If the straight lines $x+y-2=0$, $2x-y+1=0$ and $px+qy-r=0$ are concurrent, then what is the slope of the member of family of lines $2px+3qy+4r=0$ which is farthest from origin?
I wrote the coefficients of the variables of the given lines in a determinant, equated it to $0$ and got $p+5q-3r=0$. Then, I substituted $r$ in the equation of the family of lines. What do I do next? How to find the line farthest from the origin?
Apply the distance formula for the distance between a line and a plane to get $$d=\frac{r}{\sqrt{p^2+q^2}}$$ Clearly r=$\frac{p+5q}{3}$
Substitute in the above equation to get $$d=\frac{\frac{p+5q}{3}}{\sqrt{p^2+q^2}}$$ Dividing numerator and denominator by p,we get d in terms of q/p=$\lambda$. Now q/p is the slope of the line. Differentiate d w.r.t $\lambda$ and equate it to 0 to find the maximum value of the slope.