Exercise-7.2, Class 10th, Maths, Chapter 7, NCERT

1. Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Using section formula (ratio $2:3$):

$$(\frac{2\cdot 4+3\cdot (-1)}{2+3},\frac{2\cdot (-3)+3\cdot 7}{2+3})=(\frac{8-3}{5},\frac{-6+21}{5})=(1,3)\mathrm{.}$$

Answer: $(1,3)$

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2. Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Points of trisection divide the segment into three equal parts, so use ratios $1:2$ and $2:1$

For ratio $1:2$

$$(\frac{1\cdot (-2)+2\cdot 4}{3},\frac{1\cdot (-3)+2\cdot (-1)}{3})=(\frac{-2+8}{3},\frac{-3-2}{3})=(2,-5\mathrm{/}3)\mathrm{.}$$

For ratio $2:1$

$$(\frac{2\cdot (-2)+1\cdot 4}{3},\frac{2\cdot (-3)+1\cdot (-1)}{3})=(\frac{-4+4}{3},\frac{-6-1}{3})=(0,-7\mathrm{/}3)\mathrm{.}$$

Answer: Points are $(2,-5\mathrm{/}3)$ and $(0,-7\mathrm{/}3)$ 

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3. Text summary: rectangular school ground ABCD has parallel chalk lines spaced 1 m apart; 100 flower pots are placed 1 m apart along AD. Niharika runs $\frac{1}{4}$ of AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$ of AD on the 8th line and posts a red flag. Find the distance between the two flags. If Rashmi must post a blue flag halfway along the segment joining the two flags, where should she post it?

• The 100 pots, 1 m apart, are placed along AD; thus the distance from the first pot to the last pot = $99$ m. I assume this means $AD=99$ m.

• Chalk lines are parallel to AD and are 1 m apart; the “2nd line” is 1 m from the 1st line, the “8th line” is 7 m from the 1st line, so the perpendicular (line-to-line) distance between the 2nd and 8th lines is $7-1=6$ m.

• Distances along each line are measured from the same end (say point $A$).

With these assumptions:

• Niharika’s horizontal position (along AD) = $\frac{1}{4}\cdot AD=\frac{1}{4}\cdot 99=24.75$ from $A$ (on the 2nd line, i.e. at perpendicular distance $=1$ from reference line).

• Preet’s horizontal position = $\frac{1}{5}\cdot AD=\frac{1}{5}\cdot 99=\mathrm{19.8 }$m from $A$ (on the 8th line, perpendicular distance = $7$ m from reference line).

• Horizontal difference $=24.75-19.8=4.95$ m. Vertical (perpendicular) difference $=7-1=6$ m.

Distance between flags:

$$\sqrt{(4.95{)}^2+6^2}=\sqrt{24.5025+36}=\sqrt{60.5025}\approx 7.774\text{ m}\mathrm{.}$$

Midpoint (where Rashmi should post the blue flag) — average coordinates (along AD, perp):
Horizontal: ${\displaystyle \frac{24.75+19.8}{2}}=22.275$ m from $A$. Perpendicular: ${\displaystyle \frac{1+7}{2}}=4$ m from reference line, i.e. on the 5th line (since lines are integer metre spacing, the 5th line is 4 m away).

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4. Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).
Let the ratio be $m:1$ (using the book’s section formula convention). Then coordinates:

$$(\frac{m\cdot 6+1\cdot (-3)}{m+1},\frac{m\cdot (-8)+1\cdot 10}{m+1})=(-1,6)\mathrm{}$$

From the x-coordinate:

$$\frac{6m-3}{m+1}=-1\Rightarrow 6m-3=-m-1\Rightarrow 7m=2\Rightarrow m=\frac{2}{7}\mathrm{.}$$

So the ratio $m:1={\displaystyle \frac{2}{7}}:1$ → multiply by 7 → $2:7$ (You can check the y-coordinate gives $6$ too.)

