ChemistryQ1–1 of 1 questions
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1
ChemistryMCQIIT JEE · 1963
The area of the triangle formed by the points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$ is equal to:
A
$abc$
B
$a^2 + b^2 + c^2$
C
$ab + bc + ca$
D
$0$
Solution
(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$:
Area $= \frac{1}{2} |a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)|$
$= \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)|$
$= \frac{1}{2} |ac - ab + ba - bc + cb - ca|$
$= \frac{1}{2} |0| = 0$.
Since the area is $0$,the points are collinear.
Substituting the given points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$:
Area $= \frac{1}{2} |a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)|$
$= \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)|$
$= \frac{1}{2} |ac - ab + ba - bc + cb - ca|$
$= \frac{1}{2} |0| = 0$.
Since the area is $0$,the points are collinear.
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