Division of Line Segment in Given Ratio - Class X
Coordinate geometry problems involving division of line segments in given ratios, including internal and external division, division by coordinate axes, and 3D coordinate applications.
Questions
ABC is a triangle, the point P is on side BC such that $3\bar{BP}=2\bar{PC}$, the point Q is on the line $\bar{CA}$ such that $4\bar{CQ}=\bar{QA}$. If R is the common point $\bar{AP}$ & $\bar{BQ}$, then the ratio in which the fine joining CR divides $\bar{AB}$ is?
- $2:5$
- $3:8$
- $4:1$
- $6:1$
The line joining points $(3,5)$ and $(2,7)$ is divided by $X-$ axis in the ratio.
- $5:7$
- $3:2$
- $-5:7$
- $-3:2$
The point $(\dfrac{7}{4},\dfrac{7}{8})$ divides the line segment joining the points (4,-1) and (-2,4) internally in the ratio 3 : 5.
- True
- False
The ratio in which the point (4, 7) divides the line segment joining (1, 4) and (11, 14) is
- 2 : 7
- 3 : 7
- 4 : 5
- 3 : 8
A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point $M$ of $CD$. The crease will divide $BC$ in the ratio :
- $7:4$
- $5:3$
- $8:5$
- $4:1$
Perpendicular from the origin to the line joining the points $(c , cos \alpha, c , sin \alpha)$ and $(c , cos \beta , , c , sin \beta)$ divides it in the ratio
- 2 : 1
- 1 : 2
- 1 : 1
- none of these
The join of $(4, 5)$ and $(1,2)$ is divided by y-axis in the ratio
- $-1:4$
- $-4:1$
- $-2:1$
- $-5:1$
The point which divides the line segment joining $(-2, 4), (2, 7)$ in the ratio $2:1$ externally is
- $(6, 10)$
- $(2, \dfrac{10}{3})$
- $(\dfrac{-4}{3}, \dfrac{2}{3})$
- $( \dfrac{2}{3} ,6)$
Find the points $A(a, b), B(-a, -b)$ and $P(a^2, ab)$ are collinear then the ratio in which p divides $\overline{AB}$ is
- 1 + a : 1 - a
- 1 : a
- a : 1
- 1 - a : 1 + a
The plane XOZ divides the join of (1, -1, 5) and (2, 3, 4) in the ratio $\lambda : 1$, then $\lambda$ is
- $-3$
- $\dfrac{-1}{3}$
- $3$
- $\dfrac{1}{3}$
In $\triangle ABC$ $PQR$ $\overline { BC } .\overline { CA } .\overline { AB } $ respectively dividing them in the ratio $1:4,3:2$ and $3:7$. The point $S$ divides $AB$ in the ratio $1:3$ Then $\dfrac { \left| \overline { AP } +\overline { BQ } +\overline { CR } \right| }{ \left| CS \right| } =$
- $\dfrac {1}{5}$
- $\dfrac {2}{5}$
- $\dfrac {5}{2}$
- $\dfrac {7}{10}$
A straight line through the origin O meets the parallel lines 4x+2y=9 and 2x+y+6=0 at point P and Q respectively. Then the point O divides the segment PQ in the ratio
- 1:2
- 3:4
- 2:1
- 4:3
The ratio in which the line segment joining the points $\left(3,-4\right)$ and $\left(-5,6\right)$ is divided by the $x-$ axis, is
- $2:3$
- $3:2$
- $6:4$
- $none\ of\ these$
The ratio in which the point $(x _{1} \sin^{2} \theta, y _{1} \cos^{2} \theta)$ divides the line joining $(x _{1}, 0)$ and $(0, y _{1})$ is -
- $\tan^{2} \theta : \cot^{2} \theta$
- $\cos \theta : \sin \theta $
- $\cos^{2} \theta : \sin^{2} \theta$
- $(1-\cos \theta) : (1-\sin \theta)$
A point which divides the joint of $(1,2)$ and $(3,4)$ externally in the ratio $1:1$
- Lies in the first quadrant
- Lies in the second quadrant
- Lies in third quadrat
- Cannot be found
If the ratio in which the line segment joining the points (6,4) and (x,-7) divided internally by y-axis is 6: 1, then x equals
- 2
- 3
- -1
- -2