Question

A variable line with negative slope is drawn through the fixed point $A(3,4)$. If it cuts the coordinate axes at B and C, find the minimum value of the area of $\Delta BCG$, where G is the centroid of the $\Delta OBC$. (O is origin)

Answer 1

Let slope of line = $-tan\theta$
Area of $\Delta BGC$ where $A=(3,4)$
$=\frac{1}{3}$ area of $\Delta OBC$
$=\frac{1}{6}(3+4cos\theta )(4+3tan\theta )$
$=\frac{1}{6}(24+16cot\theta +9tan\theta )$
$\frac{16cot\theta +9tan\theta }{2}\ge \sqrt{16\times 9}=12$
$\therefore$ minimum area of $\Delta BCG=\frac{1}{6}(24+24)=8$