$\mathrm{a}$ and $\mathrm{b}$ are the intercepts made by a line on the co-ordinate axes. If $3…
$\mathrm{a}$ and $\mathrm{b}$ are the intercepts made by a line on the co-ordinate axes. If $3 \mathrm{a}=\mathrm{b}$ and the line passes through $(1,3)$, then the equation of the line is
$x+3 y=10$
$3 x+y=6$
$x-3 y+8=0$
$3 x-2 y+3=0$
Solution
Let the equation of the line be
$\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1$
This line passes through the point $(1,3)$
$\begin{aligned}
\therefore \quad & \frac{1}{a}+\frac{3}{b}=1 \\
& \Rightarrow b+3 a=a b \\
& \Rightarrow 2 b=a b \\
& \Rightarrow a=2 \\
\therefore \quad & b=3 a=6
\end{aligned}$
Equation (i) becomes,
$\begin{aligned}
& \frac{x}{2}+\frac{y}{6}=1 \\
& 3 x+y=6
\end{aligned}$