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Area of Triangle when coordinates of three vertices are given
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27 Jul 2017, 09:17
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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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Area of Triangle when coordinates of three vertices are given
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27 Jul 2017, 11:22
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 33996 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 ycoordinates at points B and C) W = 4 (count, or subtract xcoordinates at points A and B) Rectangle area = 242. 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 ycoordinates) 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 triangle24  \(\frac{33}{2}\) = \(\frac{48}{2}  \frac{33}{2} = \frac{15}{2}\) Answer B Hope it helps.
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Re: Area of Triangle when coordinates of three vertices are given
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27 Jul 2017, 10:57
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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Re: Area of Triangle when coordinates of three vertices are given
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12 Aug 2017, 03:00
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=2x2 & 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.



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Re: Area of Triangle when coordinates of three vertices are given
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03 Nov 2017, 00:48
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
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10 Nov 2017, 03:33
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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Area of Triangle when coordinates of three vertices are given
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20 Mar 2018, 10:24
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 NDNDThe area of triangle with vertices(A,B,C) having following coordinates Attachment:
triangle.png [ 3.16 KiB  Viewed 28536 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
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13 May 2019, 01:51
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Re: Area of Triangle when coordinates of three vertices are given
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