Question

Let two straight lines drawn from the origin \( O \) intersect the line \(3x + 4y = 12\) at the points \( P \) and \( Q \) such that \( \triangle OPQ \) is an isosceles triangle and \( \angle POQ = 90^\circ \).  If \( l = OP^2 + PQ^2 + QO^2 \), then the greatest integer less than or equal to \( l \) is:

• A. 44
• B. 48
• C. 46
• D. 42

Answer 1

The Correct Option is C

 

The line \( 3x + 4y = 12 \) intersects points \( P \) and \( Q \) at coordinates:

\( P(r \cos \theta, r \sin \theta) \) and \( Q(r \cos (90^\circ + \theta), r \sin (90^\circ + \theta)) = (-r \sin \theta, r \cos \theta) \)

Substituting these into the line equation:

\[ 3(r \cos \theta) + 4(r \sin \theta) = 12 \implies r(3 \cos \theta + 4 \sin \theta) = 12 \tag{1} \]

\( 3(r \sin \theta) + 4(r \cos \theta) = 12 \implies r(-3 \sin \theta + 4 \cos \theta) = 12 \tag{2} \)

To find \( r \), we square both sides and add equations (1) and (2):

\[ \left( \frac{12}{r} \right)^2 + \left( \frac{12}{r} \right)^2 = (3 \cos \theta + 4 \sin \theta)^2 + (-3 \sin \theta + 4 \cos \theta)^2 \]

\[ 2 \left( \frac{12}{r} \right)^2 = 25 \implies r^2 = \frac{288}{25} \implies r = \sqrt{\frac{288}{25}} = \frac{12 \sqrt{2}}{5} \]

Now, we find \( \ell = OP^2 + PQ^2 + OQ^2 \):

\[ OP^2 = r^2, \quad OQ^2 = r^2 \]

To find \( PQ^2 \), we use the distance formula between \( P(r \cos \theta, r \sin \theta) \) and \( Q(-r \sin \theta, r \cos \theta) \):

\[ PQ^2 = (r \cos \theta + r \sin \theta)^2 + (r \sin \theta - r \cos \theta)^2 \]

\[ = r^2(\cos \theta + \sin \theta)^2 + r^2(\sin \theta - \cos \theta)^2 = r^2(1 + 1) = 2r^2 \]

Therefore:

\[ \ell = OP^2 + PQ^2 + OQ^2 = r^2 + 2r^2 + r^2 = 4r^2 \]

Substituting \( r^2 = \frac{288}{25} \):

\[ \ell = 4 \cdot \frac{288}{25} = \frac{1152}{25} = 46.08 \]

The greatest integer less than or equal to \( \ell \) is: \[ \lfloor 46.08 \rfloor = 46 \]