A straight line through the origin O(0, 0) me | vedclass

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(A) Let the equation of the line passing through the origin be $y = mx$. Substituting this into the first line $4x + 3y - 10 = 0$,we get $4x + 3(mx) =

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$4: 1$

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(A) Let the equation of the line passing through the origin be $y = mx$. Substituting this into the first line $4x + 3y - 10 = 0$,we get $4x + 3(mx) = 10$,so $x_A = \frac{10}{4+3m}$ and $y_A = \frac{10m}{4+3m}$. Thus,$OA = \sqrt{x_A^2 + y_A^2} = \frac{10\sqrt{1+m^2}}{|4+3m|}$. Substituting $y = mx$ into the second line $8x + 6y + 5 = 0$,we get $8x + 6(mx) = -5$,so $x_B = \frac{-5}{8+6m}$ and $y_B = \frac{-5m}{8+6m}$. Thus,$OB = \sqrt{x_B^2 + y_B^2} = \frac{5\sqrt{1+m^2}}{|8+6m|} = \frac{5\sqrt{1+m^2}}{2|4+3m|}$. The ratio $OA : OB = \frac{10\sqrt{1+m^2}}{|4+3m|} : \frac{5\sqrt{1+m^2}}{2|4+3m|} = 10 : \frac{5}{2} = 20 : 5 = 4 : 1$. Since $O$ lies between $A$ and $B$ (as the lines are on opposite sides of the origin),$O$ divides $AB$ internally in the ratio $4:1$.

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

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