Division of a Line Segment Questions in English

(N/A) line segment of length $7.6 \, cm$ can be divided in the ratio of $5: 8$ as follows:
$1.$ Draw a line segment $AB$ of $7.6 \, cm$ and draw a ray $AX$ making an acute angle with line segment $AB$.
$2.$ Locate $13 (= 5 + 8)$ points,$A_1, A_2, A_3, \dots, A_{13}$,on $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_{12}A_{13}$.
$3.$ Join $BA_{13}$.
$4.$ Through the point $A_5$,draw a line parallel to $BA_{13}$ (by making an angle equal to $\angle AA_{13}B$) at $A_5$ intersecting $AB$ at point $C$.
$C$ is the point dividing line segment $AB$ of $7.6 \, cm$ in the required ratio of $5: 8$.
The lengths of $AC$ and $CB$ can be measured. They are $2.9 \, cm$ and $4.7 \, cm$ respectively.
Justification:
By construction,we have $A_5C \parallel A_{13}B$. By applying the Basic Proportionality Theorem $(BPT)$ for the triangle $AA_{13}B$,we obtain:
$\frac{AC}{CB} = \frac{AA_5}{A_5A_{13}}$
From the construction,$AA_5$ contains $5$ equal divisions and $A_5A_{13}$ contains $8$ equal divisions.
Therefore,$\frac{AA_5}{A_5A_{13}} = \frac{5}{8}$.
Thus,$\frac{AC}{CB} = \frac{5}{8}$. This justifies the construction.

(A) To divide a line segment $AB$ in the ratio $p:q$,we draw a ray $AX$ making an acute angle with $AB$.
We then mark $p+q$ points on the ray $AX$ at equal distances,say $A_1, A_2, \dots, A_{p+q}$,such that $AA_1 = A_1A_2 = \dots = A_{p+q-1}A_{p+q}$.
We join the last point $A_{p+q}$ to $B$.
Then,we draw a line parallel to $A_{p+q}B$ through the point $A_p$ to intersect $AB$ at a point $P$.
This point $P$ divides the line segment $AB$ in the ratio $p:q$.
Therefore,the minimum number of points to be marked on the ray $AX$ is $p+q$.

(C) We know that,to divide a line segment $AB$ in the ratio $m:n,$ first draw a ray $AX$ which makes an acute angle $\angle BAX$,then mark $m+n$ points at equal distances.
Here,$m=5$ and $n=7$.
So,the minimum number of these points $= m+n = 5+7 = 12$.

(D) To divide a line segment in the ratio $m:n,$ we first draw a ray making an acute angle with the segment.
Then,we mark $m+n$ points at equal distances on the ray.
Here,the ratio is $4:7,$ so the total number of points required is $4+7 = 11$.
Therefore,we mark points $A_1, A_2, \dots, A_{11}$ on the ray $AX$.
The last point $A_{11}$ is joined to point $B$ to complete the construction.

(A) To divide a line segment $AB$ in the ratio $m: n$ (here $5: 6$),we follow these steps:
$1$. Draw a ray $AX$ making an acute angle $\angle BAX$ with $AB$.
$2$. Draw a ray $BY$ parallel to $AX$ by making $\angle ABY = \angle BAX$.
$3$. Locate $m$ points $(A_{1}, A_{2}, \ldots, A_{5})$ on $AX$ and $n$ points $(B_{1}, B_{2}, \ldots, B_{6})$ on $BY$ such that $AA_{1} = A_{1}A_{2} = \ldots = A_{4}A_{5}$ and $BB_{1} = B_{1}B_{2} = \ldots = B_{5}B_{6}$.
$4$. Join the point $A_{m}$ (which is $A_{5}$) to the point $B_{n}$ (which is $B_{6}$). The line segment $A_{5}B_{6}$ intersects $AB$ at point $C$,dividing $AB$ in the ratio $5: 6$.

