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Applications of Cross Product

A line l1 whi...

Question

A line $l_{1}$ which makes an acute angle $β$ with the positive direction of the $x - axis$ is drawn through point $P (2, 4)$ to meet the line $l_{2} : x = 4$ at $R$ and $l_{3} : y = 8$ at $S$ . $l_{2} & l_{3}$ meet at $Q$ .

P R = P S

β = {tan}^{- 1} (2)

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P R = P S

β = {tan}^{- 1} (1)

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Area (Δ Q R S) = 2

β = \frac{π}{4}

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Area (Δ Q R S) = 0

P S = P R

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Solution

The correct options are
A $P R = P S$ at $β = {tan}^{- 1} (2)$
C $Area (Δ Q R S) = 2$ at $β = \frac{π}{4}$
D $Area (Δ Q R S) = 0$ If $P S = P R$
Equation of the line passing through $P (2, 4)$ and making $β$ angle with positive direction of the $x -$ axis is
$l_{1} : y - 4 = tan β (x - 2)$
Coordinates of $R : x = 4 \Rightarrow y = 2 tan β + 4 R (4, 4 + 2 tan β) P R = 2 sec β$
Coordinates of $S : y = 8 \Rightarrow x = \frac{4}{tan β} + 2 S (\frac{4}{tan β} + 2, 8) P S = 4 cosec β$
$Q (4, 8) ∴ Q R = | 4 - 2 tan β | Q S = \frac{4}{tan β} - 2 ∵ Q R ⊥ Q S ∴ a r e a (Δ Q R S) = \frac{1}{2} Q R \times Q S = \frac{2 (2 - tan β)^{2}}{tan β}$

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A line l1 which makes an acute angle β with the positive direction of the x−axis is drawn through point P(2,4) to meet the line l2:x=4 at R and l3:y=8 at S. l2 & l3 meet at Q.

A line $l_{1}$ which makes an acute angle $β$ with the positive direction of the $x - axis$ is drawn through point $P (2, 4)$ to meet the line $l_{2} : x = 4$ at $R$ and $l_{3} : y = 8$ at $S$ . $l_{2} & l_{3}$ meet at $Q$ .