A straight line makes on the coordinate axes positive intercepts whose sum is 5. If the line passes through the point P(-3, 4), find its equation.

Given,

A straight line makes on the coordinate axes positive intercepts whose sum is 5.

Let intercept be a on x-axis and b on y-axis.

a + b = 5

a = 5 - b ………(1)

Intercept form of equation : $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Substituting values we get :

$\Rightarrow \dfrac{-3}{a} + \dfrac{4}{b} = 1 \\[1em] \Rightarrow \dfrac{-3b + 4a}{ab} = 1 \\[1em] \Rightarrow \dfrac{4a - 3b}{ab} = 1 \\[1em] \Rightarrow 4a - 3b = ab$

Substituting value of a from equation (1) in above equation :

$\Rightarrow 4(5 - b) - 3b = (5 - b)b \\[1em] \Rightarrow 20 - 4b - 3b = 5b - b^2 \\[1em] \Rightarrow 20 - 7b = 5b - b^2 \\[1em] \Rightarrow b^2 - 7b - 5b + 20 = 0 \\[1em] \Rightarrow b^2 - 12b + 20 = 0 \\[1em] \Rightarrow b^2 - 10b - 2b + 20 = 0 \\[1em] \Rightarrow b(b - 10) - 2(b - 10) = 0 \\[1em] \Rightarrow (b - 2)(b - 10) = 0 \\[1em] \Rightarrow b = 2 \text{ or } b = 10.$

Substituting values of b in equation (1), we get :

When b = 2,

a = 5 - b = 5 - 2 = 3

When b = 10,

a = 5 - b = 5 - 10 = -5.

Since, both intercepts are positive,

So, a ≠ -5.

∴ a = 3 and b = 2

Substituting value of a and b in $\dfrac{x}{a} + \dfrac{y}{b} = 1$, we get :

$\Rightarrow \dfrac{x}{3} + \dfrac{y}{2} = 1 \\[1em] \Rightarrow \dfrac{2x + 3y}{6} = 1 \\[1em] \Rightarrow 2x + 3y = 6.$

Hence, required equation is 2x + 3y = 6.