NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

Mathematics is one of the subjects that has its own uses. All students need to have a foundational understanding of Mathematics, no matter if they are from the Arts, Sciences, or Commerce stream. Understanding certain fundamental concepts in Mathematics will benefit students in the long run, in addition to helping them do well on their board examinations. However, many students are not interested in studying Mathematics these days because of its difficulty.

Mathematics is a subject that has applications in both academic work and daily life. It might be difficult to get students interested in this subject as it is a challenging one for them to grasp since it requires a lot of mental effort. Students frequently struggle with the concept of Mathematics.

Counting is the foundation or starting point of Mathematics, where humans learned to count the things that are visible to the eye. Pure Mathematics (numerical systems, geometry, matrices, algebra, combinatorics, topology, and calculus) and Applied Mathematics are the two main categories of Mathematics (Engineering, Chemistry, Physics, numerical analysis, etc). Numerous formulas in Mathematics are based on various ideas. By repeatedly answering questions based on these formulas, one can learn them. It is possible to solve some problems quickly using mathematical techniques.

One of the most important topics in Mathematics is called Coordinate Geometry. By using graphs with curves and lines, Coordinate Geometry (also known as Analytic Geometry) explains how Geometry and Algebra are related. Students can answer geometrical problems because it gives them geometrical aspects of Algebra. In this area of Geometry, a pair of ordered numbers is used to describe the location of points on a plane. The study of Geometry using coordinate points is known as Coordinate Geometry. It is possible to determine the distance between two points by dividing a line into m:n segments, locating a line’s midpoint, determining the area of a triangle in the Cartesian plane, and performing other calculations using Coordinate Geometry.

Mathematics has different concepts and formulas that students are not familiar with. It is also considered a complex subject, so students often face problems regarding the same. The National Council of Educational Research and Technology was introduced by the Central Board of Secondary Education. It was created to promote a united educational system for the nation with a national character as well as  enable and encourage the various cultural traditions that are practised across the board. Conducting, promoting, and coordinating research in fields connected to school education is one of NCERT’s primary objectives, as well as that of its units. NCERT provides information about the world and addresses all important subjects. Its content is clear and simple for students to understand.

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

In Coordinate Geometry, any point, figure, or object’s position is specified using two or more numbers. The location of the object can be specified in a variety of ways, including a line, a plane, three dimensions, etc. The Cartesian coordinate system is broadly used. It is one type of many that use a pair of numbers (coordinates), which are the signed distances between each point and two fixed perpendicular lines known as the axes, to represent each point. There are many different concepts in the chapter Coordinate Geometry which includes graphs, lines, etc. Therefore, students face difficulty in understanding the chapter clearly, as NCERT  textbooks do not contain detailed solutions. Students can refer to the Extramarks website for reliable and concise study material.

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On the Extramarks website, students will find the entire set of notes for Class 10 Maths Exercise 7.4 Solutions. Students can learn the position of a point on a plane using Coordinate Geometry. When the coordinates of the points are given, students will learn how to use the distance formula to calculate the distance between two points and the area of a triangle.

Students who read the in-depth  NCERT Solutions offered by our subject-matter experts will be able to score higher marks. It follows the  NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, which help students prepare in the right way. From the viewpoint of the examination, it comprises all the important questions. It helps students  earn good marks in board exams for Mathematics.Students can complete and go over all of the questions in Class 10 Maths Chapter 7 Exercise 7.4 by using the NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.4.

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Access NCERT Solutions for Class-10 Maths Chapter 7 – Coordinate Geometry

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4 

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Q.1

Determine the ratio in which the line 2x+y4=0 divides the line segment joining the points A(2, 2)and B(3, 7). 

Ans

Let the given line divide the line segment joining the pointsA(2, 2) and B(3, 7) in the ratio k:1.Coordinates of the point of division=(3k+2k+1,  7k2k+1)This point of division lies on the line 2x+y4=0.    23k+2k+1+7k2k+14=0or 6k+4+7k24k4=0or 9k2=0or k=29Therefore, the required ratio is 2:9

Q.2

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

Ans

Points (x1, y1), (x2, y2) and (x3, y3) are collinear if x1(y2y3)+x2(y3y1)+x3(y1y2)=0Therefore, Points (x, y), (1, 2) and (7, 0) are collinear if x(20)+1(0y)+7(y2)=0 or 2xy+7y14=0 or 2x+6y14=0 or x+3y7=0

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

Ans

Let O(x, y) be the centRE of the circle and A(6, 6), B(3, 7)and (3, 3) are points on the circle.Then OA, OB and OC are radii of the circle and OA=OB=OC.Therefore, by distance formula, we have (x6)2+(y+6)2=(x3)2+(y+7)2or x212x+36+y2+12y+36=x26x+9+y2+14y+49or 6x2y+14=0or 3xy+7=0or y=3x+7 ...(1)Also, OA=OCor (x6)2+(y+6)2=(x3)2+(y3)2or x212x+36+y2+12y+36=x26x+9+y26y+9or 6x+18y+54=0or x+3y+9=0or x+3(3x+7)+9=0 [From (1), y=3x+7]or x9x+21+9=0or 10x+30=0or x+3=0or x=3On putting this value of x in (1), we get y=3×(3)+7=2Therefore, coordinates of the centre are (3, 2).

Q.4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.


