Question

Find the equation of the line which passes through the point $(3,4)$ and makes intercepts on the axes such that their sum is $14$

Answer 1

Equation of the line is $\frac{x}{a}+\frac{y}{b}=1$

Sum of the intercepts$=14$

$\Rightarrow a+b=14$

$\Rightarrow b=14-a$

Equation of the line is $\frac{x}{a}+\frac{y}{14-a}=1$

This line passes throught $(3,4)$

$\Rightarrow \frac{3}{a}+\frac{4}{14-a}=1$

$\Rightarrow 3(14-a)+4a=a(14-a)$

$\Rightarrow 42-3a+4a=14a-a^2$

$\Rightarrow 42+a-14a+a^2=0$

$\Rightarrow a^2-13a+42=0$

$\Rightarrow a^2-6a-7a+42=0$

$\Rightarrow a(a-6)-7(a-6)=0$

$\Rightarrow (a-6)(a-7)=0$

$\therefore a=6,7$

Substituting $a=6$ in $\frac{x}{a}+\frac{y}{14-a}=1$ we get

$\frac{x}{6}+\frac{y}{14-6}=1$

$\Rightarrow \frac{x}{6}+\frac{y}{8}=1$

$\Rightarrow 8x+6y=48$

or $4x+3y=24$

Substituting $a=7$ in $\frac{x}{a}+\frac{y}{14-a}=1$ we get

$\frac{x}{7}+\frac{y}{14-7}=1$

$\Rightarrow \frac{x}{7}+\frac{y}{7}=1$

$\Rightarrow 7x+7y=1$

Equation of the intercepts are $4x+3y=24$ and $7x+7y=1$