A variable line L passes through the point (3,5) and intersects the positive coordinate ax…

Question

A variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$, where $O$ is the origin, is :

Accepted Answer

$$\begin{aligned}
& \frac{x}{a}+\frac{y}{b}=1 \
& \frac{3}{a}+\frac{5}{b}=1 \
& 3 b+5 a=a b \
& 5 a-a b=-3 b \
& a(5-b)=-3 b \
& a=\frac{3 b}{b-5}
\end{aligned}$$

$$\begin{aligned}
& \text { Area of triangle }=\left|\frac{1}{2} \times a \times b\right| \
& =\frac{1}{2} \times \frac{3 b}{b-5} \times b \
& \Rightarrow f(b)=\frac{3 b^2}{2 b-10} \
& \Rightarrow f^{\prime}(b)=0,(2 b-10) 6 b-2\left(3 b^2\right)=0 \
& 12 b^2-60 b-6 b^2=0 \
& 6 b^2-60 b=0 \
& b^2-10 b=0 \
& b(b-10)=0 \
& b=0 \text { or } b=10
\end{aligned}$$

So for minimum area, $b=10$

then $$\frac{1}{2} \times a \times b=30$$

Solutions · 1