Ans

Let ABCD be a square having A(1, 2) and C(3, 2) as oppositevertices. Let other two opposite vertices are B(x, y) and D(x1, y1).We know that sides of a square are equal to each other. AB=BCor (x+1)2+(y2)2=(x3)2+(y2)2or x2+2x+1+(y2)2=x26x+9+(y2)2or 8x=8or x=1We know that all interior angles are of 90° in a square.Therefore, using Pythagoras theorem in ΔABC, we have AB2+BC2=AC2or (x+1)2+(y2)2+(x3)2+(y2)2=(3+1)2+(22)2or x2+2x+1+y24y+4+x26x+9+y24y+4=16or 2x2+2y24x+28y=0or 2+2y24+28y=0 [x=1]or 2y28y=0or y24y=0or y(y4)=0i.e., y=0 or y=4We know that in a square, the diagonals are of equal lengthand bisect each other at 90°. Let O be the mid-point of AC.Therefore, it is also the mid-point of BD.Coordinates of point O=(1+32, 2+22)=(x+x12, y+y12)i.e., (x+x12, y+y12)=(1, 2)i.e., x+x12=1 and y+y12=2i.e., x+x1=2 and y+y1=4Putting the values of x and y, we get x1=1 and y1=0 or y1=4.Therefore, the required coordinates of other two opposite verticesof the given square are (1, 0) and (1, 4).

Q.5 The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown
in the following figure. 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.
(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?


Ans

(i) We take A as origin and AD as x-axis and AB as y-axis.From the given figure, we observe that the coordinatesof point P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.Now,Area of ΔPQR=12[x1(y2y3)+x2(y3y1)+x3(y1y2)] =12[4(25)+3(56)+6(62)] =12[123+24]                              =92 square units(ii) We take C as origin and CB as x-axis and CD as y-axis.From the given figure, we observe that the coordinatesof point P, Q and R are (12, 2), (13, 6) and (10, 3) respectively.Now,Area of ΔPQR=12[x1(y2y3)+x2(y3y1)+x3(y1y2)] =12[12(63)+13(32)+10(26)] =12[36+1340]                              =92 square unitsWe observe that area of the triangle in both cases is same.

Q.6

The vertices of a Δ 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 ADAB=AEAC=14. Calculate the area of the Δ ADE and compare it with the area of Δ ABC.(Recall converse of basic proportionality theorem and Theorem 6.6 related to the ratio of the areas of twosimilar triangles).

Ans

Given that,           ADAB=AEAC=14or ADAD+DB=AEAE+EC=14or AD+DBAD=AE+ECAE=4or AD+DBAD1=AE+ECAE1=41or DBAD=ECAE=3or ADDB=AEEC=13Therefore, D and E are two points on side AB and AC respectivelysuch that they divide side AB and AC in the ratio 1:3.Coordinates of point D=(1×1+3×41+3, 1×5+3×61+3) =(134, 234)Coordinates of point E=(1×7+3×41+3, 1×2+3×61+3) =(194, 204)Area of a triangle=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]Area of ΔADE  =12[4(234204)+134(2046)+194(6234)]                              =12[3134+1916]                            =12[4852+1916]                           =1532 square unitsArea of ΔABC  =12[4(52)+1(26)+7(65)]                              =12[124+7]                            =152  square unitsRatio between the areas of ΔADE and ΔABC is 1:16.

Q.7

Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC. (i) The median from A meets BC at D. Find the coordinates of the point D. (ii) Find the coordinates of the point P on AD such that AP:PD=2:1 (iii) 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. (iv) What do yo observe? [Note: The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.] (v) If A(x1,  y1), B(x2,  y2) and C(x3,  y3) are the vertices of Δ ABC, find the coordinates of the centroid of the triangle.

Ans

(i) Median AD of ΔABC bisects BC. So, D is the mid-point of side BC. Coordinates of D=(6+12, 5+42)=(72, 92)(ii) Point P divides the side AD in the ratio 2:1.Coordinates of P=(2×72+1×42+1, 2×92+1×22+1)=(113, 113)(iii)Median BE of the triangle will divide the side AC in two equal parts.Therefore, E is the mid-point of the side AC.Coordinates of E=(4+12, 2+42)=(52, 3)Point Q divides the side BE in the ratio 2:1.Coordinates of Q=(2×52+1×62+1, 2×3+1×52+1)=(113, 113)Median CF of the triangle will divide the side AB in twoequal parts.Therefore, F is the mid-point of the side AB.Coordinates of F=(4+62, 2+52)=(5, 72)Point R divides the side CF in the ratio 2:1.Coordinates of R=(2×5+1×12+1, 2×72+1×42+1)=(113, 113)(iv)We observe that coordinates of point P, Q and R are same.Therefore, all these are representing the same point onthe plane i.e., the centroid of the triangle.(v)We consider ΔABC having vertices A(x1,  y1), B(x2,  y2) andC(x3,  y3). Let AD is median and so D bisects side BC.Coordinates of D=(x2+x32, y2+y32)Let O is the centroid of ΔABC. So, O divides median AD inthe ratio 2:1.Coordinates of centroid O=(x1+2(x2+x32)2+1, y1+2(y2+y32)2+1) =(x1+x2+x33,y1+y2+y33)

Q.8 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.

Ans

P is the mid-point of side AB.Coordinates of P=(112, 1+42)=(1, 32)Similarly, coordinates of Q, R and S are (2, 4), (5, 32)and (2, 1) respectively.Now, we find lengths of each side and each diagonal of quadrilateral PQRS by using distance formula.PQ=(12)2+(324)2=9+254=614QR=(25)2+(432)2=9+254=614RS=(52)2+(32+1)2=9+254=614SP=(2+1)2+(132)2=9+254=614PR=(15)2+(3232)2=36+0=6QS=(22)2+(4+1)2=0+25=5We find that all sides of the quadrilateral PQRS are of equallength, but diagonals are of different length. Therefore,quadrilateral PQRS is a rhombus.

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