If a - b - c- it follows that

The correct option is C a ! b + cStatement givena gt;b lt;cOptions givenA. c lt;b gt;a. This is falseB. a gt;b lt;c. This is TrueC. b gt;a gt;c ...

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If a × b ...

Question

If a $\times$ b $Δ$ c, it follows that

c + b ! a

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a ! b + c

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b ! a

\times

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ϕ

Δ

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Solution

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A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

A, B, C

\frac{1}{2} | Δ |

Δ = ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣

a = B C, b = C A, c = A B

2 s = a + b + c

Δ^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + a b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3. (b - c) (c - a) (a - b) (a + b + c) (a b + b c + c a)

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + c b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 (b - c) (c - a) (a - b) (a + b + c) (b c + c a + a b)

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ

a, b

c

Δ = ∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ = 0

a + b + c = 0

a = b = c

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