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 x_{1}y_{2}, x_{2}y_{3} and x_{3}y_{1} as shown in the dark arrows.
Also add the diagonal products x_{2}y_{1}, x_{3}y_{2} and x_{1}y_{3} 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
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+a^{2}+ba+b^{2}+cb+c^{2}-b^{2}-bc-c^{2}-ac-a^{2}-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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