Question

If $x+y-2=0,$ $2x-y+1=0$ and $px+qy-r=0$ are concurrent lines, then the slope of the member in the family of lines $2px+3qy+4r=0$ which is farthest from the origin is

• A. $-\frac{1}{2}$
• B. $-2$
• C. $-\frac{2}{3}$
• D. $-\frac{3}{10}$

Answer 1

The correct option is D $-\frac{3}{10}$

Given $x+y-2=0,2x-y+1=0$ and $px+qy-r=0$ are concurrent lines, so
$\left|\begin{array}{ccc}1& 1& -2\\ 2& -1& 1\\ p& q& -r\end{array}\right|=0\Rightarrow p+5q-3r=0\cdots \left(1\right)$

Now, $2px+3qy+4r=0$
$\Rightarrow 2px+3qy+\frac{4}{3}(p+5q)=0\Rightarrow p(2x+\frac{4}{3})+q(3y+\frac{20}{3})=0\Rightarrow (2x+\frac{4}{3})+\frac{q}{p}(3y+\frac{20}{3})=0$
Point of intersection of the family of lines is
$x=-\frac{2}{3},y=-\frac{20}{9}$

Line farthest from the origin is the line which is perpendicular to the line joining origin and $(-\frac{2}{3},-\frac{20}{9})$

Therefore, the slope of required line
$=-\frac{-\frac{2}{3}-0}{-\frac{20}{9}-0}=-\frac{3}{10}$