(B) False.
The ratio $2 \sqrt{3} : 2 \sqrt{3}$ simplifies to $1 : 1$.
However,the standard geometrical construction method for dividing a line segment in a given ratio $m : n$ requires $m$ and $n$ to be positive integers.
Since $2 \sqrt{3}$ is an irrational number and not a positive integer,it is not possible to divide a line segment in this ratio using the standard ruler and compass construction method.

(A) True.
Given ratio $= \sqrt{3} : \frac{1}{\sqrt{3}}$.
To simplify this ratio,multiply both terms by $\sqrt{3}$:
$\sqrt{3} \times \sqrt{3} : \frac{1}{\sqrt{3}} \times \sqrt{3} = 3 : 1$.
Since $3 : 1$ is a ratio of two positive integers,it is always possible to divide a line segment in this ratio using a ruler and compass by constructing $3 + 1 = 4$ equal parts on an auxiliary ray.
Therefore,the statement is True.

(N/A) $1.$ Draw a line segment $AB = 7 \, cm$.
$2.$ Draw a ray $AX$ making an acute angle $\angle BAX$.
$3.$ Along $AX,$ mark $3 + 5 = 8$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7 = A_7A_8$.
$4.$ Join $A_8B$.
$5.$ From $A_3,$ draw $A_3P \parallel A_8B$ meeting $AB$ at $P$ (by making an angle equal to $\angle BA_8A$ at $A_3$).
Then,$P$ is the point on $AB$ which divides it in the ratio $3: 5.$
Thus,$AP: PB = 3: 5.$
Justification:
Let $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = \dots = A_7A_8 = x.$
In $\triangle ABA_8,$ we have $A_3P \parallel A_8B.$
By Basic Proportionality Theorem,$\frac{AP}{PB} = \frac{AA_3}{A_3A_8} = \frac{3x}{5x} = \frac{3}{5}.$
Hence,$AP: PB = 3: 5.$

(N/A) Data: $\overline{ AB }$ of length $5.9\,cm$ is given.
To construct: $\overline{ AB }$ is to be divided in the ratio $5: 7$ from $A$.
Steps of construction:
$(1)$ In different half-planes of $\overleftrightarrow{ AB }$ containing $\overline{ AB }$,draw $\overrightarrow{ AX }$ and $\overrightarrow{ BY }$ such that $\angle XAB$ and $\angle YBA$ are congruent acute angles.
$(2)$ With some appropriate radius and centre $A$,draw an arc to intersect $\overrightarrow{ AX }$ at $A_{1}$. Similarly,with centre $A_{1}$ and the same radius,draw an arc to intersect $\overrightarrow{ AX }$ at $A_{2}$ such that $A-A_{1}-A_{2}$. Similarly,draw an arc with the same radius and centre $A_{k}$ to intersect $\overrightarrow{ AX }$ at $A_{k+1}$ such that $A_{k-1}-A_{k}-A_{k+1}$,where $k=2, 3, 4$. Thus,we obtain five points $A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$ on $\overrightarrow{ AX }$ such that $AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}$.
$(3)$ Now,with the same radius and beginning with point $B$ as centre,obtain seven points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ and $B_{7}$ on $\overrightarrow{ BY }$ such that $BB_{1} = B_{1}B_{2} = \dots = B_{6}B_{7}$.
$(4)$ Draw $\overline{A_{5}B_{7}}$ to intersect $\overline{ AB }$ at $M$.
Thus,$M \in \overline{ AB }$ is the point which divides $\overline{ AB }$ in the ratio $5: 7$.
Justification: Here $\frac{AA_{5}}{BB_{7}} = \frac{5}{7}$. Since $\overrightarrow{AX} \parallel \overrightarrow{BY}$,the correspondence $AA_{5}M \leftrightarrow BB_{7}M$ between $\Delta AA_{5}M$ and $\Delta BB_{7}M$ is a similarity.
$\therefore \frac{AM}{BM} = \frac{AA_{5}}{BB_{7}} = \frac{5}{7}$.

