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

Let the required straight line passing through the origin be written in the slope-intercept form

$$y = mx,$$

where $$m$$ is an arbitrary real number (the slope). Because this line contains the origin $$O(0,0),$$ every point on it can be written as $$\bigl(t,\;mt\bigr)$$ for some real parameter $$t$$. Our task is to find the points $$A$$ and $$B$$ at which this line meets the two given lines and then compare the distances $$OA$$ and $$OB$$ along the same straight line.

Intersection with the first given line. The first line is

$$3y = 10 - 4x.$$

Substituting $$y = mx$$ from the line through the origin, we get

$$3(mx) \;=\; 10 - 4x.$$

Simplifying,

$$3mx + 4x \;=\; 10,$$

$$x\,(3m + 4) \;=\; 10,$$

$$x_A \;=\; \dfrac{10}{\,3m + 4\,}.$$

Since $$y = mx,$$ the corresponding ordinate is

$$y_A \;=\; m\,x_A \;=\; \dfrac{10m}{\,3m + 4\,}.$$

Thus

$$A\Bigl(\dfrac{10}{3m + 4},\;\dfrac{10m}{3m + 4}\Bigr).$$

Intersection with the second given line. The second line is

$$8x + 6y + 5 = 0.$$

Again putting $$y = mx, $$ we obtain

$$8x + 6(mx) + 5 \;=\; 0,$$

$$x\,(8 + 6m) \;=\; -5,$$

$$x_B \;=\; -\dfrac{5}{\,8 + 6m\,}.$$

Hence

$$y_B \;=\; m\,x_B \;=\; -\dfrac{5m}{\,8 + 6m\,},$$

and

$$B\Bigl(-\dfrac{5}{8 + 6m},\;-\dfrac{5m}{8 + 6m}\Bigr).$$

Expressing the distances $$OA$$ and $$OB$$. Because $$A,\,O,\,B$$ all lie on the same line $$y = mx,$$ the vector from the origin to any point on this line is a scalar multiple of the direction vector $$(1,m).$$ If a point has coordinates $$(t,mt),$$ its distance from the origin is

$$\sqrt{t^{2} + (mt)^{2}} \;=\; |t|\sqrt{\,1 + m^{2}}.$$

Thus the distance from the origin is directly proportional to the absolute value of the $$x$$-coordinate. Therefore,

$$\dfrac{OA}{OB} \;=\; \dfrac{|x_A|}{|x_B|}.$$

Calculating the ratio.

$$\dfrac{OA}{OB} \;=\; \dfrac{\Bigl|\dfrac{10}{\,3m + 4\,}\Bigr|}{\Bigl|-\dfrac{5}{\,8 + 6m\,}\Bigr|}$$

$$\;\;=\; \dfrac{10}{5}\;\cdot\;\dfrac{|8 + 6m|}{|3m + 4|}$$

$$\;\;=\; 2\,\cdot\,\dfrac{|8 + 6m|}{|3m + 4|}.$$

Notice that the two factors $$8 + 6m$$ and $$3m + 4$$ always have the same sign (they are both positive or both negative for every real $$m$$), because

$$8 + 6m \;=\; 2\,(4 + 3m)$$

and

$$3m + 4 \;=\; 3m + 4.$$

The linear expressions differ only by the constant factor 2, so their signs coincide; hence the absolute values cancel the need for the sign, and we simply have

$$|8 + 6m| \;=\; 2\,|3m + 4|.$$

Substituting this into the ratio,

$$\dfrac{OA}{OB} \;=\; 2 \cdot \dfrac{2\,|3m + 4|}{|3m + 4|} \;=\; 2 \cdot 2 \;=\; 4.$$

Thus $$OA : OB = 4 : 1.$$ The origin $$O$$ therefore divides the segment $$AB$$ internally in the ratio $$4 : 1$$.

Hence, the correct answer is Option C.