Question

hi guys........

The area of triangle with vertices (a,b+c), (b,c+a) and (c,a+b) is.........????​

Answer 1

heya....your answer is here....concept : if three points A(x_1,y_1) , B(x_2,y_2) and C(x_3,y_3) form a triangle ABC.

then, area of triangle ABC = \frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)

here given, three points (a, b + c), (b, c + a) and (c, a + b)

then, area of triangle formed by these points = 1/2 [ a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)]

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

= 1/2 [ ac - ab + ba - bc + cb - ac ]

= 0

hence, area of triangle whose vertices are (a, b + c), (b, c + a) and (c, a + b) = 0

[important point : if area of triangle becomes zero, it means vertices of triangle is collinear. I mean, they lie in same line . ]

Answer 2

Answer:

0

Step-by-step explanation:

Given points are (a, b + c), (b, c + a), (c, a + b).

Area of triangle ABC = (1/2)[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

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

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

= (1/2)[0]

= 0

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