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.

• A. $2x+y-7=0$
• B. $x+y-7=0$
• C. $x-y-7=0$
• D. $x+y+7=0$

Answer 1

The correct option is B

$x+y-7=0$

Given $a+b=14$,
Coordinates are $A=(0,b),B=(a,0)$
Since $A=(0,b)=(0,14-a)$
So equation of AB
$y-(14-a)=\frac{0-(14-a)}{(a-0)}(x-0)\Rightarrow y-14+a=\frac{a-14}{a}\left(x\right),\dots \left(i\right)$
It passes through (3, 4) So,
$4-14+a=\frac{a-14}{a}\times 3\Rightarrow a-10=\frac{3a-42}{a}\Rightarrow a^2-10a-3a+42=0\Rightarrow a^2-13a+42=0\Rightarrow (a-7)(a-6)=0$
So, $a=7ora=6$
Put $a=6$ equation (i), we get
$y-14+a=\frac{a-14}{a}\left(x\right)\Rightarrow y-14+6=\frac{6-14}{6}x\Rightarrow y-8=\frac{-4}{3}x\Rightarrow 3y-24=-4x\Rightarrow 4x+3y-24=0$
Put $a=7$ equation (i), we get
$y-14+a=\frac{a-14}{a}\left(x\right)\Rightarrow y-14+7=\frac{7-14}{7}x\Rightarrow y-7=\frac{-1}{1}x\Rightarrow x+y-7=0$