NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry, created by experts at Careers360, provide comprehensive study material for students preparing for the CBSE Class 10 board exam. Class 10 maths ncert solutions chapter 7 cover all exercises in the NCERT textbook and are easily accessible for download. The step-by-step answers for various types of questions in the NCERT textbook can help students achieve mastery of the topics in the coordinate geometry chapter. The Class 10 maths chapter 7 NCERT solutions can be of great help for the students. If students want NCERT solutions for other classes and subjects, they can get the solutions by clicking on the above link. Students must refer to the NCERT Class 10 Maths books and read all the topics without fail.

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Coordinate Geometry Class 10 NCERT Solutions
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry FDF Free Download
Coordinate Geometry Class 10 - Important Formulae
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Intext Questions and Exercise)
Summary Of Class 10 maths NCERT solutions chapter 7
Key Features of class 10 maths ncert solutions chapter 7
NCERT Solutions for Class 10 Maths - Chapter Wise
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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry FDF Free Download

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Coordinate Geometry Class 10 - Important Formulae

Distance Formulae:

For a line defined by two points A(x₁, y₁) and B(x₂, y₂), the distance between these points can be calculated using the formula:
Distance AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

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Section Formula:

When a point P divides a line AB, with A(x₁, y₁) and B(x₂, y₂) as endpoints, in a ratio of m:n, the coordinates of point P can be found using the formula:
Point P = {(mx₂ + nx₁)/(m + n) ,(my₂ + ny₁)/(m + n)}

Midpoint Formula:

The midpoint of a line AB, defined by A(x₁, y₁) and B(x₂, y₂), can be determined using the following formula:
Midpoint P = {(x₁+ x₂)/2, (y₁+ y₂)/2}

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Area of a Triangle:

Consider a triangle formed by points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). The area of this triangle can be calculated using the formula:

Area ∆ABC = (1/2)|x₁(y₂ − y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|

Free download NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Intext Questions and Exercise)

Class 10 maths ch 7 solutions Exercise: 7.1

Q1 (i) Find the distance between the following pairs of points : (2, 3), (4, 1)

Answer:

Given points: (2, 3), (4, 1)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$D= \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$

Q1 (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)

Answer:

Given points: (– 5, 7), (– 1, 3)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$D= \sqrt{(-1+5)^2+(3-7)^2} = \sqrt{16+16} = 4\sqrt{2}$

Q1 (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)

Answer:

Given points: (a, b), (– a, – b)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$D= \sqrt{(-a-a)^2+(-b-b)^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2+b^2}$

Q2 Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Answer:

Given points: (0, 0) and (36, 15)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$D= \sqrt{(36-0)^2+(15-0)^2} =\sqrt{1296+225} = \sqrt{1521} = 39$

The distance between the two towns A and B is, thus 39 km for given towns location

$(0,0)$ and $(36,15)$ .

Q3 Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Answer:

Let the points (1, 5), (2, 3) and (– 2, – 11) be representing the vertices A, B, and C of the given triangle respectively.

$A = (1,5),\ B = (2,3),\ C = (-2,-11)$

Therefore,

$AB = \sqrt{(1-2)^2+(5-3)^2} = \sqrt{5}$

$BC = \sqrt{(2-(-2))^2+(3-(-11))^2} = \sqrt{4^2+14^2} = \sqrt{16+196} = \sqrt{212}$ $CA = \sqrt{(1-(-2))^2+(5-(-11))^2} = \sqrt{3^2+16^2} = \sqrt{9+256} = \sqrt{265}$ Since these are not satisfied.

$AB+BC \neq CA$

$BA+AC \neq BC$

$BC+CA \neq BA$

As these cases are not satisfied.

Hence the points are not collinear.

Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Answer:

The distance between two points $A(x_{1},y_{1})\ and\ B(x_{2},y_{2})$ is given by:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangle A, B, and C respectively.

$AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}$

$BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}$

$CA = \sqrt{(5-7)^2+(-2+2)^2} = \sqrt{4+0} = 2$

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

Q5 In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

1638427088941

Answer:

The coordinates of the points:

$A(3,4),\ B(6,7),\ C(9,4),\ and\ D(6,1)$ are the positions of 4 friends.

The distance between two points $A(x_{1},y_{1}), \ and\ B(x_{2},y_{2})$ is given by:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Hence,

$AB = \sqrt{(3-6)^2+(4-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$BC = \sqrt{(6-9)^2+(7-4)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$CD = \sqrt{(9-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$AD= \sqrt{(3-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

And the lengths of diagonals:

$AC = \sqrt{(3-9)^2+(4-4)^2} =\sqrt{36+0} = 6$

$BD = \sqrt{(6-6)^2+(7-1)^2} =\sqrt{36+0} = 6$

So, here it can be seen that all sides of quadrilateral ABCD are of the same lengths and diagonals are also having the same length.

