CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

Explore the NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium), with detailed explanations for Exercise 7.2. This resource is designed to simplify complex Coordinate Geometry, helping Class 10 Maths students grasp the subject effortlessly. For any queries regarding NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2, feel free to leave a comment, and we’ll respond as soon as possible. || Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)
Exercise - 7.2

CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

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

Answer

If $ \left(x_{1}, y_{1}\right) $ and $ \left(x_{2}, y_{2}\right) $ are points that are divided in ratio $ \mathrm{m}: \mathrm{n} $ then,
$ (x, y)=\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n} $
Let $ \mathrm{D}(x, y) $ be the required point.
Now,
Using the section formula, we get
$ x=\frac{2 \times 4+3 \times-1}{2+3}$
$x=\frac{8-3}{5}$
$x=1$
$y=\frac{2 \times-3+3 \times 7}{2+3}$
$y=\frac{-6+21}{5}$
$y=3 $
Therefore, the point is $ (1,3) $.

2. Find the coordinates of the points of trisection of the line segment joining $ (4,-1) $ and $ (-2,-3) $.

Answer

The points of trisection means that the points which divide the line in three equal parts. From the figure C , and D are these two points. Let C $ \left(x_{1}, y_{1}\right) $ and $ \mathrm{D}\left(x_{2}, y_{2}\right) $ are the points of trisection of the line segment joining the given points i.e., $ \mathrm{BC}=\mathrm{CD}=\mathrm{DA} $
Let $ \mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{k}$
$Point \mathrm{C} $ divides the BC and CA as
$ : \mathrm{BC}=\mathrm{kCA} $
$ =\mathrm{CD}+\mathrm{DA}=\mathrm{k}+\mathrm{k}=2 \mathrm{k} $
Hence ratio between BC and CA is:
$ \frac{B C}{C A}=\frac{k}{2 k}=\frac{1}{2} $
Therefore, point C divides BA internally in the ratio 1:2
then by section formula we have that if a point $ \mathrm{P}(x, y) $ divide two points $ \mathrm{P}\left(x_{1}, y_{l}\right) $ and $ \mathrm{Q}\left(x_{2}, y_{2}\right) $ in the ratio m:n then,
the point $ (x, y) $ is given by $ (x, y)=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right) $
Therefore $ \mathrm{C}(x, y) $ divides $ \mathrm{B}(-2,-3) $ and $ \mathrm{A}(4,-1) $ in the ratio $ 1: 2 $,
then $ C(x, y)=\frac{(1 \times 4)+(2 \times-2)}{1+2}, \frac{(1 \times-1)+(2 \times-3)}{1+2} $
$ C(x, y)=\frac{4-4}{1+2}, \frac{-1-6}{1+2} $
$ C(x, y)=0, \frac{-7}{3} $
Point D divides the BD and DA as: $ \mathrm{DA}=\mathrm{kBD}=\mathrm{BC}+\mathrm{CD} =\mathrm{k}+ $ $ \mathrm{k}=2 \mathrm{k} $
Hence ratio between BD and DA is:
$ \frac{B D}{D A}=\frac{2 k}{k}=\frac{2}{1} $
The point D divides the line BA in the ratio $ 2: 1 $
So now applying section formula again we get
$ D(x, y)=\frac{(2 \times 4)+(1 \times-2)}{2+1}, \frac{(2 \times-1)+(1 \times-3)}{2+1}$
$D(x, y)=\frac{8-2}{3}, \frac{-2-3}{3}$
$D(x, y)=\frac{6}{3}, \frac{-5}{3}$
$D(x, y)=2, \frac{-5}{3} $

CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

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3. To conduct Sports Day activities, in your rectangular shaped school ground $ A B C D $, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Fig. 7.12. Niharika runs $ \frac{1}{4} $ th thedistance AD on the 2 nd line and posts a green flag. Preet runs $ \frac{1}{5} $ the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Answer

