Q. If a line intercepted between the coordinate axes is trisected at a point $A(4, 3)$, which is nearer to $x$-axis, then its equation is :
Solution:
$\Rightarrow 4=\left(\frac{1\times0+2\times a}{1+2}\right)=\frac{2a}{3}$
$\Rightarrow a=6 \Rightarrow $ coordinate of $B$ is $B \left(6, 0\right)$
$3=\left(\frac{1\times a+2\times 0}{1+2}\right)=\frac{b}{3}$
$\Rightarrow b=9$ and $C\left(0, 9\right)$
Slope of line passing through $\left(6,0\right), \left(0,9\right)$
slope, $m=\frac{9}{-6}=-\frac{3}{2}$
Equation of line $y-0= \frac{-3}{2}\left(x- 6\right)$
$2y = -3x + 18$
$3x + 2y = 18$
