Let $A (z_{1}), B (z_{2}), C (z_{3}) a n d D (z_{4})$ be the vertices of a trapezium in a Argand plane such that AB||CD. Let $| z_{1} - z_{2} | = 4 | z_{3} - z_{4} | = 10$ and the diagonals AC and BD intersect at P. It is given that $| A r g (\frac{z_{4} - z_{2}}{z_{3} - z_{1}}) | = \frac{π}{2}$ and $| A r g (\frac{z_{3} - z_{2}}{z_{4} - z_{1}}) | = \frac{π}{4}$ .
Which of the following option(s) is/are correct?

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$P C : A C = 5 : 7$

Area of quadrilateral ABCD $= \frac{140}{3} s q . u n i t s$

value of $| C P - D P |$ is equal to $\frac{10}{\sqrt{21}}$

$P C : A C = 2 : 3$

answer is A, B, C.

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Detailed Solution

$Δ A B P a n d Δ C D P$ are similar if AP=2x,
BP=2y then CP=5x, DP=5y
Area of trapezium $A B C D = \frac{49}{2} x y$
$\tan α = \frac{2 x}{5 y}, \tan β = \frac{2 y}{5 x}$ also $α + β = 45^{0}$
$\Rightarrow x y = \frac{10}{21} (x^{2} + y^{2})$
Also $A B^{2} = A P^{2} + B P^{2} \Rightarrow x^{2} + y^{2} = 4 \Rightarrow x y = \frac{40}{21}$
$A r (Δ P C D) = 5 x y$
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