Find the area of the triangle formed by the straight lines whose equations are
$y = a x - b c, y = b x - c a$ , and $y = c x - a b$ .

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Solution

$y = a x - b c . . . . . . . (i) y = b x - c a . . . . . . (i i) y = c x - a b . . . . . . . (i i i)$
solving $(i)$ and $(i i)$
$\Rightarrow x = - c, y = - c (a + b)$
So the coordinates of $A$ are $(- c, - c (a + b))$
solving $(i i)$ and $(i i i)$
$\Rightarrow x = - a, y = - a (b + c)$
So the coordinates of $B$ are $(- a, - a (b + c))$
Solving $(i)$ and $(i i i)$
$\Rightarrow x = - b, y = - b (a + c)$
So the coordinates of $C$ are $(- b, - b (a + c))$
Area of triangle $= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$Δ = \frac{1}{2} | - c (- a b - a c + a b + b c) - a (- a b - b c + a c + b c) - b (- a c - b c + a b + a c) | Δ = \frac{1}{2} | - c (b c - a c) - a (a c - a b) - b (a b - b c) | Δ = \frac{1}{2} (a c^{2} - b c^{2} + b a^{2} - c a^{2} + c b^{2} - a b^{2}) Δ = \frac{1}{2} (a - b) (b - c) (c - a)$

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