A straight line through origin O meets the lines `3y=10-4x& 8x+6y+5=0` at points A and B respectively. Then O divides the segment AB in the ratio:

Options

A. `3:4`
B. `1:2`
C. `2:3`
D. `4:1`

Answer

D

Explanation

Clearly, origin O is between the lines `4x+3y-10=0 and 8x+6y+5=0` . Equation of a line passing through the origin O is `(x)/(cos theta)=(y)/(sin theta)`. Let OP =`r_(1) and OQ=r_(2)` . Then, the coordinates of P and Q are `(r_(1)cos theta , r_(1) sin theta)` and `(r_(2)cos theta, r_(2)sintheta)` respectively. These points lie on the lines `8x+6y+5=0 and 4x+3y-10=0` respectively.
`therefore r_(2)(8cos theta+6sin theta)+5=0 and r_(2)(4costheta+3sin theta)-10=0`
`implies r_(1)-(-5)/(2(4costheta+3sin theta)) and r_(2)=(10)/(4cos theta+3 sin theta)`

`implies r_(1)=(r_(2))/(4)`
`implies (r_(2))/(r_(1))=-(4)/(1)`
Hence , the required ratio is 4:1