

is given by
A = \(\frac{{|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|}}{2}\)

https://www.mathopenref.com/coordtrianglearea.html
Senior Manager
Senior Manager
Joined: 01 Dec 2020
Posts: 480
Own Kudos [?]: 373 [0]
Given Kudos: 359
GMAT 1: 680 Q48 V35
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
Distance b/w points B and C = √(x2-x1)^2 + (y2-y1)^2 = √(3-0)^2 + (4+2)^2 = √45 = 3√5

Distance b/w points A and D = √(x2-x1)^2 + (y2-y1)^2 = √(-1-1)^2 + (1-0)^2 = √5

Area of triangle = 1/2 * Base * Height = 1/2 * 3√5 * √5 = 15/2 (B)
Attachments

CG.jpg
CG.jpg [ 48.19 KiB | Viewed 4127 times ]

User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32726
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3 [#permalink]
Moderators:
Math Expert
92977 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne