Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that is an isosceles triangle and . If , then the greatest integer less than or equal to is: [2024]

Question

Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that $△ O P Q$ is an isosceles triangle and $∠ P O Q = 90 °$ . If $l = O P^{2} + P Q^{2} + Q O^{2}$ , then the greatest integer less than or equal to $l$ is: [2024]

Smart Achievers · Accepted Answer

(3)

Let $P (r \cos θ, r \sin θ)$ and $Q (- r \sin θ, r \cos θ)$

$∴ O P^{2} = O Q^{2} = r^{2}$

Now, in $△ P O Q, ∠ P O Q = 90 °, P Q^{2} = 2 r^{2}$

Now, $l = O P^{2} + P Q^{2} + O Q^{2} = 4 r^{2}$

$∵$ P and Q lies on 3x + 4y = 12

$∴ 3 (r \cos θ) + 4 (r \sin θ) = 12$

$\Rightarrow 3 \cos θ + 4 \sin θ = \frac{12}{r}$ ... (i)

and $3 (- r \sin θ) + 4 (r \cos θ) = 12$

$\Rightarrow - 3 \sin θ + 4 \cos θ = \frac{12}{r}$ ... (ii)

From (i) and (ii), we have

$25 = \frac{288}{r^{2}} \Rightarrow r^{2} = \frac{288}{25}$

Now, $l = 4 r^{2} = 4 \times \frac{288}{25} = 46.08$

So, $[l] = 46$ .

Solution

Q.
Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that $△ O P Q$ is an isosceles triangle and $∠ P O Q = 90 °$ . If $l = O P^{2} + P Q^{2} + Q O^{2}$ , then the greatest integer less than or equal to $l$ is: [2024]

1 44

2 42

3 46

4 48

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