NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Coordinate Geometry is an integral part of Mathematics. This chapter can be found in NCERT class 10 Maths chapter 7 and it deals with geometrical figures and coordinates like the x-axis and y-axis to locate the exact position of the points in various types of planes. This chapter consists of an introduction phase into coordinate geometry and gradually makes students aware of topics like distance formula and section formula.

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1. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry FDF Free Download
2. Coordinate Geometry Class 10 - Important Formulae
3. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Questions and Exercise)
4. Key Features provided for class 10 maths NCERT solutions chapter 7
5. NCERT Solutions for Class 10 Maths - Chapter Wise
6. NCERT Books and NCERT Syllabus
7. Benefits of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
8. NCERT solutions of class 10 subject-wise
9. NCERT Exemplar solutions - Subject-wise

NCERT class 10 maths ch 7 solutions can be often challenging. Students often get confused during the class 10 chapter 7 NCERT solutions. This article is for every student to understand all the Chapter 7 Maths class 10 NCERT solutions. Students can also read Chapter 7 maths class 10 pdf solutions for a quick revision. 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. NCERT class 10 maths ch 7 solutions

Also Read,

• NCERT Exemplar Solutions for Class 10 Maths Chapter Coordinate Geometry
• Coordinate Geometry Class 10 Mathematics Notes

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 = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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 = $\{\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\}$

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 = $\{\frac{x_1+x_2}{2},\frac{(y_1+y_2}{2}\}$

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 = $\frac{1}{2}$|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 (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 $(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

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 $(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

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 $(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

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 $(x_2,y_2)$

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

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

$AB+BC\ne CA$

$BA+AC\ne BC$

$BC+CA\ne BA$

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 triangles 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 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 the distance formula, find which of them is correct.

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,

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

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

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

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

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 length 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:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

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

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

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

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:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

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

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

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

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like a 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 $(4,5),(7,6),(4,3),(1,2)$ represent 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 $A(2,-5)andB(-2,9)$ 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 in length.

Hence, we have,

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

Squaring both sides we get,

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

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

$\Rightarrow (-8x)=56$

$\Rightarrow x=-7$

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 $y=3or-9$.

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: $R(4,6)orR(-4,6.)$

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

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

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

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

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

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: $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

Substituting the values in the formula:

Here, $m_1:m_2=2:3$

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

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

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

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: $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

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;

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

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

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

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

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

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

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

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

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 $\frac{1}{4}$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth line and posts a red flag. What is the distance between both 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:

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

$\frac{1}{4}\times 100m=25m$ 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.,

$\frac{1}{5}\times 100m=20m$ 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,

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

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

$\Rightarrow x=5,y=22.5$

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:

$P(x,y)=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})$

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

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

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

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

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:

$P(x,y)=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})$

$\Rightarrow \frac{k{y}_2+y_1}{k+1}=0$

$\Rightarrow k=-\frac{y_1}{y_2}$

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 the section formula.

$P(x,0)=(\frac{x_2+x_1}{2},0)$

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

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 the mid-point of AC.

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

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

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

Since both coordinates are of the same point O.

Therefore,

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

Or,

$x=6andy=3$

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,

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

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

$\Rightarrow x=3andy=-10$ .

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:

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:

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

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

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

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

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:

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 Points C, D, and E.

Section Formula:

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

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

$D(x_2,y_2)=(\frac{-2+2}{2},\frac{2+8}{2})$

$\Rightarrow D(x_2,y_2)=(0,5)$

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

$C(x_1,y_1)=(\frac{-2+0}{2},\frac{2+5}{2})$

$\Rightarrow C(x_2,y_2)=(-1,\frac{7}{2})$

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

$E(x_1,y_1)=(\frac{2+0}{2},\frac{8+5}{2})$

$\Rightarrow E(x_2,y_2)=(1,\frac{13}{2})$

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:

Let the vertices of the rhombus be:

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

Area of the rhombus ABCD is given by:

$=\frac{1}{2}\times$ Product of length 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

• Coordinate Geometry Class 10 Exercise 7.1
• Coordinate Geometry Class 10 Exercise 7.2

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The NCERT class 10 CoordMaths Geometry chapter includes several key features to aid students in their understanding and practice of the material provided in Class 10 Chapter 7 Example 9 and answer all exercise questions given in the NCERT textbook

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• Chapter 7 Maths Class 10 pdf solutions help to familiarise students with important formulas and standards.

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• NCERT Books Class 10
• NCERT Syllabus Class 10

How to use NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry?

Students can follow the steps provided below to use the class 10 maths ncert solutions chapter 7 in the best way possible.

• Go through the concepts and examples given in the textbook to understand the basics of the chapter.
• Memorize the important formulae and their uses to answer some directly asked questions.
• Practice various types of questions to get accustomed to every topic in coordinate geometry chapter 7.
• During the Maths Chapter of exercises, you can strategically use NCERT solutions in Chapter 7.
• Now, students can practise the previous year's questions to understand the question pattern and important topics.

Benefits of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Coordinate geometry is not only a useful chapter in class 10 but also for higher studies and competitive exams. If your basic concepts are clear right now, then later you will not face any difficulties solving coordinate geometry questions. These are some further benefits listed below.

• Students will be able to study strategically after accessing Class 10 Maths NCERT Solutions chapter 7.
• Class 10 Maths Chapter 7 NCERT solutions are solved by subject-matter experts and are very reliable at the same time.
• NCERT solutions for class 10 maths chapter 7 Coordinate Geometry is designed to give the students step-by-step solutions for a particular question.
• In Coordinate Geometry, students will study the algebraic forms of geometric figures.

NCERT solutions of class 10 subject-wise

• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Science

NCERT Exemplar solutions - Subject-wise

• NCERT Exemplar Class 10th Maths
• NCERT Exemplar Class 10th Science

Good luck and keep on reading.