It can be observed that Niharika posted the green flag at $ \frac{1}{4} $ of the distance AD i.e., $ \frac{1}{4} \times 100=25 \mathrm{~m} $ from the starting point of 2 nd line.
Therefore, the coordinates of this point G is $ (2,25) $
Similarly, Preet posted red flag at $ \frac{1}{5} $ of the distance AD i.e., $ \frac{1}{5} \times 100=20 \mathrm{~m} $ from the starting point of 8 th line.
Therefore, the coordinates of this point $ R $ are $ (8,20) $
Now we have the positions of posts by Preet and Niharika According to distance formula, distance between points $ \mathrm{A}\left(x_{1}, y_{1}\right) $ and $ B\left(x_{2}, y_{2}\right) $ is given by
$ D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} $
Distance between these flags by using distance formula, D
$ =\left[(8-2)^{2}+(25-20)^{2}\right]^{\frac{1}{2}}$
$=(36+25)^{\frac{1}{2}}$
$=\sqrt{61} \mathrm{~m} $
The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be $ \mathrm{A}(x, y) $
Now by midpoint formula,
$ (x, y)=\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}$
$x=\frac{2+8}{2}=5$
$y=\frac{25+20}{2}=22.5 $
Hence, $ \mathrm{A}(x, y)=(5,22.5) $
Therefore, Rashmi should post her blue flag at 22.5 m on 5 th line.

4. Find the ratio in which the line segment joining the points $(-3 , 10) $ and $ (6,-8) $ is divided by $ (-1,6) $.

Answer

Let the ratio in which the line segment joining $ (-3,10) $ and $ (6,-8) $ is divided by point $ (-1,6) $ be $ k: 1 $
Using section formula
i.e. the coordinates of the points $ \mathrm{P}(x, y) $ which divides the line segment joining the points $ A\left(x_{1}, y_{1}\right) $ and $ B\left(x_{2}, y_{2}\right) $, internally in the ratio $ \mathrm{m}: \mathrm{n} $ are
$ \left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$
$(-1,6)=\left(\frac{k(6)+1(-3)}{k+1}, \frac{k(-8)+1(10)}{k+1}\right) $
Therefore,
$ -1=\frac{6 k-3}{k+1}$
$\Rightarrow-\mathrm{k}-1=6 \mathrm{k}-3$
$\Rightarrow-\mathrm{k}-6 \mathrm{k}=-3+1$
$\Rightarrow-7 \mathrm{k}=-2$
$\Rightarrow 7 \mathrm{k}=2$
$\Rightarrow k=\frac{2}{7} $
Therefore, the required ratio is $ 2: 7 $.

CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

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5. Find the ratio in which the line segment joining $ A(1,-5) $ and $ \mathrm{B}(-4,5) $ is divided by the $x$ -axis. Also, find the coordinates of the point of division.

Answer

The diagram for the question is

Let the ratio in which the line segment joining $ \mathrm{A}(1,-5) $
and $ \mathrm{B}(-4 $, 5 ) is divided by $ x $-axis be k : 1
Therefore, the coordinates of the point of division is $ \left(\frac{-4 k+1}{k+1}, \frac{5 k-5}{k+1}\right) $
We know that $ y $-coordinate of any point on $ x $-axis is 0
Therefore,
$ \frac{5 k-5}{k+1}=0$
$\mathrm{k}=1 $
Therefore, $ x $-axis divides it in the ratio 1:1.
Division point $ =\left(\frac{-4(1)+1}{1+1}, \frac{5(1)-5}{1+1}\right) $
$ =\left(\frac{-4+1}{2}, \frac{5-5}{2}\right) $
$ =\left(\frac{-3}{0}, 0\right) $

6. If $ (1,2),(4, y),(x, 6) $ and $ (3,5) $ are the vertices of a parallelogram taken in order, find $ x $ and $ y $.

Answer

Let $ (1,2),(4, y),(x, 6) $, and $ (3,5) $ are the coordinates of A, $ B, C, D $ vertices of a parallelogram $ A B C D $.
Intersection point O of diagonal AC and BD also divides these diagonals.
Therefore, O is the mid-point of AC and BD .

By mid point formula, If $ (x, y) $ is the midpoint of the line joining points $ \mathrm{A}\left(x_{1}, y_{1}\right) $ and $ \mathrm{B}\left(x_{2}, y_{2}\right) $ Then,
$ (x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) $
If $ O $ is the mid-point of $ A C $, then the coordinates of $ O $ are:
$ \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{x+1}{2}, 4\right) $
If $ O $ is the mid-point of $ B D $, then the coordinates of $ O $ are:
$ \left(\frac{4+3}{2}, \frac{5+y}{2}\right)=\left(\frac{7}{2}, \frac{5+y}{2}\right) $
Since both the coordinates are of the same point $ O $
Comparing the $x$ coordinates we get,
$ \frac{x+1}{2}=\frac{7}{2}$
$x+1=7$
$x=6 $
And, Comparing the y coordinates we get,
$ \frac{5+y}{2}=4$
$5+y=8$
$y=3 $
Hence, $ x=6 $ and $ y=3 $.

CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

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7. Find the coordinates of a point $ A $, where $ A B $ is the diameter of a circle whose centre is $ (2,-3) $ and $ B $ is $ (1,4) $

Answer

For points $ \mathrm{A}\left(x_{1}, y_{1}\right) $ and $ \mathrm{B}\left(x_{2}, y_{2}\right) $, the midpoints $ (x, y) $ is given by
$ (x, y)=\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} $
Let the coordinates of point A be $ (x, y) $
Mid-point of AB is $ (2,-3) $, which is the center of circle and point B is $ (1,4) $
Therefore,
$ (2,-3)=\left(\frac{x+1}{2}, \frac{y+4}{2}\right) $
Equating the coordinates we get,
$ \frac{x+1}{2}=2, \frac{y+4}{2}=-3$
$x=4-1, y=-6-4$
$x=3, y=-10 $
Therefore point is $ \mathrm{A}(3,-10) $
Therefore, the coordinates.

8. If $ A $ and $ B $ are $ (-2,-2) $ and $ (2,-4) $, respectively, find the coordinates of P such that $ A P=\frac{3}{7} A B $ and P lies on the line segment AB

Answer

To find: The coordinates of P
Given: Points $ \mathrm{A}(-2,-2) $ and $ \mathrm{B}(2,-4) $ and ratio $ \mathrm{AP}: \mathrm{AB}=3: 7 $
The coordinates of point A and B are $ (-2,-2) $ and $ (2,-4) $ respectively
$ \mathrm{AP}=\frac{3}{7} A B $

Therefore, $ \mathrm{AP}: \mathrm{PB}=3: 4 $
Point P divides the line segment AB in the ratio 3:4
By section formula,
If a point divides the point $ \left(x_{1}, y_{1}\right) $ and $ \left(x_{2}, y_{2}\right) $ in the ratio m:n
Then, $ (X, Y)=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right) $
$ (X, Y)=\left(\frac{3 \times 2+4 \times(-2)}{3+4}, \frac{3 \times(-4)+4 \times(-2)}{3+4}\right)$
$(X, Y)=\left(\frac{-2}{7}, \frac{-20}{7}\right) $
$ (\frac{-2}{7},\frac{-20}{7}) $ is the point which divides line in the ratio of $ 3: 4 $.

9. Find the coordinates of the points which divide the line segment joining $ \mathrm{A}(-2,2) $ and $ \mathrm{B}(2,8) $ into four equal parts.

Answer

It can be observed from the figure that points P, Q, R are dividing the line segment in a ratio $ 1: 3,1: 1,3: 1 $ respectively

Coordinate of $ X=\left(\frac{1 * 2+3 *-2}{1+3}, \frac{1 * 8+3 * 2}{1+3}\right) $
$ =\left(-1, \frac{7}{2}\right) $
Coordinate of $ Y=\left(\frac{2 \pm 2}{2}, \frac{2+8}{2}\right) $
$ =(0,5) $
Coordinates of $ Z=\left(\frac{3 * 2+1 *-2}{3+1}, \frac{3 * 8+1 * 2}{3+1}\right) $
$ =\left(1, \frac{13}{2}\right) $

10. Find the area of a rhombus if its vertices are $ (3,0),(4,5),(-1 $, $ 4) $ and $ (-2,-1) $ taken in order.

Answer

Let $ (3,0),(4,5),(-1,4) $ and $ (-2,-1) $ are the vertices A, B, $ \mathrm{C}, \mathrm{D} $ of a rhombus ABCD

Length of diagonal $ \mathrm{AC}=\left[(3+1)^{2}+(0-4)^{2}\right]^{\frac{1}{2}} $
$ =\sqrt{16+16}$
$=4 \sqrt{2} $
Length of diagonal $ \mathrm{BD}=\left[(4+2)^{2}+(5+1)^{2}\right]^{\frac{1}{2}} $
$ =\sqrt{36+36}$
$=6 \sqrt{2}$
$\text { Therefore, } $
Area of rhombus $ =\frac{1}{2} $ (Product of the length of the diagonals) Therefore,
Area of rhombus $ \mathrm{ABCD}=\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}=\frac{1}{2} \times 48 $
$ =24 $ square units

CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

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CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 || NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (English Medium)

Download the Math Ninja App Now