Therefore, quadrilateral ABCD is a square and Champa is saying right.

Q6 (i) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

Answer:

Let the given points $(-1,-2),\ (1,0),\ (-1,2),\ and\ (-3,0)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$AB= \sqrt{(-1-1)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$BC= \sqrt{(1+1)^2+(0-2)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$CD= \sqrt{(-1+3)^2+(2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$AD= \sqrt{(-1+3)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

Finding the length of the diagonals:

$AC= \sqrt{(-1+1)^2+(-2-2)^2} =\sqrt{0+16} = 4$

$BD= \sqrt{(1+3)^2+(0-0)^2} =\sqrt{16+0} = 4$

It is clear that all sides are of the same lengths and also the diagonals have the same lengths.

Hence, the given quadrilateral is a square.

Q6 (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)

Answer:

Let the given points $(-3,5),\ (3,1),\ (0,3),\ and\ (-1,-4)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$AB= \sqrt{(-3-3)^2+(5-1)^2} =\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

$BC= \sqrt{(3-0)^2+(1-3)^2} =\sqrt{9+4} = \sqrt{13}$

$CD= \sqrt{(0+1)^2+(3+4)^2} =\sqrt{1+49} = \sqrt{50} = 5\sqrt2$

$AD= \sqrt{(-3+1)^2+(5+4)^2} =\sqrt{4+81} = \sqrt{85}$

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like square, rectangle, etc.

Q6 (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

Let the given points $(4,5),\ (7,6),\ (4,3),\ (1,2)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$AB= \sqrt{(4-7)^2+(5-6)^2} =\sqrt{9+1} = \sqrt{10}$

$BC= \sqrt{(7-4)^2+(6-3)^2} =\sqrt{9+9} = \sqrt{18}$

$CD= \sqrt{(4-1)^2+(3-2)^2} =\sqrt{9+1} = \sqrt{10}$

$AD= \sqrt{(4-1)^2+(5-2)^2} =\sqrt{9+9} = \sqrt{18}$

And the diagonals:

$AC =\sqrt{(4-4)^2+(5-3)^2} = \sqrt{0+4} = 2$

$BD =\sqrt{(7-1)^2+(6-2)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

Here we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

Q7 Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Answer:

Let the point which is equidistant from $A(2,-5)\ and \ B(-2,9)$ be $X(x,0)$ as it lies on X-axis.

Then, we have

Distance AX: $= \sqrt{(x-2)^2+(0+5)^2}$

and Distance BX $= \sqrt{(x+2)^2+(0+9)^2}$

According to the question, these distances are equal length.

Hence we have,

$\sqrt{(x-2)^2+(0+5)^2}$ $= \sqrt{(x+2)^2+(0+9)^2}$

Solving this to get the required coordinates.

Squaring both sides we get,

$(x-2)^2+25 = (x+2)^2+81$

$\Rightarrow (x-2+x+2)(x-2-x-2)= 81 - 25 = 56$

$\Rightarrow (-8x)= 56$

Or, $\Rightarrow x = -7$

Hence the point is $X(-7,0)$ .

Q8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Answer:

Given the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.

The distance formula :

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

So, given $PQ = 10\ units$

$PQ= \sqrt{(10-2)^2+(y-(-3))^2} = 10$

After squaring both sides

$\Rightarrow (10-2)^2+(y-(-3))^2 = 100$

$\Rightarrow (y+3)^2 = 100 - 64$

$\Rightarrow y+3 = \pm 6$

$\Rightarrow y = 6 - 3\ or\ y = -6-3$

Therefore, the values are $y = 3\ or -9$ .

Q9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also, find the distances QR and PR.

Answer:

Given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$ .

Then, the distances $PQ = RQ$ .

Distance $PQ = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}$

Distance $RQ = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}$

$\Rightarrow \sqrt{x^2+25} = \sqrt{41}$

Squaring both sides, we get

$\Rightarrow x^2 = 16$

$\Rightarrow x = \pm 4$

The points are: $R(4,6)\ or\ R(-4,6.)$

CASE I: when R is $(4,6)$

The distances QR and PR.

$QR = \sqrt{(0-4)^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$

$PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1^2+(-9)^2} = \sqrt{1+81} = \sqrt{82}$

CASE II: when R is $(-4,6)$

The distances QR and PR.

