Question

The line x + y = a, meets the axis of x and y at A and B respectively. A triangle AMN is inscribed in the triangle OAB, O being the origin, with right angle at N. M and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN

=______________

• A. 2 : 3
• B. 3 : 4
• C. 3 : 1
• D. None of these

Answer 1

Answer: (C)

3 : 1

Answer 2

Let $\frac{AN}{BN}$ = l, then the coordinates of N are

$\left( \frac{a}{1 + \lambda},\frac{\lambda a}{1 + \lambda} \right)$

Where (a, 0) and (0, a) are the coordinates of A and B respectively. Now equation of MN perpendicular to AB is

y –$\frac{\lambda a}{1 + \lambda}$ = x –$\frac{a}{1 + \lambda}$.

or x – y = $\frac{1 - \lambda}{1 + \lambda}$a.

So the coordinates of M are $\left( 0,\frac{\lambda - 1}{\lambda + 1}a \right)$

Therefore, area of the triangle AMN is

= $\frac{1}{2}\left| \left\lbrack a\left( \frac{- a}{\lambda + 1} \right) + \frac{1 - \lambda}{(1 + \lambda)^{2}}a^{2} \right\rbrack \right|$ = $\frac{\lambda a^{2}}{(1 + \lambda)^{2}}$

Also area of the triangle OAB = a²/2.

So that according to the given condition.

$\frac{\lambda a^{2}}{(1 + \lambda)^{2}}$ = $\frac{3}{8} \cdot \frac{1}{2}$a²

Ž 3l² – 10l + 3 = 0

Ž l = 3 or l = 1/3.

For l = 1/3, M lies outside the segment OB and hence the required value of l is 3.