Question

Find the equation of the straight line which passes through the point (−3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7.

Answer 1

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.

Here, a + b = 7
$\Rightarrow$b = 7 − a ... (1)

The line passes through (−3, 8).

∴ $\frac{-3}{a}+\frac{8}{b}=1$ ... (2)

Substituting b = 7 − a in (2) we get,

$$\begin{gathered} \frac{-3}{a}+\frac{8}{7-a}=1 \\ \Rightarrow -3\left(7-a\right)+8a=7a-a^2 \\ \Rightarrow a^2+4a-21=0 \\ \Rightarrow \left(a-3\right)\left(a+7\right)=0 \\ \Rightarrow a=3,a\ne -7\left(\because a\mathrm{is}\mathrm{positive}\right) \end{gathered}$$

Substituting a = 3 in (1) we get,

b = 7 − 3 = 4

Hence, the equation of the line is $\frac{x}{3}+\frac{y}{4}=1$ or 4x + 3y = 12