$QR = \sqrt{(0-(-4))^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$

$PR = \sqrt{(5-(-4))^2+(-3-6)^2} = \sqrt{9^2+(-9)^2} = \sqrt{81+81} = 9\sqrt{2}$

Q10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Answer:

Let the point $P(x,y )$ is equidistant from $A(3,6)$ and $B(-3,4)$ .

Then, the distances $AP =BP$

$AP = \sqrt{(x-3)^2+(y-6)^2}$ and $BP = \sqrt{(x-(-3))^2+(y-4)^2}$

$\Rightarrow \sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x-(-3))^2+(y-4)^2}$

Squaring both sides: we obtain

$\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2$

$\Rightarrow (2x)(-6)+(2y-10)(-2)= 0$ $\left [\because a^2-b^2 = (a+b)(a-b) \right ]$

$\Rightarrow -12x-4y+20 = 0$

$\Rightarrow 3x+y-5 = 0$

Thus, the relation is $3x+y-5 = 0$ between x and y.

Class 10 maths ch 7 solutions Exercise: 7.2

Q1 Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Answer:

Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$ , internally, in the ratio $m_{1}:m_{2}$ then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Substituting the values in the formula:

Here, $m_{1}:m_{2} = 2:3$

$\Rightarrow \left (\frac{2(4)+3(-1)}{2+3} , \frac{2(-3)+3(7)}{2+3} \right )$

$\Rightarrow \left (\frac{5}{5} , \frac{15}{5} \right )$

Hence the coordinate is $P \left (1 , 3 \right )$ .

Q2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Answer:

Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$

Then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

By observation point, P divides AB internally in the ratio $1:2$ .

Hence, $m:n = 1:2$

Substituting the values in the equation we get;

$\Rightarrow P\left (\frac{1(-2)+2(4)}{1+2} , \frac{1(-3)+2(-1)}{1+2} \right )$

$\Rightarrow P \left (\frac{-2+8}{3} , \frac{-3-2}{3} \right )$

$\Rightarrow P \left (2 , \frac{-5}{3} \right )$

And by observation point Q, divides AB internally in the ratio $2:1$

Hence, $m:n = 2:1$

Substituting the values in the equation above, we get

$\Rightarrow Q\left (\frac{2(-2)+1(4)}{2+1} , \frac{2(-3)+1(-1)}{2+1} \right )$

$\Rightarrow Q \left (\frac{-4+4}{3} , \frac{-6-1}{3} \right )$

$\Rightarrow Q\left (0 , \frac{-7}{3} \right )$

Hence, the points of trisections are $P \left (2 , \frac{-5}{3} \right )$ and $Q\left (0 , \frac{-7}{3} \right )$

Q3 To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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

1638427152215

Answer:

Niharika posted the green flag at the distance P, i.e.,

$\frac{1}{4}\times100\ m = 25\ m$ from the starting point of $2^{nd}$ line.

Therefore, the coordinates of this point $P$ are $(2,25).$

Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,

$\frac{1}{5}\times100\ m = 20\ m$ from the starting point of $8^{th}$ line.

Therefore, the coordinates of this point Q are $(8,20)$ .

The distance $PQ$ is given by,

$PQ = \sqrt{(8-2)^2+(25-20)^2} = \sqrt{36+25} = \sqrt{61} m$

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$ .

Then, by Section Formula,

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$x = \frac{2+8}{2},\ y = \frac{25+20}{2}$

$x = 5,\ y = 22.5$

Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.

Q4 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 be : $k:1$

Then, By section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

Given point $P(x,y) = (-1,6)$

$-1 = \frac{6k-3}{k+1}$

$\Rightarrow -k-1 = 6k-3$

$\Rightarrow k = \frac{7}{2}$

Hence, the point $P$ divides the line AB in the ratio $2:7$ .

Q5 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.

Answer:

Let the point on the x-axis be $P(x,0)$ and it divides it in the ratio $k:1$ .

Then, we have

Section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

$\implies \frac{ky_{2}+y_{1}}{k+1} = 0$

$k =-\frac{y_{1}}{y_{2}}$

Hence, the value of k will be: $k =-\frac{-5}{5}= 1$

Therefore, the x-axis divides the line in the ratio $1:1$ and the point will be,

Putting the value of $k=1$ in section formula.

$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$

$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$

Q6 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 the given points $A(1,2),\ B(4,y),\ C(x,6),\ D(3,5)$ .

Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

The coordinates of the point O when it is mid-point of AC.

