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Coordinate Geometry Class 10 Maths NCERT Solutions Chapter 7 is concerned with the understanding of various aspects of coordinate geometry. Students preparing for chapter 7 maths class 10 will be able to clear all their concepts on coordinate geometry at the root level. It allows us to learn geometry with algebra and understand algebra with geometry. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

Class 10 Maths chapter 7 NCERT Solutions Coordinate Geometry of the section, concentrates on the essential concepts such as the distance computation between two locations in the cartesian plane, internal and external division of a line segment in a specific ratio, area calculation of a triangle, and real-world applications of these formulas. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Class 10 Coordinate Geometry Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The NCERT Solutions for Class 10 Maths Chapter 7 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Exercise 7.1

Question 1

Find the distance between the following pairs of points :

i) $(2,3),(4,1)$

ii) $(−5,7),(−1,3)$

iii) $(a,b),(−a,−b)$

i) $(2,3),(4,1)$

ii) $(−5,7),(−1,3)$

iii) $(a,b),(−a,−b)$

Solution

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$(i)$D $=(4−2)_{2}+(3−1)_{2} =22 $ units

$(ii)$D $=(7−3)_{2}+(−5+1)_{2} =42 $ units

$(iii)$D $=(a+a)_{2}+(b+b)_{2} =2a_{2}+b_{2} $ units

Question 2

Find the distance between the points $(0,0)$ and $(36,15)$. Can you now find the distance between the two towns $A$ and $B$ .

Solution

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$LetthetwopointsbeP(0,0)andQ(36,15)Now,PQ=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} =(36−0)_{2}(15−0)_{2} =(36)_{2}+(15)_{2} =1296+225 =1521 =39.Now,ItisgiventhatBislocated36kmeastand15kmnorthoftownA.LetustakeAasorigin.Now,AB=(36−0)_{2}+(15−0)_{2} =12596+225 =1521 =39.Hence,thedistancebetweenthetwotownsis39km. $

Question 3

Determine if the points $(1,5),(2,3)$ and $(−2,−11)$ are collinear , by distance formula.

Solution

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$A(1,5),B(2,3)andC(−2,−11)$

Now, $AB=(1_{2})+(2_{2}) =5 BC=(4_{2})+(14_{2}) =212 CA=(1_{2})+(16_{2}) =257 $

As,$CA>BC>AB$

If points $A,BandC$ are collinear then $AB+BC=CA$

But $5 +212 =257 $

So they are not collinear.

Question 4

Check whether $(5,−2)$, $(6,4)$ and $(7,−2)$ are the vertices of an isosceles triangles.

Solution

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Let ABC be the triangle $A(5,−2),B(6,4),C(7,−2)$

By distance formula

$AB=(5−6)_{2}+(−2−4)_{2} =(−1)_{2}+(−6)_{2} 37 $

$BC=(6−7)_{2}+(4−(−2)_{2} =(−1)_{2}+(6)_{2} =37 $

$CA=(7−5)_{2}+(−2−(−2))_{2} =(2)_{2}+0 =2$

Since $AB=BC=37 $

Hence $(5,−2),(6,4)$ & $(7,−2)$ are vertices of isosceles triangle.

Question 5

In a classroom, $4$ friends are seated at the points , $A,B,C$ and $D$ as shown in the following figure. 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.

Solution

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From the figure co-ordinates of points can be expressed as shown below,

$A(3,4),B(6,7),C(9,4),D(6,1)$$AB=(6−3)_{2}+(7−4)_{2} =3_{2}+3_{2} $

$=9+9 =18 =32 $

$BC=(9−6)_{2}+(6−7)_{2} =3_{2}+(−3)_{2} $

$=9+9 =18 =32 $

$CD=(6−9)_{2}+(1−4)_{2} =(−3)_{2}+(−3)_{2} $

$=9+9 =18 =32 $

$DA=(6−3)_{2}+(1−4)_{2} =3_{2}+(−3)_{2} $

$=9+9 =18 =32 $

Four sides of the quadrilateral $ABCD$ are equal

$DiagonalAC=DiagonalBD$

Therefore $ABCD$ is a square

Therefore, Champa is correct between two

$AB=BC=CD=DA=32 $

$DiagonalAC=(9−3)_{2}+(4−4)_{2} =6$

$DiagonalBD=(6−6)_{2}+(1−7)_{2} =6$

$DiagonalAC=DiagonalBD$

Therefore $ABCD$ is a square

Therefore, Champa is correct between two

Question 6

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) $(−1,−2),(1,0),(−1,2),(−3,0)$

