A straight line through origin ‘O’ meets the line 3y = 10 – 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then ‘O’ divides the segment AB in the ratio 

Given lines

$\begin{array}{l}4x+3y-10=0--\left(1\right)\\ 8x+6y+5=0--\left(2\right)\\ \left(1\right)and\left(2\right)areparallel\\ P_1=\perp rdis\tan cefrom\left(0,0\right)to\left(1\right)\\ =\frac{\left|-10\right|}{\sqrt{16+9}}\Rightarrow \frac{10}{5}\Rightarrow 2\\ P_2=\perp rdis\tan cefrom\left(0,0\right)to\left(2\right)\\ =\frac{\left|5\right|}{\sqrt{64+36}}=\frac{5}{10}=\frac{1}{2}\\ \mathrm{Re}quiredratio=P_1:P_2=2:\frac{1}{2}=4:1\end{array}$