Question

The portion of a line intercepted between the coordinate axes is divided by the point $(2,-1)$ in the ration $3:2$. The equation of the line is :

• A. $5x-2y-20=0$
• B. $2x-y+7=0$
• C. $x-3y-5=0$
• D. $2xy+y-4=0$

Answer 1

Answer: (C)

$x-3y-5=0$

$\text{We}\text{have}$

$\text{given}\text{points}\text{are}(2,-1)$

$\text{ratio}=m_1:m_2=3:2$

$\text{Let}\text{the}\text{equation}\text{of}\text{line}\text{be}$

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

$\text{The}\text{line}\text{meets}\text{the}\text{coordinates}a\text{xes}\text{at}$

$A(a,0)\text{and}B(0,b)respectively.$

$m_1:m_2=3:2$

$\text{So}$

$\text{using}\text{section}\text{formula}$

$(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$\Rightarrow (\frac{3\times 0+2\times a}{3+2},\frac{3\times b+2\times 0}{3+2})$

$\Rightarrow (\frac{2a}{5},\frac{3b}{6})$

$\text{It}\text{is}\text{given}\text{that}\text{point}\text{(2,}\text{-1)}\text{divides}$

$\text{ratio}3:2$

$\text{Now},\frac{2a}{5}=2,\frac{3b}{5}=-1$

$a=5,b=\frac{-5}{3}$

$\text{Hence,}\text{the}\text{equation}\text{of}\text{line}$

$\frac{x}{5}+\frac{y}{\frac{-5}{3}}=1\Rightarrow x-3y=5$

$x-3y-5=0.$

$\text{Hence,}\text{this}\text{is}\text{the}\text{answer}\text{.}$