If a line intercepted between the coordinate axes is trisected at a point $\mathrm{A}(4,3)$, which is nearer…
- $4 x-3 y=7$
- $3 \mathrm{x}+2 \mathrm{y}=18$
- $3 x+8 y=36$
- $x+3 y=13$
Solution

$ \text { A divides } \mathrm{CB} \text { in } 2: 1 $ $ \begin{aligned} &\Rightarrow 4=\left(\frac{1 \times 0+2 \times a}{1+2}\right)=\frac{2 a}{3} \\ &\Rightarrow a=6 \Rightarrow \text { coordinate of B is B }(6,0) \\ &3=\left(\frac{1 \times b+2 \times 0}{1+2}\right)=\frac{b}{3} \\ &\Rightarrow b=9 \text { and } \mathrm{C}(0,9) \end{aligned} $ Slope of line passing through $(6,0),(0,9)$ slope, $m=\frac{9}{-6}=-\frac{3}{2}$ Equation of line $y-0=\frac{-3}{2}(x-6)$ $ \begin{aligned} &2 y=-3 x+18 \\ &3 x+2 y=18 \end{aligned} $
Asked in: JEE Main 2014 (12 Apr Online)