Find the equation of the lines

Question:

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Solution:

The equation of line in intercept form is

$\frac{x}{a}+\frac{y}{b}=1$

Where $\mathrm{a}$ and $\mathrm{b}$ are the intercepts on the axis. Given that $\mathrm{a}+\mathrm{b}=14$

The above equation can be written as

$\Rightarrow \mathrm{b}=14-\mathrm{a}$

Substituting the value of $a$ and $b$ in equation 1 we get So, equation of line is

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

Taking LCM

$\Rightarrow \frac{x(14-a)+a y}{(a)(14-a)}=1$

$\Rightarrow 14 x-a x+a y=14 a-a^{2} \ldots \ldots 2$

$\Rightarrow \frac{x(14-a)+a y}{(a)(14-a)}=1$

$\Rightarrow 14 x-a x+a y=14 a-a^{2} \ldots \ldots .2$

$\Rightarrow \frac{x(14-a)+a y}{(a)(14-a)}=1$

$\Rightarrow 14 x-a x+a y=14 a-a^{2} \ldots \ldots 2$

If equation 2 passes through the point $(3,4)$ then

$14(3)-a(3)+a 4)=14 a-a^{2}$

$\Rightarrow 42-3 a+4 a-14 a+a^{2}=0$

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

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

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

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

$\Rightarrow a-6=0$ or $a-7=0$

$\Rightarrow a=6$ or $a=7$

If $a=6$, then

$\Rightarrow b=14-6=8$

If $a=7$, then

$7+b=14$

$\Rightarrow b=14-7$

$=7$

If $a=6$ and $b=8$, then equation of line is

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

$\Rightarrow \frac{4 x+3 y}{24}=1$

$\Rightarrow 4 x+3 y=24$

If $a=7$ and $b=7$, then equation of line is

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

$\Rightarrow x+y=7$