Question

The line $x+y=a$ meets the axes of x and y at A and B respectively. A $\Delta$AMN is inscribed in the $\Delta$OAB, O being the origin, with right angle at N. M and N lie on OB and AB respectively. If the area of the $\Delta$AMN is $\frac{3}{8}$ of the area of the $\Delta$OAB, then $\frac{AN}{BN}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{3},3$
• C. $\frac{2}{3},3$
• D. $3$

Answer 1

Answer: (D)

$3$

Let $\frac{AN}{BN}=\frac{\lambda }{1}$

According to the section formula $N(x,y)=\frac{0\times \lambda +a\times 1}{\lambda +1},\frac{a\times \lambda }{\lambda +1}$

$N(x,y)=\frac{a}{\lambda +1},\frac{a\lambda }{\lambda +1}$

Slope of $AB=-1$

$AB$ is perpendicular to $MN$

So , Slope of $MN=1$

Equation of $MN=y-\frac{a\lambda }{\lambda +1}=x-\frac{a}{\lambda +1}$

When $x=0$ , then $y=\frac{a(\lambda -1)}{\lambda +1}$

$M(x,y)=(0,\frac{a(\lambda -1)}{\lambda +1})$

By the help of distance formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AN=\frac{\sqrt{2}a\lambda }{1+\lambda }$

$MN=\frac{\sqrt{2}a}{1+\lambda }$

Area of $\Delta AMN=\frac{1}{2}\times AN\times BN=\frac{a^2\lambda }{{(\lambda +1)}^2}$

Area of $\Delta OAB=\frac{1}{2}\times a^2$

Area of $\Delta AMN=\frac{3}{8}\times$ area of $\Delta OAB$

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

$\lambda =\frac{1}{3},3$

$\lambda =\frac{1}{3}$ , In this $M$ lies outside. Thus this can't be true.

$\lambda =3$