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Senior Manager  P
Joined: 17 Mar 2014
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Area of Triangle when coordinates of three vertices are given  [#permalink]

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21 00:00

Difficulty:   35% (medium)

Question Stats: 77% (02:31) correct 23% (02:40) wrong based on 156 sessions

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Hi Bunuel,

How to calculate area of triangle when coordinates of three vertices are given. For example, how to solve following question.

On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3, 4). What is the area of the triangle connecting the 3 points?

(A) 15/4
(B) 15/2
(C) 15
(D) 13/4
(E) 13/2

Regards,
Ammu
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Joined: 22 May 2016
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Area of Triangle when coordinates of three vertices are given  [#permalink]

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ammuseeru wrote:
Hi Bunuel,

How to calculate area of triangle when coordinates of three vertices are given. For example, how to solve following question.

On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3, 4). What is the area of the triangle connecting the 3 points?

(A) 15/4
(B) 15/2
(C) 15
(D) 13/4
(E) 13/2

Regards,
Ammu

Attachment: tri.area.grid.rectangle.jpg [ 37.89 KiB | Viewed 34335 times ]

Because this triangle has no vertical or horizontal leg, plot its three coordinates, connect them, then draw a rectangle around the triangle (see diagram).

Then find area of rectangle and subtract the areas of the three yellow triangles, whose areas are easy to find because they are right triangles (their legs are base and height).

1. (Area of rectangle) - (area of OTHER triangles) = area of THIS triangle

Area of rectangle = (L * W)
L = 6 (count on the graph, or subtract y-coordinates at points B and C)
W = 4 (count, or subtract x-coordinates at points A and B)

Rectangle area = 24

2. Find area of three triangles (yellow) ABX, BCY, and DXY (coordinates are in red - find base and height for each by counting or subtracting x- and then y-coordinates)

Area of triangle ABX = $$\frac{1}{2} b*h = \frac{(3*4)}{2}$$ = 6

Area of triangle BCY = $$\frac{(6 * 3)}{2}$$ = 9

Area of triangle DXY = $$\frac{(1*3)}{2} = \frac{3}{2}$$

Total area of yellow triangles = 6 + 9 + $$\frac{3}{2}$$ = $$\frac{33}{2}$$

3. (Rectangle area) - (total area of 3 yellow triangles) =
area of THIS triangle

24 - $$\frac{33}{2}$$ =

$$\frac{48}{2} - \frac{33}{2} = \frac{15}{2}$$

Hope it helps.
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Re: Area of Triangle when coordinates of three vertices are given  [#permalink]

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5
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)
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Manager  P
Joined: 01 Feb 2017
Posts: 244
Re: Area of Triangle when coordinates of three vertices are given  [#permalink]

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2
1
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 &

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

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

Ans B.
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Re: Area of Triangle when coordinates of three vertices are given  [#permalink]

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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
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Re: Area of Triangle when coordinates of three vertices are given  [#permalink]

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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.
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Senior Manager  V
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Area of Triangle when coordinates of three vertices are given  [#permalink]

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1
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 [ 3.16 KiB | Viewed 28870 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
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Re: Area of Triangle when coordinates of three vertices are given  [#permalink]

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_________________ Re: Area of Triangle when coordinates of three vertices are given   [#permalink] 13 May 2019, 01:51
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