$\left ( \frac{1+x}{2}, \frac{2+6}{2} \right ) \Rightarrow \left ( \frac{x+1}{2}, 4 \right )$

The coordinates of the point O when it is mid-point of BD.

$\left ( \frac{4+3}{2}, \frac{5+y}{2} \right ) \Rightarrow \left ( \frac{7}{2}, \frac{5+y}{2} \right )$

Since both coordinates are of same point O.

Therefore,

$\frac{x+1}{2} =\frac{7}{2}$ and $4 = \frac{5+y}{2}$

Or,

$x = 6\ and\ y = 3$

Q7 Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Answer:

As the centre point $C(2,-3)$ will be the mid-point of the diameter AB.

Then, the coordinates of point A will be $A(x,y)$ .

Given point $B(1,4)$ .

Therefore,

$(2,-3) = \left ( \frac{x+1}{2}, \frac{y+4}{2} \right )$

$\frac{x+1}{2} = 2\ and\ \frac{y+4}{2} = -3$

$\Rightarrow x = 3\ and\ y = -10$ .

Therefore, the coordinates of A are $(3,-10).$

Q8 If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Answer:

From the figure:

1638427203509

As $AP = \frac{3}{7}AB$

$\Rightarrow PB = \frac{4}{7}AB$ hence the ratio is 3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$P(x,y)= \left (\frac{3(2)+4(-2)}{3+4} , \frac{3(-4)+4(-2)}{3+4} \right )$

$P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$

$P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$

Q9 Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure:

1638427222556

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Point C, D, and E.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Here point D divides the line segment AB into two equal parts hence

$D(x_{2},y_{2})= \left (\frac{-2+2}{2} , \frac{2+8}{2} \right )$

$D(x_{2},y_{2})= \left (0 , 5 \right )$

Now, point C divides the line segment AD into two equal parts hence

$C(x_{1},y_{1})= \left (\frac{-2+0}{2} , \frac{2+5}{2} \right )$

$C(x_{2},y_{2})= \left (-1 , \frac{7}{2} \right )$

Also, point E divides the line segment DB into two equal parts hence

$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$

$E(x_{2},y_{2})= \left (1 , \frac{13}{2} \right )$

Q10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer:

From the figure:

1638427243029

Let the vertices of the rhombus are:

$A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)$

Area of the rhombus ABCD is given by;

$= \frac{1}{2}\times(Product\ of\ lengths\ of\ diagonals)$

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of the diagonal AC:

$AC = \sqrt{(3-(-1))^2+(0-4)^2} = \sqrt{16+16} = 4\sqrt{2}$

Length of the diagonal BD:

$BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}$

Thus, the area will be,

$= \frac{1}{2}\times (AC)\times(BD)$

$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24\ square\ units.$

Class 10 maths ch 7 solutions Exercise: 7.3

Q 1 (i) Find the area of the triangle whose vertices are : (2, 3), (–1, 0), (2, – 4)

Answer:

As we know, the area of a triangle with vertices (x1,x2) ,(y1, y2) and (z1 z2 ) is given by :

$A= \frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$

So Area of a triangle whose vertices are(2, 3), (–1, 0)and (2, – 4) is

$A= \frac{1}{2}[2(0-(-4))+(-1)(-4-3)+2(3-0)]$

$A= \frac{1}{2}[8+7+6]$

$A= \frac{1}{2}[21]$

$A= \frac{21}{2}$

$A= 10.5\:unit^2$

Hence, the area of the triangle is 10.5 per unit square.

Q1 (ii) Find the area of the triangle whose vertices are (–5, –1), (3, –5), (5, 2)

Answer:

From the figure:

1638427297678

Area of the triangle is given by:

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

Substituting the values in the above equation, we obtain

$Area = \frac{1}{2}\left [ (-5)((-5)-(-2))+3(2-(1))+5(-1-(-5)) \right ]$

$= \frac{1}{2}\left [ 35+9+20 \right ] = 32 \ square\ units.$

Q2 (i) In each of the following find the value of ‘k’, for which the points are collinear. (7, –2), (5, 1), (3, k)

Answer:

The points (7, –2), (5, 1), (3, k) are collinear if the area of the triangle formed by the points will be zero.

Area of the triangle is given by:

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0$

Substituting the values in the above equation, we obtain

$\frac{1}{2}\left [ 7(1-k)+5(k-(-2))+3(-2-1) \right ] = 0$

$\left [ 7-7k+5k+10-9 \right ] = 0$

$\Rightarrow -2k+8 = 0$

$\Rightarrow k = 4$

Hence, the points are collinear for k=4 .