(N/A) Data: $\overline{AB}$ is given.
To construct: $\overline{AB}$ is to be divided into three parts in the ratio $2: 3: 4$ from $A$.
Steps of construction:
$(1)$ In different half-planes of $\overleftrightarrow{AB}$ containing $\overline{AB}$,draw $\overrightarrow{AX}$ and $\overrightarrow{BY}$ such that $\angle XAB$ and $\angle YBA$ are congruent acute angles.
$(2)$ With some appropriate radius and centre $A$,draw an arc to intersect $\overrightarrow{AX}$ at $A_1$. Similarly,with centre $A_1$ and the same radius,draw an arc to intersect $\overrightarrow{AX}$ at $A_2$ such that $A-A_1-A_2$. Similarly,draw an arc with the same radius and centre $A_k$ to intersect $\overrightarrow{AX}$ at $A_{k+1}$ such that $A_{k-1}-A_k-A_{k+1}$,where $k=2, 3, 4, \dots, 8$. Thus,we obtain nine points $A_1, A_2, \dots, A_9$ on $\overrightarrow{AX}$ such that $AA_1 = A_1A_2 = \dots = A_8A_9$.
$(3)$ Now,with the same radius and beginning with point $B$ as centre,obtain nine points $B_1, B_2, \dots, B_9$ on $\overrightarrow{BY}$ such that $BB_1 = B_1B_2 = \dots = B_8B_9$.
$(4)$ Draw $\overline{AB_9}$,$\overline{A_2B_7}$,$\overline{A_5B_4}$,and $\overline{A_9B}$ so that $\overline{A_2B_7}$ intersects $\overline{AB}$ at $M$ and $\overline{A_5B_4}$ intersects $\overline{AB}$ at $N$.
Thus,we obtain points $M$ and $N$ which divide $\overline{AB}$ in the ratio $2: 3: 4$ from $A$,i.e.,$AM : MN : NB = 2: 3: 4$.
Justification: Here,the intercepts made on transversals $\overleftrightarrow{AX}$,$\overleftrightarrow{AB}$,and $\overleftrightarrow{BY}$ by $\overline{A_2B_7} \parallel \overline{A_5B_4}$ are proportionate.
Hence,$AM : MN : NB = AA_2 : A_2A_5 : A_5A_9 = B_9B_7 : B_7B_4 : B_4B = 2: 3: 4$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{PQ} = 8.2 \, cm$ using a ruler.
$2$. Draw an acute angle $\angle QPX$ at point $P$ below the line segment $\overline{PQ}$.
$3$. Starting from $P$,mark $3 + 7 = 10$ points $A_1, A_2, A_3, \dots, A_{10}$ on the ray $PX$ such that $PA_1 = A_1A_2 = A_2A_3 = \dots = A_9A_{10}$.
$4$. Join the point $A_{10}$ to $Q$.
$5$. Through point $A_3$,draw a line parallel to $A_{10}Q$ intersecting $\overline{PQ}$ at point $R$.
$6$. The point $R$ divides $\overline{PQ}$ in the ratio $3:7$ such that $PR:RQ = 3:7$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{AB}$ of length $10 \,cm$ using a ruler.
$2$. Draw any ray $AX$ making an acute angle with $\overline{AB}$.
$3$. Locate $3 + 8 = 11$ points $A_1, A_2, ..., A_{11}$ on ray $AX$ such that $AA_1 = A_1A_2 = ... = A_{10}A_{11}$.
$4$. Join $B$ to $A_{11}$.
$5$. Through the point $A_3$,draw a line parallel to $A_{11}B$ (by making an angle equal to $\angle AA_{11}B$) intersecting $\overline{AB}$ at point $P$.
$6$. Thus,$P$ is the point dividing $\overline{AB}$ in the ratio $3:8$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{XY}$ of length $9\, cm$ using a ruler.
$2$. Draw a ray $XP$ making an acute angle with $\overline{XY}$ at point $X$.
$3$. Locate $2 + 5 = 7$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7$ on ray $XP$ such that $XA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$.
$4$. Join $A_7$ to $Y$.
$5$. Through point $A_2$,draw a line parallel to $A_7Y$ (by making corresponding angles equal) intersecting $\overline{XY}$ at point $Z$.
$6$. Thus,point $Z$ divides $\overline{XY}$ in the ratio $2:5$.

