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

The position of a point in time $t$ is given by $x=\mathrm{a}+\mathrm{bt}-\mathrm{ct}^2, y=\mathrm{at}+\mathrm{bt}^2$. It's resultant acceleration at time $t$ in seconds is given by

Let $\overline{\mathrm{OA}}=\overline{\mathrm{a}}, \overline{\mathrm{OB}}=\overline{\mathrm{b}}$ and if the vector along the angle bisector of $\angle \mathrm{AOB}$ is given by $x \frac{\overline{\mathrm{a}}}{|\overline{\mathrm{a}}|}+y \frac{\overline{\mathrm{~b}}}{|\overline{\mathrm{~b}}|}$ then

In triangle ABC , the point P divides BC internally in the ratio $3: 4$ and Q divides CA internally in the ratio $5: 3$. If AP and BQ intersect in a point $G$, then $G$ divides $A P$ internally in the ratio