Q2 (ii) In each of the following find the value of ‘k’, for which the points are collinear. (8, 1), (k, – 4), (2, –5)

Answer:

The points (8,1), (k, -4), (2,-5) are collinear if the area of the triangle formed by these points will be zero.

Area of the triangle is given by:

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0$

Substituting the values in the above equation, we obtain

$\frac{1}{2}\left [ 8(-4-(-5))+k((-5)-1)+2(1-(-4)) \right ] = 0$

$\Rightarrow 8-6k+10 = 0$

$\Rightarrow 6k = 18$

$\Rightarrow k = 3$

Hence, the points are collinear for k = 3 .

Q3 Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer:

From the figure:

1638427329430

The coordinates of the point P, Q, and R are:

Point P is the midpoint of side AB, hence the coordinates of P are :

$P(x_{1},y_{1}) = \left (\frac{0+2}{2}, \frac{3+1}{2} \right ) = \left (1, 2 \right )$

Point Q is the midpoint of side AC, hence the coordinates of Q are :

$Q(x_{2},y_{2}) = \left (\frac{2+0}{2}, \frac{1-1}{2} \right ) = \left (1, 0 \right )$

Point R is the midpoint of side BC, hence the coordinates of R are :

$R(x_{3},y_{3}) = \left (\frac{0+0}{2}, \frac{-1+3}{2} \right ) = \left (0, 1 \right )$

Hence, the area of the triangle formed by the midpoints PQR will be,

$Area_{(PQR)} = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$= \frac{1}{2}\left [ (2-1)+1(1-0)+0(0-2) \right ]$

$=\frac{1}{2}(1+1) = 1\ square\ units.$

And the area formed by the triangle ABC will be:

$Area_{(ABC)} = \frac{1}{2}\left [ 0(1-3)+2(3-(-1))+0(-1-1) \right ]$

$= \frac{1}{2}\left [ 8 \right ] = 4\ square\ units.$

Thus, the ratio of Area of $\triangle PQR$ to the Area of $\triangle ABC$ will be $1:4$ .

Q4 Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Answer:

From the figure:

1638427363305

The coordinates are $A(-4,-2),\ B(-3,-5),\ C(3,-2)\ and\ D(2,3)$

Divide the quadrilateral into 2 parts of triangles.

Then the area will be, $ABC + ADC$

Area of the triangle formed by ABC will be,

$Area_{(ABC)} = \frac{1}{2}\left [ (-4)((-5)-(-2))+(-3)((-2)-(-2))+3((-2)-(-2)) \right ]$ $= \frac{1}{2}\left [ 12+0+9 \right ] = \frac{21}{2}\ Square\ units.$

Area of the triangle formed by ADC will be,

$Area_{(ADC)} = \frac{1}{2}\left [ (-4)((-2)-(-3))+3(3-(-2))+2((-2)-(-2)) \right ]$ $= \frac{1}{2}\left [ 20+15+0 \right ] = \frac{35}{2}\ Square\ units.$

Therefore, the area of the quadrilateral will be:

$= \frac{21}{2}+\frac{35}{2} = 28\ square\ units.$

Alternatively,

The points A and C are in the same ordinates.

Hence, the length of base AC will be $(3-(-4)) = 7\ units.$

Therefore,

Area of triangle ABC:

$= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(3)$

Area of triangle ADC:

$= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(5)$

Therefore, the area will be, $\frac{1}{2}\times(7)\times(5+3) =28\ square\ units.$

Q5 You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for D ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

Answer:

From the figure:

1638427386798

The coordinates of midpoint M of side BC is:

$M = \left ( \frac{3+5}{2}, \frac{-2+2}{2} \right ) = \left ( 4,0 \right )$

Now, calculating the areas of the triangle ABM and ACM :

Area of triangle, ABM:

$Area_{(ABM)} = \frac{1}{2}\left [ 4((-2)-0)+3(0-(-6))+4((-6)-(-2)) \right ]$

$= \frac{1}{2}\left [ -8+18-16 \right ] = 3\ Square\ units.$

Area of triangle, ACM:

$Area_{(ACM)} = \frac{1}{2}\left [ 4(0-(-2))+4(2-(-6))+5((-6)-0) \right ]$

$= \frac{1}{2}\left [ -8+32-30 \right ] = -3\ Square\ units.$

However, the area cannot be negative, Therefore, area of $\triangle ACM$ is 3 square units.

Clearly, the median AM divided the $\triangle ABC$ in two equal areas.

Chapter 7 maths class 10 ncert solutions Exercise: 7.4

Q1 Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Answer:

Let the line divide the line segment AB in the ratio $k:1$ at point C.