(N/A) Steps of construction:
$1$. Draw a line segment $AB$ of length $7 \,cm$ using a ruler.
$2$. Draw a ray $AX$ making an acute angle with $AB$.
$3$. Locate $3 + 5 = 8$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8$ on ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7 = A_7A_8$.
$4$. Join the point $A_8$ to $B$.
$5$. Through the point $A_3$,draw a line parallel to $A_8B$ (by making corresponding angles equal) to intersect $AB$ at point $M$.
$6$. Thus,$M$ is the point on $AB$ which divides it in the ratio $3 : 5$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{XY} = 7.5\, cm$ using a ruler.
$2$. Draw an acute angle $\angle YXA$ at point $X$ below the segment $\overline{XY}$.
$3$. Starting from $X$,mark $3 + 4 + 5 = 12$ points $X_1, X_2, ..., X_{12}$ on the ray $XA$ such that $XX_1 = X_1X_2 = ... = X_{11}X_{12}$ using a compass.
$4$. Join the point $X_{12}$ to $Y$.
$5$. Through points $X_3$ and $X_7$ (since $3$ and $3+4=7$),draw lines parallel to $X_{12}Y$ intersecting $\overline{XY}$ at points $P$ and $Q$ respectively.
$6$. The points $P$ and $Q$ divide the segment $\overline{XY}$ in the ratio $3:4:5$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{PQ}$ of length $8.5 \text{ cm}$ using a ruler.
$2$. Draw any ray $PY$ making an acute angle with $\overline{PQ}$.
$3$. Locate $4 + 7 = 11$ points $A_1, A_2, \dots, A_{11}$ on ray $PY$ such that $PA_1 = A_1A_2 = A_2A_3 = \dots = A_{10}A_{11}$.
$4$. Join $A_{11}$ to $Q$.
$5$. Through the point $A_4$ (since the ratio is $4:7$),draw a line parallel to $A_{11}Q$ intersecting $\overline{PQ}$ at point $X$.
$6$. Thus,$X$ is the point on $\overline{PQ}$ which divides it in the ratio $4 : 7$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{ AB }$ of length $7 \,cm$ using a ruler.
$2$. Draw an acute angle $\angle BAX$ at point $A$ with the line segment $\overline{ AB }$.
$3$. Mark $5$ points $A_1, A_2, A_3, A_4, A_5$ on the ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$ using a compass.
$4$. Join $A_5$ to $B$.
$5$. Through point $A_2$,draw a line parallel to $A_5B$ intersecting $\overline{ AB }$ at point $M$.
$6$. Thus,$M$ is the required point on $\overline{ AB }$ such that $AM : AB = 2 : 5$.

(N/A) Step $1$: Draw a line segment $\overline{AB} = 8\, cm$ using a ruler.
Step $2$: Draw an acute angle $\angle BAX$ at point $A$ with the line segment $\overline{AB}$.
Step $3$: Mark $2 + 3 + 5 = 10$ points $A_1, A_2, A_3, \dots, A_{10}$ on the ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_9A_{10}$ using a compass.
Step $4$: Join the last point $A_{10}$ to point $B$.
Step $5$: Through points $A_2$ and $A_5$ (since $2$ and $2+3=5$),draw lines parallel to $A_{10}B$ intersecting $\overline{AB}$ at points $P$ and $Q$ respectively.
Step $6$: The points $P$ and $Q$ divide the line segment $\overline{AB}$ in the ratio $2:3:5$.

(A) point $P$ lying on the line segment $\overline{AB}$ divides the segment into two parts: $AP$ and $PB$.
By definition,the division ratio of a line segment $\overline{AB}$ by a point $P$ (where $P$ is between $A$ and $B$) is the ratio of the lengths of the two segments formed,which is $\frac{AP}{PB}$.
Thus,$P$ divides $\overline{AB}$ from $A$ in the ratio $\frac{AP}{PB}$.