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

Let the intercepts of the line on the x-axis and y-axis be $a$ and $b$ respectively. The equation of the line in intercept form is $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

Given that $a + b = 5$, we have $b = 5 - a$. Substituting this into the equation, we get $\dfrac{x}{a} + \dfrac{y}{5 - a} = 1$.

Since the line passes through the point $P(-3, 4)$, we substitute $x = -3$ and $y = 4$ into the equation:

$\dfrac{-3}{a} + \dfrac{4}{5 - a} = 1$

Multiplying by $a(5 - a)$, we get $-3(5 - a) + 4a = a(5 - a)$.

$-15 + 3a + 4a = 5a - a^2$

$a^2 + 2a - 15 = 0$

Factoring the quadratic equation: $(a + 5)(a - 3) = 0$.

Since the intercepts are positive, $a = 3$. Then $b = 5 - 3 = 2$.

Substituting $a = 3$ and $b = 2$ into the intercept form equation: $\dfrac{x}{3} + \dfrac{y}{2} = 1$.

Multiplying by 6, we get $2x + 3y = 6$.

Answer: $2 x+3 y=6$