# What is the area of the triangle formed by the points (8,7), (2,3) and (1,4)?

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We have to find the area of the triangle formed by the points (8, 7), (2, 3) and (1, 4). This can be done in many ways. Here, we use the fact that the area of a triangle is given by (1/2)*base*height.

Let’s take the base formed by the points (2, 3) and (8, 7).

The distance between the points is sqrt [(8 – 2) ^2 + (7 – 3) ^2]

=> sqrt [6^2 + 4^2]

=> sqrt [36 + 16]

=> sqrt (52)

=> 2*sqrt 13

The equation of the line between these points is: y – 3 = [(7 – 3)/ (8 – 2)]*(x – 2)

=> y – 3 = (4/6)*(x – 2)

=> y – 3 = (2/3) (x – 2)

=> 3y – 9 = 2x – 4

=> 2x – 3y + 5 = 0

The distance between (1, 4) and the line 2x – 3y + 5 = 0 is given by

| 2*1 – 3*4 + 5| / sqrt (2^2 + 3^2)

=> |2 – 12 + 5| / sqrt 13

=> 5 / sqrt 13

So the height of the triangle is 5 / sqrt 13.

The area of the triangle is (1/2)*base*height

=> (1/2) * (2*sqrt 13) * (5/ sqrt 13)

=> 5

**Therefore the required area is 5.**

We know that the area A of the triangle formed by the 3 points (x1,y1), (x2,y2) and (x3,y3) is given by:

A = (1/2{(y1+y2)(x2-x1)+(y+y3)(x3-x1)+ (y3+y1)((x1-x3)}.

The given 3 points are: (8,7) , (2,3) and (1,4):

A = (1/2) {(7+3)(2-8)+(3+4)((1-2)+(4+7)(8-1)}.

A = (1/2){ 10(-6)+7(-1)+11(7}.

A = (1/2){-60 -7 +77}.

A = (1/2)(10)

A = 5 sq units is the area of the triangle formed by the points (8,7), (2,3) and (1,4).

The area of the triangle is:

|8 7 1|

S= (1/2)* |2 3 1|

|1 4 1|

We'll calculate the determinant:

S = (1/2)*(8*3*1 + 2*4*1 + 7*1*1 - 1*3*1 - 4*8*1 - 2*7*1)

S = (1/2)*(24 + 8 + 7 - 3 - 32 - 14)

S = 10/2

S = 5 square units