If - 6i -3i 1 2 3 i 4 3i -1 - -x-iy

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Quantitative Aptitude

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If | 6i -...

Question

If $∣ ∣ ∣ ∣ \begin{matrix} 6 i & - 3 i & 1 2 & 3 & i 4 & 3 i & - 1 \end{matrix} ∣ ∣ ∣ ∣ = x + i y$

x = y = 1

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x = y = 0

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x = 3, y = 1

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x = - 3, y = - 1

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Solution

The correct option is B $x = y = 0$

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Similar questions

Q. If the points

(x_{1}, y_{1}), (x_{2}, y_{2})

and

(x_{3}, y_{3})

are collinear, show that

\sum (\frac{y_{1} - y_{2}}{x_{1} x_{2}}) = 0,

i.e

\frac{y_{1} - y_{2}}{x_{1} x_{2}} + \frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} = 0

Q. If three points (x₁, y₁) (x₂, y₂), (x₃, y₃) lie on the same line, prove that

\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0

Q. Given that

x, y \in R

. Solve:

\frac{x}{1 + 2 i} + \frac{y}{3 + 2 i} = \frac{5 + 6 i}{8 i - 1}

Find the real part of $(\frac{(x + i y) + i}{(x + i y) + 2})$
Take $(| (x + 2) + i y |)$ = $\sqrt{\frac{1}{k}}$

If three points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ lie on the same line, prove that $\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0.$

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If ∣∣ ∣∣6i−3i123i43i−1∣∣ ∣∣=x+iy

The correct option is B x=y=0

If $∣ ∣ ∣ ∣ \begin{matrix} 6 i & - 3 i & 1 2 & 3 & i 4 & 3 i & - 1 \end{matrix} ∣ ∣ ∣ ∣ = x + i y$

The correct option is B $x = y = 0$