Question

Find the equations of straight lines which pass through the intersection of the lines $x-2y-5=0,7x+y=50$ and divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $2:1$

• A. $4x-3y-25=0$ or $3x+4y-25=0$
• B. $3x-4y-25=0$ or $-4x+3y-25=0$
• C. $4x+3y+25=0$ or $3x+4y+25=0$
• D. none of these

Answer 1

Answer: (A)

$4x-3y-25=0$ or $3x+4y-25=0$

Point of intersection of lines $x-2y-5=0$

and $7x+y=50$ will be $(7,1)$

$\frac{OP}{OR}=\cos {60}^o$ $=\frac{1}{2}$

$\Rightarrow$ $OP=5$

Let the equation of $PR$ be $:(y-1)=m(x-7)$

$y-mx-1+7m=0$

$OP=5=\left|\frac{-1+7m}{\sqrt{1+m^2}}\right|$

$25+25m^2=49m^2+1-14m$

$24m^2-14m-24=0$ $\Rightarrow$ $m=\frac{4}{3},-\frac{3}{4}$

$\Rightarrow$ equation will be $:(y-1)=\frac{4}{3}(x-7)$ and $(y-1)=-\frac{3}{4}(x-7)$

$\Rightarrow$ equation will be $:4x-3y=25$ and $3x+4y=25$