Question

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

Answer 1

Let the equation of the line be $\frac{x}{a}+\frac{y}{b}=1$ .....(i)

This passes through $(3,4)$ therefore $\frac{3}{a}+\frac{4}{b}=1$ ........(ii)

It is given that $a+b=14$

$\Rightarrow$ $b=14-a$

Putting $b=14-a$ in (ii)

we get,

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

$\Rightarrow a^2-13a+42=0$ $\Rightarrow$ $(a-7)(a-6)=0$ $\Rightarrow$ $a=7,6$

For $a=7,b=14-7=7$ and for $a=6,b=14-6=8$

Putting the values of a and b in (i) we get the equations of the lines

$\frac{x}{7}+\frac{y}{7}=1and\frac{x}{6}+\frac{y}{8}=1$

$\Rightarrow x+y=7$ and $4x+3y=24$