(ii) $(−3,5),(3,1),(0,3),(−1,−4)$

(iii) $(4,5),(7,6),(4,3),(1,2)$

(i) $(−1,−2),(1,0),(−1,2),(−3,0)$

(ii) $(−3,5),(3,1),(0,3),(−1,−4)$

(iii) $(4,5),(7,6),(4,3),(1,2)$

Solution

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(i) Let the given points are $A(−1,−2)$, $B(1,0)$, $C(−1,2)$and $D(−3,0)$ Then,

$AB=(1+1)_{2}+(0+2)_{2} =2_{2}+2_{2} =4+4 =8 $

$BC=(−1−1)_{2}+(2−0)_{2} =(2_{2}+2_{2}) =4+4 =8 $

$CD=((−3)−(−1))_{2}+(0−2)_{2} =2_{2}+(−2)_{2} =4+4 =8 $

$DA=(−3)−(−1))_{2}+(0−(−2))_{2} =(−2)_{2}+2_{2} =4+4 =8 $

$AC((−1)−(−1))_{2}+(2−(−2))_{2} =0+4_{2} =16 =4$

$BD=(−3−1)_{2}+(0−0)_{2} =−4_{2} =16 =4$

Since the four sides $AB,BC,CD$ and $DA$ are equal and the diagonals $AC$ and $BD$ are equal .

$∴$ Quadrilateral $ABCD$ is a square.

(ii)Let the given points are $A(−3,5)$,$B(3,1)$,$C(0,3)$ and $D(−1,−4)$Then

$AB=(−3−3)_{2}+(5−1)_{2} =(−6)_{2}+4_{2} =36+16 =52 $

$BC=(3−0)_{2}+(1−3)_{2} =(3_{2}+(−2)2_{2}) =9+4 =11 $

$CD=(0−(−1))_{2}+(3−(−4))_{2} =1_{2}+(7)_{2} =1+49 =50 $

$DA=(−1)−(−3))_{2}+((−4)−5))_{2} =(2)_{2}+(−9)_{2} =4+81 =85 $

Here $AB=BC=CD=DA$

$∴$ it is a quadrilateral.

(iii)Let the given points are $A(4,5)$,$B(7,6)$,$C(4,3)$ and $D(1,2)$Then

$AB=(7−4)_{2}+(6−5)_{2} =3_{2}+1_{2} =9+1 =10 $

$BC=(4−7)_{2}+(3−6)_{2} =((−3)_{2}+(−3)_{2}) =9+9 =18 $

$CD=(1−4)_{2}+(2−3)_{2} =(−3)_{2}+(−1)_{2} =9+1 =10 $

$DA=(1−4)_{2}+(2−5)_{2} =(−3)_{2}+(−3)_{2} =9+9 =18 $

$AC(4−4)_{2}+(3−5)_{2} =0+(−2)_{2} =4 =2$

$BD=(1−7)_{2}+(2−6)_{2} =(−6)_{2}+(−4)_{2} =36+16 =52 $

Here $AB=CD,BC=DA$ . But $AC=BD$

Hence the pairs of opposite sides are equal but diagonal are not equal so it is a parallelogram.

Question 7

Point on the $x$-axis which is equidistant from $(2,−5)$ and $(−2,9)$ is:

Solution

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Since point on $x−$axis, then coordinate of the point is $(x,0)$.

According to the question this point $(x,0)$ is equidistant from the points $(2,−5)$ and $(−2,9)$.

That is, distance from $(x,0)$ and $(2,−5)$ $=$ distance from $(x,0)$ and $(−2,9)$.

$(2−x)_{2}+(−5−0)_{2} =(−2−x)_{2}+(9−0)_{2} ⇒(2−x)_{2}+(−5)_{2}=(−2−x)_{2}+(9)_{2}⇒4−4x+x_{2}+25=4+4x+4+81⇒8x=25−81⇒8x=−56⇒x=−7$

So, point is $(−7,0)$.

According to the question this point $(x,0)$ is equidistant from the points $(2,−5)$ and $(−2,9)$.

That is, distance from $(x,0)$ and $(2,−5)$ $=$ distance from $(x,0)$ and $(−2,9)$.

$(2−x)_{2}+(−5−0)_{2} =(−2−x)_{2}+(9−0)_{2} ⇒(2−x)_{2}+(−5)_{2}=(−2−x)_{2}+(9)_{2}⇒4−4x+x_{2}+25=4+4x+4+81⇒8x=25−81⇒8x=−56⇒x=−7$

So, point is $(−7,0)$.

