Find the equation of a straight line parallel to $2x + 3y = 10$ and which is such that the sum of its intercepts on the axes is 15.
Equation of the given line is: $\hspace{1em}$ $2x + 3y = 10$
i.e. $y = - \dfrac{2}{3} x + \dfrac{10}{3}$
$\therefore$ $\;$ Slope of the given line $= m = - \dfrac{2}{3}$
Since the required line is parallel to the given line,
$\therefore$ $\;$ slope of the required line $= m = - \dfrac{2}{3}$ $\;\;\; \cdots \; (1)$
Let the equation of the required line be: $\hspace{1em}$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$ $\;\;\; \cdots \; (2)$
where $a$ and $b$ are the intercepts on the X and the Y axes respectively.
Given: Sum of intercepts on the axes $= a + b = 15$
i.e. $a = 15 - b$ $\;\;\; \cdots \; (3)$
Substituting the value of $a$ from equation $(3)$ in equation $(2)$ gives
$\dfrac{x}{15 - b} + \dfrac{y}{b} = 1$
i.e. $b \; x + \left(15 - b\right)y = b \left(15 - b\right)$
i.e. $y = \left(\dfrac{-b}{15 - b}\right) \; x + b$
$\therefore$ $\;$ Slope of the required line is $= \dfrac{b}{b - 15}$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ We have from equations $(1)$ and $(4)$,
$\dfrac{b}{b - 15} = - \dfrac{2}{3}$
i.e. $3 b = - 2 b + 30$ $\implies$ $b = 6$ $\;\;\; \cdots \; (5)$
Substituting the value of $b$ from equation $(5)$ in equation $(3)$ we have,
$a = 15 - 6 = 9$ $\;\;\; \cdots \; (6)$
$\therefore$ The required equation of line is [from equation $(2)$]:
$\dfrac{x}{9} + \dfrac{y}{6} = 1$
i.e. $2x + 3y - 18 = 0$