Find the equations of straight lines passing through $\left(8,3\right)$ and having intercepts whose sum is 1.
Let the intercept on the X axis be $= a$
and the intercept on the Y axis be $= b$.
Let the equation of the required line be: $\hspace{1em}$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$ $\;\;\; \cdots \; (1)$
Given: $\hspace{1em}$ $a + b = 1$ $\implies$ $a = 1 - b$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ becomes
$\dfrac{x}{1 - b} + \dfrac{y}{b} = 1$ $\;\;\; \cdots \; (3)$
Now, equation $(3)$ passes through $\left(8,3\right)$.
$\therefore$ $\;$ We have: $\hspace{1em}$ $\dfrac{8}{1 - b}+ \dfrac{3}{b} = 1$
i.e. $8 b + 3 - 3b = b \left(1 - b\right)$
i.e. $5 b + 3 = b - b^2$
i.e. $b^2 + 4 b + 3 = 0$
i.e. $\left(b + 3\right) \left(b + 1\right) = 0$
$\implies$ $b = - 3$ or $b = -1$
Substituting the value of $b$ in equation $(2)$ gives
when $b = - 3$, $a = 1 - \left(-3\right) = 4$
when $b = -1$, $a = 1 - \left(-1\right) = 2$
$\therefore$ $\;$ Substituting the values of $a$ and $b$ in equation $(1)$ gives the equations of lines as
$\dfrac{x}{4} + \dfrac{y}{-3} = 1$ $\;\;\;$ i.e. $- 3x + 4y = -12$ $\;\;\;$ i.e. $\;$ $3x - 4y = 12$
and $\dfrac{x}{2} + \dfrac{y}{-1} = 1$ $\;\;\;$ i.e. $\;$ $- x + 2y = -2$ $\;\;\;$ i.e. $\;$ $x - 2y = 2$