Then, the coordinates of point C will be:

$C(x,y) = \left ( \frac{3k+2}{k+1},\frac{7k-2}{k+1} \right )$

Point C will also satisfy the given line equation $2x + y - 4 = 0$ , hence we have

$\Rightarrow 2\left ( \frac{3k+2}{k+1} \right )+\left (\frac{7k-2}{k+1} \right ) - 4 = 0$

$\Rightarrow \frac{6k+4+7k-2-4k-4}{k+1} = 0$

$\Rightarrow 9k-2 = 0$

$\Rightarrow k=\frac{2}{9}$

Therefore, the ratio in which the line $2x + y - 4 = 0$ divides the line segment joining the points $A(2,-2)$ and $B(3,7)$ is $2:9$ internally.

Q2 Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Answer:

If the points $(x, y), (1, 2)\ and\ (7, 0)$ are collinear then, the area formed by these points will be zero.

The area of the triangle is given by,

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0$

Substituting the values in the above equation, we have

$Area = \frac{1}{2}\left [ x(2-0)+1(0-y)+7(y-2) \right ]= 0$

$\Rightarrow 2x-y+7y-14= 0$

Or,

$\Rightarrow x+3y-7= 0$

Hence, the required relation between x and y is $x+3y-7= 0$ .

Q3 Find the center of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

Answer:

From the figure:

1638427419480

Let the center point be $O(x,y)$ .

Then the radii of the circle $OA,\ OB,\ and\ OC$ are equal.

The distance OA:

$OA = \sqrt{(x-6)^2+(y+6)^2}$

The distance OB:

$OB = \sqrt{(x-3)^2+(y+7)^2}$

The distance OC:

$OC = \sqrt{(x-3)^2+(y-3)^2}$

Equating the radii of the same circle.

When equating, $OA = OB$

$\sqrt{(x-6)^2+(y+6)^2}= \sqrt{(x-3)^2+(y+7)^2}$

Squaring both sides and applying $a^2-b^2 = (a+b)(a-b)$

$\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y+7)(y+6-y-7) = 0$

$\Rightarrow (2x-9)(-3) + (2y+13)(-1) = 0$

$\Rightarrow -6x+27-2y-13 = 0$ or

$\Rightarrow 3x+y -7= 0$ ...................................(1)

When equating, $OA = OC$

$\sqrt{(x-6)^2+(y+6)^2}= \sqrt{(x-3)^2+(y-3)^2}$

Squaring both sides and applying $a^2-b^2 = (a+b)(a-b)$

$\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y-3)(y+6-y+3) = 0$

$\Rightarrow (2x-9)(-3) + (2y+3)(9) = 0$

$\Rightarrow -3x+9y+27 = 0$ ...................................(2)

Now, adding the equations (1) and (2), we get

$\Rightarrow 10y = -20$

$\Rightarrow y = -2$ .

From equation (1), we get

$\Rightarrow 3x-2 = 7$

$\Rightarrow 3x =9$

$\Rightarrow x =3$

Therefore, the centre of the circle is $(3,-2)$ .

Q4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

Answer:

From the figure:

1638427461655

We know that the sides of a square are equal to each other.

Therefore, AB = BC

So,

$\sqrt{(x-1)^2+(y-2)^2} = \sqrt{(x-3)^2+(y-2)^2}$

Squaring both sides, we obtain

$\implies (x-1)^2+(y-2)^2 = (x-3)^2+(y-2)^2$

Now, doing $\left ( a^2-b^2 = (a+b)(a-b) \right )$

We get

$\implies (x-1+x-3)(x-1-x+3) = 0$

Hence $x = 2$ .

Applying the Pythagoras theorem to find out the value of y.

$AB^2+BC^2 = AC^2$

$(\sqrt{(2-1)^2+(y-2)^2})^2 + (\sqrt{(2-3)^2+(y-2)^2})^2 = (\sqrt{(3+1)^2+(2-2)^2})^2$

$\Rightarrow \left (\sqrt{1+(y-2)^2} \right )^2 + \left (\sqrt{1+(y-2)^2} \right )^2 = \left (\sqrt{16} \right )^2$

$\Rightarrow (y-2)^2 = 7$

Q5 (i) The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar is planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in Fig. 7.14. The students are to sow seeds of flowering plants on the remaining area of the plot. (i) Taking A as origin, find the coordinates of the vertices of the triangle.

