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[[1]] is the equation of the line AB in the following figure, given $OP = 3 / 2$ .

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Detailed Solution

https://www.vedantu.com/question-sets/e02a4e79-46fb-46a0-9ee7-a78fb241a10f984959542949603058.png

We have ∠RPB=∠OPQ=30∘ because the both angles are vertical opposite to each other. We have the right angle ∠POQ=90∘. We use the property that sum of internal angles in a triangle is 180∘ to find ∠PQO=θ. So we have

∠ POQ + ∠ PQO + ∠ OPQ = 180 °

⇒90

°

+θ+30

°

=180

°

⇒θ=60

°

tanθ = \tan (60 °) = 3 - \sqrt{tanθ} = \tan (60 °) = \sqrt{3}

We know the equation of the line in slope point form with slope mm and a point on the line (x1,y1) as

y - y 1 = m (x - x 1)

The slope of the line AB is m=

\sqrt{3}

and the point

P (0, \frac{3}{2})

lies on it. So the equation of the line AB is

y - \frac{3}{2} = 3 - \sqrt{(x - 0)}

\Rightarrow 2 y - 3 = 2 \sqrt{3} x

\Rightarrow 2 \sqrt{3} x + 2 y - 3 = 0

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