AREA OF TRIANGLE WITH THREE VERTICES

Here, we are going to see, how to find area of a triangle when coordinates of the three vertices are given. 

Let us consider the triangle given below. 

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

And the diagonal products x₁y₂, x₂y₃ and x₃y₁ as shown in the dark arrows. 

Also add the diagonal products x₂y₁, x₃y₂ and x₁y₃ as shown in the dotted arrows.

Now, subtract the latter product from the former product to get area of the triangle ABC.

So, area of the triangle ABC is

Practice Problems

Problem 1 :

Find the area of the triangle whose vertices are 

(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

Problem 2 :

Find the area of the triangle whose vertices are 

(-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

Problem 3 :

Determine whether the sets of points are collinear.

(-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

Because the area of the triangle is zero, the given points are collinear.

Problem 4 :

Determine whether the sets of points are collinear.

(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

Problem 5 :

Find the value of p in each case. 

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

Solution (ii) :

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

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