Question

Equation of the line passing through the point of intersection of the lines $x-y=1$ and $2x+3y=4$ and equidistant from the point $(2,-3)$ and $(1,1)$ is

• A. $7x-2y=0$
• B. $14x+y-20=0$
• C. $2x-2y=5$
• D. $Noneofthese$

Answer 1

The correct option is B $14x+y-20=0$

$\begin{array}{c}Letequationofliney=mx+c\to \left(i\right)\\ Bysolvingn-y=1and2x+3y=4\\ x-y-1=0\\ 2x+3y-4=0\\ \Rightarrow \frac{x}{7}=\frac{y}{2}=\frac{1}{5}\\ x=\frac{7}{5}andy=\frac{2}{5}\\ distancefromapoints\left(2,-3\right)and\left(h^1\right)online\\ y=mx+caresame\\ \Rightarrow \frac{\left|-3-2m-c\right|}{\sqrt{1+m^2}}=\frac{\left|1-mx-c\right|}{\sqrt{1+m^2}}\\ \Rightarrow 3=2m+c=m+c-1\\ m=-4\\ Putx=\frac{7}{5},y=\frac{2}{5}andm=-4inequation\left(i\right)\\ \Rightarrow \frac{2}{5}=-4\times \frac{2}{5}+c,c=2\\ Hence,equationofliney=-4x+2\end{array}$