Find the equation of the line passing through the point $\left(1,5\right)$ and which divides the coordinate axes in the ratio $3 : 10$.
Let the required line cut the X axis at $A \left(h,0\right)$ and the Y axis at $B \left(0,k\right)$.
Then, X intercept $= OA = h$, Y intercept $= OB = k$
Given: $\dfrac{OA}{OB}= \dfrac{h}{k}=\dfrac{3}{10}$ $\implies$ $h = \dfrac{3 k}{10}$ $\;\;\; \cdots \; (1)$
Let the equation of the required line be: $\hspace{1em}$ $\dfrac{x}{h}+ \dfrac{y}{k} = 1$ $\;\;\; \cdots \; (2)$
The required line passes through $P \left(1,5\right)$
$\therefore$ $\;$ We have from equation $(2)$, $\hspace{1em}$ $\dfrac{1}{h} + \dfrac{5}{k} = 1$ $\;\;\; \cdots \; (3)$
Substituting the value of $h$ from equation $(1)$ in equation $(3)$ gives
$\dfrac{10}{3k} + \dfrac{5}{k} = 1$
i.e. $25 = 3k$ $\implies$ $k = \dfrac{25}{3}$ $\;\;\; \cdots \; (4a)$
Substituting the value of $k$ in equation $(1)$ gives
$h = \dfrac{3}{10} \times \dfrac{25}{3} = \dfrac{5}{2}$ $\;\;\; \cdots \; (4b)$
$\therefore$ $\;$ In view of equations $(4a)$ and $(4b)$, equation $(2)$ becomes
$\dfrac{x}{5/2} + \dfrac{y}{25/3} = 1$
i.e. $\dfrac{2x}{5} + \dfrac{3y}{25} = 1$
i.e. $10 x + 3 y = 25$
$\therefore$ $\;$ The equation of the required line is: $\hspace{1em}$ $10x + 3 y = 25$