Answer: the point divides the segment internally in the ratio $2:7$

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5. Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let ratio be $m:1$ (A to B). A point on x-axis has y = 0. Using y-coordinate:

$$\frac{m\cdot 5+1\cdot (-5)}{m+1}=0\Rightarrow 5m-5=0\Rightarrow m=1.$$

So ratio $1:1$ (mid-point). Mid-point coordinates:

$$(\frac{1+(-4)}{2},\frac{-5+5}{2})=(-\frac{3}{2},0)\mathrm{.}$$

Answer: ratio 1:1 (midpoint); point is $\boxed{{\displaystyle (-\frac{3}{2},0)}}$

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6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Midpoint of diagonal joining first and third vertices equals midpoint of diagonal joining second and fourth. So midpoint of (1,2) and $(x,6)$ is $(\frac{1+x}{2},4)$. Midpoint of $(4,y)$ and (3,5) is $(\frac{7}{2},\frac{y+5}{2})$. Equate:

$$\frac{1+x}{2}=\frac{7}{2}\Rightarrow 1+x=7\Rightarrow x=6.$$
$$4=\frac{y+5}{2}\Rightarrow y+5=8\Rightarrow y=3.$$

Answer: $x=6,y=3$

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7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
Centre = midpoint of diameter AB. Let A = $(x,y)$. Then

$$\frac{x+1}{2}=2\Rightarrow x+1=4\Rightarrow x=3,$$
$$\frac{y+4}{2}=-3\Rightarrow y+4=-6\Rightarrow y=-10.$$

Answer: $A=(3,-10)$ 

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8. If A and B are (−2, −2) and (2, −4), respectively, find the coordinates of P such that $AP={\displaystyle \frac{3}{7}}AB$ and P lies on the line segment AB.
Vector $\overrightarrow{AB}=(4,-2)$. Then $\overrightarrow{AP}=\frac{3}{7}\overrightarrow{AB}=(\frac{12}{7},-\frac{6}{7})$. So

$$P=A+\overrightarrow{AP}=(-2+\frac{12}{7},-2-\frac{6}{7})=(-\frac{2}{7},-\frac{20}{7})\mathrm{}$$

Answer: $P=(-\frac{2}{7},-\frac{20}{7})$

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9. Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
Vector $\overrightarrow{AB}=(4,6)$. Points at $1\mathrm{/}4$, $2\mathrm{/}4$, $3\mathrm{/}4$ from A:

• $t=\frac{1}{4}:(-2,2)+\frac{1}{4}(4,6)=(-1,2+1.5)=(-1,7\mathrm{/}2)\mathrm{}$

• $t=\frac{1}{2}:(-2,2)+\frac{1}{2}(4,6)=(0,5)\mathrm{}$
  

• $t=\frac{3}{4}:(-2,2)+\frac{3}{4}(4,6)=(1,13\mathrm{/}2)\mathrm{}$
  

Answer: ${\displaystyle (-1,\frac{7}{2}),(0,5),(1,\frac{13}{2})\mathrm{}}$

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10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order.
For a rhombus, area $={\displaystyle \frac{1}{2}}$(product of diagonals). Diagonals are between opposite vertices:

• Diagonal 1: between (3,0) and (−1,4): length $d_1=\sqrt{(-1-3{)}^2+(4-0{)}^2}=\sqrt{16+16}=4\sqrt{2}\mathrm{.}$
  

• Diagonal 2: between (4,5) and (−2,−1): length $d_2=\sqrt{(-2-4{)}^2+(-1-5{)}^2}=\sqrt{36+36}=6\sqrt{2}\mathrm{.}$
  

Area $=\frac{1}{2}{d}_1{d}_2=\frac{1}{2}\cdot (4\sqrt{2})\cdot (6\sqrt{2})=\frac{1}{2}\cdot 48=24.$

Answer: Area $=24$ square units.