Question

A line passes through the point of intersection of the lines $100x+50y-1=0$ and $75x+25y+3=0$ and makes equal intercepts on the axes. Its equation is

• A. $25x+25y-1=0$
• B. $5x-5y+3=0$
• C. $25x+25y-4=0$
• D. $25x-25y+6=0$

Answer 1

Answer: (C)

$25x+25y-4=0$

$Formulaused$
$linein\mathrm{intercept}form$
$\frac{x}{a}+\frac{y}{b}=1$
$Given$
$100x+50y-1=0.........\left(i\right)$
$75x+25y+3=0.........\left(ii\right)$
$solving\left(i\right)and\left(ii\right)forpo\mathrm{int}of\mathrm{int}er\sec tion$
$x=-7/50,y=15/50$
$Hencepo\mathrm{int}is(-7/50,15/50)$
$lettheequationoflinebe$
$\frac{x}{a}+\frac{y}{a}=1$
$\Rightarrow x+y=a$
$linepassesthrough(-7/50,15/50)$
$Hence$
$\frac{-7}{50}+\frac{15}{50}=a$
$\Rightarrow a=\frac{8}{50}=\frac{4}{25}$
$Hencetherequiredlineis$
$x+y=\frac{4}{25}$
$\Rightarrow 25x+25y-4=0$