1638427489636

Answer:

Taking A as origin then, the coordinates of P, Q, and R can be found by observation:

Coordinates of point P is $(4,6).$

Coordinates of point Q is $(3,2).$

Coordinates of point R is $(6,5).$

The area of the triangle, in this case, will be:

$Area =\frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$=\frac{1}{2}\left [ 4(2-5)+3(5-6)+6(6-2) \right ]$

$=\frac{1}{2}\left [ -12-3+24\right ] = \frac{9}{2}\ Square\ units.$

Q5 (ii) The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar is planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in Fig. 7.14. The students are to sow seeds of flowering plants on the remaining area of the plot. (ii) What will be the coordinates of the vertices of D PQR if C is the origin? Also, calculate the areas of the triangles in these cases. What do you observe?

1638427504689

Answer:

Taking C as origin, then CB will be x-axis and CD be y-axis.

The coordinates fo the vertices P, Q, and R are: $(12,2),\ (13,6),\ (10,3).$ respectively.

The area of the triangle, in this case, will be:

$Area =\frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$=\frac{1}{2}\left [ 12(6-3)+13(3-2)+10(2-6) \right ]$

$=\frac{1}{2}\left [ 36-13+40 \right ] = \frac{9}{2}\ Square\ units.$

It can be observed that in both cases the area is the same so, it means that the area of any figure does not depend on the reference which you have taken.

Q6 The vertices of a $\Delta ABC$ are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that $\frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4}$ Calculate the area of the $\Delta ADE$ and compare it with the area of $\Delta ABC$ .

Answer:

From the figure:

1638427528989

Given ratio:

$\frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4}$

Therefore, D and E are two points on side AB and AC respectively, such that they divide side AB an AC in the ratio of $1:3$ .

Section formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Then, coordinates of point D:

$D(x_{1},y_{1})= \left (\frac{1\times1+3\times 4}{1+3} , \frac{1\times 5+3\times 6}{1+3} \right )$

Coordinates of point E:

$E(x_{2},y_{2})= \left (\frac{1\times7+3\times 4}{1+3} , \frac{1\times 2+3\times 6}{1+3} \right )$

$= \left ( \frac{19}{4}, \frac{20}{4} \right )$

Then, the area of a triangle:

$= \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

Substituting the values in the above equation,

$Area\ of\ \triangle ADE = \frac{1}{2}\left [ 4\left ( \frac{23}{4} - \frac{20}{4}\right )+\frac{13}{4}\left ( \frac{20}{4} - 6 \right )+\frac{19}{4}\left (6-\frac{23}{4} \right )\right ]$ $= \frac{1}{2}\left [ 3-\frac{13}{4} +\frac{19}{16}\right ] = \frac{1}{2}\left [ \frac{48-52+19}{16} \right ] = \frac{15}{32}\ square\ units.$

$Area\ of\ \triangle ABC = \frac{1}{2}\left [ 4(5-2)+1(2-6)+7(6-5) \right ]$

$= \frac{1}{2}\left [ 12-4+7 \right ] = \frac{15}{2}\ Square\ units.$

Hence the ratio between the areas of $\triangle ADE$ and $\triangle ABC$ is $1:16.$

Q7 (1) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of $\triangle ABC$ . The median from A meets BC at D. Find the coordinates of the point D.

Answer:

From the figure:

1638427550821

Let AD be the median of the triangle

Then, D is the mid-point of BC

Coordinates of Point D:

$\left ( \frac{6+1}{2},\frac{5+4}{2} \right ) = \left ( \frac{7}{2}, \frac{9}{2} \right )$

Q7 (ii) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. Find the coordinates of the point P on AD such that AP: PD = 2: 1

Answer:

From the figure,

1638427579264

The point P divides the median AD in the ratio, AP: PD = 2: 1

Hence using the section formula,

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$P(x,y)= \left (\frac{2\times\frac{7}{2}+1\times4}{2+1} , \frac{2\times\frac{9}{2}+1\times2}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

Q7 (iii) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. Find the coordinates of points Q and R on medians BE and CF respectively such that BQ: QE = 2: 1 and CR: RF = 2: 1

Answer:

From the figure,

1638427609335

$\Rightarrow$ The point Q divides the median BE in the ratio, BQ : QE = 2 : 1

Hence using the section formula,

$Q(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$Q(x,y)= \left (\frac{2\times\frac{5}{2}+1\times6}{2+1} , \frac{2\times3+1\times5}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

$\Rightarrow$ The point R divides the median CF in the ratio, CR: RF = 2: 1

Hence using the section formula,

$R(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$R(x,y)= \left (\frac{2\times 5+1\times1}{2+1} , \frac{2\times\frac{7}{2}+1\times4}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

Q7 (iv) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. What do you observe?