Question 8

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

Solution

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$PQ=10$

$⇒(10−2)_{2}+(y+3)_{2} =10$

$64+y_{2}+9+6y =10$

Squaring both sides,

$y_{2}+6y−27=0$

$⇒y_{2}+9y−3y−37=0$

$⇒y(y+9)−3(y+9)=0$

$⇒(y+9)(y−3)=0$

$∴y=−9,3$

$⇒(10−2)_{2}+(y+3)_{2} =10$

$64+y_{2}+9+6y =10$

Squaring both sides,

$y_{2}+6y−27=0$

$⇒y_{2}+9y−3y−37=0$

$⇒y(y+9)−3(y+9)=0$

$⇒(y+9)(y−3)=0$

$∴y=−9,3$

Question 9

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 $PR$.

Solution

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As given $Q(0,1)$ is equidistant from $P(5,−3)$ and $R(x,6)$

$⇒PQ=QR$

$⇒(0−5)_{2}+(1−(−3))_{2} $$=(x−0)_{2}+(1−6)_{2} $ [ By using disatnce formula $=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $ ]

$⇒25+16=x_{2}+25$

$⇒41=x_{2}+25$

$⇒x_{2}=16$

$⇒x=4$

Distance between $Q(0,1)$ and $R(4,6)$

$QR=(4−0)_{2}+(6−1)_{2} =4_{2}+5_{2} =16+25 =41 $

Distance between $P(5,−3)$ and $R(4,6)$

$PR=(4−5)_{2}+(6+3)_{2} =1+81 =82 $

Question 10

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

Solution

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

$∴PA=PB$

By using the distance formula

$=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $ we have

$⇒(x−3)_{2}+(y−6)_{2} =(x+3)_{2}+(y−4)_{2} $

$⇒(x−3)_{2}+(y−6)_{2}=(x+3)_{2}+(y−4)_{2}$

$⇒x_{2}−6x+9+y_{2}−12y+36=x_{2}+6x+9+y_{2}−8y+16$

$⇒−6x−6x−12y+8y+36−16=0$

$⇒−12x−4y+20=0$

$⇒3x+y−5=0$

Hence this is a relation between $x$ and $y$.

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Practice more questions

The x-coordinate, or abscissa, of a point, is its distance from the y-axis. The y-coordinate, or ordinate, of a point, is its distance from the x-axis. A point on the x-axis has coordinates of the form (x, 0), while a point on the y-axis has coordinates of the form (0, y).

Coordinate geometry is the branch of mathematics that uses algebraic methods to solve geometrical issues. It is the study of cartesian coordinates. This system is typically used to alter the equations of two-dimensional shapes such as triangles, squares, circles, and so on.

Coordinate geometry is significant in mathematics because it allows us to identify points on any given plane. Coordinate geometry is also known as the study of graphs. It also has numerous applications in trigonometry, dimensional geometry, calculus, and other sciences.

The distance between any two points, P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) in the plane are given by:

Also, the distance between P(x_{1}, y_{1}) and the origin is:

The section formula assists us in determining the coordinates of the point that divides the line in the ratio m:n. If P is the point splitting the line AB internally in the ratio m:n where A(x_{1}, y_{1}) and B(x_{2}, y_{2}).

P's coordinates will be:

If P is the point splitting the line AB externally in the ratio m:n where A(x_{1}, y_{1}) and B(x_{2}, y_{2}).

P's coordinates will be:

The area of ∆ABC formed by the vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) is given by:

Area of Triangle = ½ x [x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} – y_{2})]

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Q1. What are the most important topics in NCERT Class 10 Chapter 7 Coordinate Geometry?

Answer: The distance formula, the section formula, and the area of the triangle are all important concepts in class 10 chapter 7 Coordinate Geometry. Questions on all three topics carry a considerable weightage in tests, therefore students should practice them thoroughly.

Q2. How many exercises are there in Chapter 7 of Class 10 Maths?

Answer: There are four exercises in total. Class 10 Maths Chapter 7 Coordinate Geometry has 33 problems, 20 of which are pretty easy, 7 of which are moderately tough, and 6 of which are long answer intricate sums. These questions are spread among four exercises in this chapter. They are both theoretical and practical in character, based on the cartesian plane and the associated formulas.

Q3. What is the purpose of Coordinate Geometry?

Answer: Coordinate geometry has many uses in everyday life. It is an important part of mathematics that helps us locate points on a plane. Furthermore, it has several applications in trigonometry, calculus, dimensional geometry, and other sciences. A few examples of coordinate geometry applications are: GPS positions in digital maps to pinpoint exact locations, Coordinate geometry is also employed in the construction of highways, buildings, and other structures, and Coordinate geometry is also employed in the cartographic representation of 2D and 3D objects.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability