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Evaluation of a Determinant

Area of trian...

Question

Area of triangle whose vertices are $(a, a^{2}), (b, b^{2}), (c, c^{2})$ is 1, and area of another triangle whose vertices are $(p, p^{2}), (q, q^{2})$ and $(r, r^{2})$ is 4 then the value of $∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + a p)^{2} & (1 + b p)^{2} & (1 + c p)^{2} (1 + a q)^{2} & (1 + b q)^{2} & (1 + c q)^{2} (1 + a r)^{2} & (1 + b r)^{2} & (1 + c r)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

A

4

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B

8

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C

16

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D

32

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Solution

The correct option is D 32
Let
$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + a p)^{2} & (1 + b p)^{2} & (1 + c p)^{2} (1 + a q)^{2} & (1 + b q)^{2} & (1 + c q)^{2} (1 + a r)^{2} & (1 + b r)^{2} & (1 + c r)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + 2 a p + a^{2} p^{2}) & (1 + 2 b p + b^{2} p^{2}) & (1 + 2 c p + c^{2} p^{2}) (1 + 2 a q + a^{2} q^{2}) & (1 + 2 b q + b^{2} q^{2}) & (1 + 2 c q + c^{2} q^{2}) (1 + 2 a r + a^{2} r^{2}) & (1 + 2 b r + b^{2} r^{2}) & (1 + 2 c r + c^{2} r^{2}) \end{matrix} ∣ ∣ ∣ ∣ ∣$

Then,

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 a & a^{2} 1 & 2 b & b^{2} 1 & 2 c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 p & q & r p^{2} & q^{2} & r^{2} \end{matrix} ∣ ∣ ∣ ∣$
$= 2 \times (2 Δ_{1}) \times (2 Δ_{2}) = 2 \times (2 \times 1) \times (2 \times 4) = 32$

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Similar questions

Q. Area of triangle whose vertices are

(a, a^{2}), (b, b^{2}), (c, c^{2})

is 1, and area of another triangle whose vertices are

(p, p^{2}), (q, q^{2})

(r, r^{2})

is 4 then the value of

∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + a p)^{2} & (1 + b p)^{2} & (1 + c p)^{2} (1 + a q)^{2} & (1 + b q)^{2} & (1 + c q)^{2} (1 + a r)^{2} & (1 + b r)^{2} & (1 + c r)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ \begin{matrix} {(a - p)}^{2} & {(a - q)}^{2} & {(a - r)}^{2} {(b - p)}^{2} & {(b - q)}^{2} & {(b - r)}^{2} {(c - p)}^{2} & {(c - q)}^{2} & {(c - r)}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(1 + a p)}^{2} & {(1 + a q)}^{2} & {(1 + a r)}^{2} {(1 + b q)}^{2} & {(1 + b q)}^{2} & {(1 + b r)}^{2} {(1 + c p)}^{2} & {(1 + c q)}^{2} & {(1 + c r)}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1

\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0

\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}}

a, b, c, p, q, r

are thre non-zero complex numbers such that

\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 + i

\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0,

\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}}

a, b, c, p, q, r

are three non-zero complex number such that

\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 + i

\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0

, then value of

\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}}

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