Question

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

• A. $x+y-7=0$
• B. $x-y+1=0$
• C. $4x+3y-24=0$
• D. $4x-3y=0$

Answer 1

Answer: (A), (C)

$4x+3y-24=0$

Let the equation of the line be
$\frac{x}{a}+\frac{y}{b}=\mathrm{1...}\left(i\right)$
This line passes through $(3,4),$ therefore
$\frac{3}{a}+\frac{4}{b}=1...\left(ii\right)$
It is given that $a+b=14$
$\Rightarrow b=14-a$
Putting $b=14-a$ in equation $\left(ii\right)$, we get
$\frac{3}{a}+\frac{4}{14-a}=1$
$\Rightarrow 3(14-a)+4a=a(14-a)$
$\Rightarrow a^2-13a+42=0$
$\Rightarrow (a-7)(a-6)=0$
$\Rightarrow a=7,6$
For $a=7,b=14-7=7$
and for $a=6,b=14-6=8$
Putting the values of $a$ and $b$ in $\left(i\right)$, we get the equation of the lines as
$\frac{x}{7}+\frac{y}{7}=1$ and $\frac{x}{6}+\frac{y}{8}=1$
$\Rightarrow x+y=7$ and $4x+3y=24$