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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.
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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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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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pushpitkc
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 38947 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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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 5376 times ]

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