Answer:

We observed that the coordinates of P, Q, and R are the same. Therefore, all these are representing the same point on the plane. i.e., the centroid of the triangle.

Q7 (v) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of $\triangle ABC$ . If $A(x_1, y_1), B(x_2, y_2)\ and\ C(x_3, y_3)$ are the vertices of $\triangle ABC$ , find the coordinates of the centroid of the triangle.

Answer:

From the figure, triangle-median-d

Let the median be AD which divides the side BC into two equal parts.

Therefore, D is the mid-point of side BC.

Coordinates of D:

$= \left ( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} \right )$

Let the centroid of this triangle be O.

Then, point O divides the side AD in a ratio 2:1.

Coordinates of O:

$= \left ( \frac{2\times\frac{x_{2}+x_{3}}{2}+1\times x_{1}}{2+1}, \frac{2\times\frac{y_{2}+y_{3}}{2}+1\times y_{1} }{2+1} \right )$

$= \left ( \frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3} \right )$

Q8 ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Answer:

From the figure:

1638427884180

P is the mid-point of side AB.

Therefore, the coordinates of P are, $\left ( \frac{-1-1}{2}, \frac{-1+4}{2} \right ) = \left ( -1, \frac{3}{2} \right )$

Similarly, the coordinates of Q, R and S are: $\left ( 2,4 \right ),\ \left ( 5, \frac{3}{2} \right ),\ and\ \left ( 2,-1 \right )$ respectively.

The distance between the points P and Q:

$PQ = \sqrt{(-1-2)^2+\left ( \frac{3}{2} -4 \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

and the distance between the points Q and R:

$QR = \sqrt{(2-5)^2+\left ( 4-\frac{3}{2} \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points R and S:

$RS = \sqrt{(5-2)^2+\left ( \frac{3}{2}+1 \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points S and P:

$SP = \sqrt{(2+1)^2+\left ( -1-\frac{3}{2} \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points P and R the diagonal length:

$PR = \sqrt{(-1-5)^2+\left ( \frac{3}{2}-\frac{3}{2} \right )^2} = 6$

Distance between points Q and S the diagonal length:

$QS = \sqrt{(2-2)^2+\left ( 4+1 \right )^2} = 5$

Hence, it can be observed that all sides have equal lengths. However, the diagonals are of different lengths.

Therefore, PQRS is a rhombus.

Summary Of Class 10 maths NCERT solutions chapter 7

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NCERT Solutions for Class 10 Maths - Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

NCERT Books and NCERT Syllabus

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The chapter 7 maths class 10 ncert solutions are designed to align with the current CBSE syllabus and include detailed explanations for each question. Practising these NCERT solutions for Class 10 Maths will help students improve their exam skills and time management, making them more likely to score well on the exam. Additionally, working through these solutions of maths chapter 7 class 10 can help students become more familiar with the exam format, giving them more confidence when taking the test.

2. Mention the topics that are covered in NCERT Solutions for Class 10 Maths Chapter 7.

NCERT coordinate geometry class 10 ncert solutions covers several key topics, including Introduction to Coordinate Geometry, Distance Formula, Section Formula, and Area of the Triangle. Through a thorough study of these solutions of class 10 chapter 7 maths, students will gain the skills necessary to easily solve complex problems related to these topics.

3. Is it necessary to practise all the exercises presented in class 10 maths chapter 7 solutions?

Yes, It is essential to practise all the exercises in coordinate geometry class 10 chapter 7 Maths because they include a variety of questions that may appear in the CBSE board examination. By solving these exercises of chapter 7 class 10 maths, students will increase their chances of success on the final exams and feel more confident when taking the test. You can also study NCERT solutions for class 10 maths chapter 7 in hindi medium if you are facing issue with english medium.

4. How to master the CBSE class 10 Maths?

First, you should try to solve every problem on your own including examples given in the NCERT textbook. Once you complete the entice syllabus, you must solve CBSE previous years paper in order to get familiar with the exam. For ease students can download chapter 7 Maths class 10 pdf solutions and study both online and offline mode.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

hindi 10th class mp board sample paper

Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

Hope you get it !

Thanking you

Read Complete Answer

Shivanshu 04 February,2024

CBSE class 10th date sheet 11/1/2024 2024

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 12 January,2024

iam in 7 th plz send 7th previous paper

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

Read Complete Answer

Sajal Trivedi 25 October,2023

my son's dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

Read Complete Answer

Shubhangi 29 July,2022

can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

Read Complete Answer

Aditi Singh 22 July,2022

View All

Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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