# AREA OF TRIANGLE WITH THREE VERTICES

Area of Triangle with Three Vertices

Here we are going to see, how to find the area of triangle with given vertices. To find the area of a triangle, the following steps may be useful.

(i) Plot the points in a rough diagram.

(ii) Take the vertices in counter clock-wise direction. Otherwise the formula gives a negative value.

(iii)  Use the formula given below ## Area of triangle with three vertices - Questions

Question 1 :

Find the area of the triangle formed by the points

(i) (1,–1), (–4, 6) and (–3, –5)

Solution : Area of triangle ABC  = =  (1/2)[(6 + 20 + 3) - (4 - 18 - 5)]

=  (1/2)[29 - (-19)]

=  (1/2)[29 + 19]

=  (1/2) 48

=  24 square units.

(ii)  (-10, -4) (-8, -1) and (-3, -5)

Solution :  =  (1/2)[(32 + 50 + 3) - (10 + 12 + 40)]

=  (1/2)[85 - 62]

=  23/2

=  11.5 square units

## How to Determine If the Points are Collinear ?

Question 1 :

Determine whether the sets of points are collinear?

(i)  (-1/2, 3), (-5, 6) and (-8, 8)

Solution : =  (1/2)[(-3 - 40 - 24) - (-15 - 48 - 4)]

=  (1/2)[(-67) - (-67)]

=  (1/2)(-67 + 67)

=  0

Hence the given points are collinear.

(ii) (a, b+c), (b, c+a) and (c, a+b)

Solution : =  (1/2)[a(c+a)+b(a+b)+c(b+c)-b(b+c)-c(c+a)-a(a+b)]

=  (1/2)[ac+a2+ba+b2+cb+c2-b2-bc-c2-ac-a2-ab]

=  0

## How to Determine the Missing Coordinate When area of Triangle is Given ?

Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p ’. Solution :

(i)  Area of triangle  =  20 square units (1/2)[(0 + 2p + 0) - (0 + 48 + 0)]  =  20

2p - 48  =  40

2p  =  40 + 48

2p  =  88

p  =  44

Hence the value of p is 44.

(ii)

Solution :

Area of triangle = 32 square units (1/2)[(6p - 10 + 5p) - (5p + 30 - 2p)]  =  32

(11p - 10) - (3p + 30)  =  64

11p - 3p - 10 - 30  =  64

8p  =  64 + 40

8p  =  104

p  =  104/8

p  =  13

Hence the value of p is 13. After having gone through the stuff given above, we hope that the students would have understood, "Area of Triangle with Three Vertices".

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