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

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

Edited By Safeer PP | Updated on Jul 07, 2022 05:02 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Maths exercise 7.4 is the final and optional exercise in this chapter, and it covers all of the concepts from the entire chapter, including the distance formula, section formula, and triangle area. Coordinate Geometry is the study of geometry using coordinate points. The distance formula used to find the distance between two points in a two-dimensional plane is known as the Euclidean distance formula. The section formula is used to get the coordinates of the point that splits a line segment into a ratio either externally or internally. We can utilize the section when a point divides a line segment in some ratio either externally or internally.

This Story also Contains
  1. Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1
  2. More About NCERT Solutions for Class 10 Maths Exercise 7.4:
  3. Benefits of NCERT Solutions for Class 10 Maths Exercise 7.4 :

NCERT solutions for Class 10 Maths chapter 7 exercise 7.4 consists of 8 questions in which 6 of them are long answer questions, 1 is a practical based question and the remaining 1 is a reasoning question. In NCERT book Class 10 Maths chapter 7 exercise 7.4, the ideas linked to the solution of coordinate geometry are well discussed. The following activities are included along with NCERT syllabus Class 10 Maths chapter 7 exercise 7.4.

Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1

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)=(3k+2k+1,7k2k+1)

Point C will also satisfy the given line equation 2x+y4=0 , hence we have

2(3k+2k+1)+(7k2k+1)4=0

6k+4+7k24k4k+1=0

9k2=0

k=29

Therefore, the ratio in which the line 2x+y4=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=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]=0

Substituting the values in the above equation, we have

Area=12[x(20)+1(0y)+7(y2)]=0

2xy+7y14=0

Or,

x+3y7=0

Hence, the required relation between x and y is x+3y7=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=(x6)2+(y+6)2

The distance OB:

OB=(x3)2+(y+7)2

The distance OC:

OC=(x3)2+(y3)2

Equating the radii of the same circle.

When equating, OA=OB

(x6)2+(y+6)2=(x3)2+(y+7)2

Squaring both sides and applying a2b2=(a+b)(ab)

(x6+x3)(x6x+3)+(y+6+y+7)(y+6y7)=0

(2x9)(3)+(2y+13)(1)=0

6x+272y13=0 or

3x+y7=0 ...................................(1)

When equating, OA=OC

(x6)2+(y+6)2=(x3)2+(y3)2

Squaring both sides and applying a2b2=(a+b)(ab)

(x6+x3)(x6x+3)+(y+6+y3)(y+6y+3)=0

(2x9)(3)+(2y+3)(9)=0

3x+9y+27=0 ...................................(2)

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

10y=20

y=2 .

From equation (1), we get

3x2=7

3x=9

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,

(x1)2+(y2)2=(x3)2+(y2)2

Squaring both sides, we obtain

(x1)2+(y2)2=(x3)2+(y2)2

Now, doing (a2b2=(a+b)(ab))

We get

(x1+x3)(x1x+3)=0

Hence x=2 .

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

AB2+BC2=AC2

((21)2+(y2)2)2+((23)2+(y2)2)2=((3+1)2+(22)2)2

(1+(y2)2)2+(1+(y2)2)2=(16)2

(y2)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=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

=12[4(25)+3(56)+6(62)]

=12[123+24]=92 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=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

=12[12(63)+13(32)+10(26)]

=12[3613+40]=92 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 ΔABC 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 ΔADE and compare it with the area of ΔABC .

Answer:

From the figure:

1638427528989

Given ratio:

ADAB=AEAC=14

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)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

Then, coordinates of point D:

D(x1,y1)=(1×1+3×41+3,1×5+3×61+3)

Coordinates of point E:

E(x2,y2)=(1×7+3×41+3,1×2+3×61+3)

=(194,204)

Then, the area of a triangle:

=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

Substituting the values in the above equation,

Area of ADE=12[4(234204)+134(2046)+194(6234)] =12[3134+1916]=12[4852+1916]=1532 square units.

Area of ABC=12[4(52)+1(26)+7(65)]

=12[124+7]=152 Square units.

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

Q7 (1) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of 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:

(6+12,5+42)=(72,92)

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)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

P(x,y)=(2×72+1×42+1,2×92+1×22+1)=(113,113)

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

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

Hence using the section formula,

Q(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

Q(x,y)=(2×52+1×62+1,2×3+1×52+1)=(113,113)

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

Hence using the section formula,

R(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

R(x,y)=(2×5+1×12+1,2×72+1×42+1)=(113,113)

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 ABC . If A(x1,y1),B(x2,y2) and C(x3,y3) are the vertices of ABC , find the coordinates of the centroid of the triangle.

Answer:

From the figure, 1656326442871

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:

=(x2+x32,y2+y32)

Let the centroid of this triangle be O.

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

Coordinates of O:

=(2×x2+x32+1×x12+1,2×y2+y32+1×y12+1)

=(x1+x2+x33,y1+y2+y33)

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, (112,1+42)=(1,32)

Similarly, the coordinates of Q, R and S are: (2,4), (5,32), and (2,1) respectively.

The distance between the points P and Q:

PQ=(12)2+(324)2=9+254=614

and the distance between the points Q and R:

QR=(25)2+(432)2=9+254=614

Distance between points R and S:

RS=(52)2+(32+1)2=9+254=614

Distance between points S and P:

SP=(2+1)2+(132)2=9+254=614

Distance between points P and R the diagonal length:

PR=(15)2+(3232)2=6

Distance between points Q and S the diagonal length:

QS=(22)2+(4+1)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.

More About NCERT Solutions for Class 10 Maths Exercise 7.4:

The midpoint and area of the quadrilateral were also covered in the NCERT solutions for Class 10 Maths exercise 7.4. The questions in exercise 7.3 Class 10 Maths are based on concepts such as the distance formula, section formula, and triangle area. When a point P(x,y) divides the line segment into two segments, with marked points as A(x1,y1) and B(x2,y2) the method used to find the coordinates of that point is known as the section formula that is covered in the Class 10 Maths chapter 7 exercise 7.4.

Also Read| Coordinate Geometry Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 7.4 :

• NCERT solutions for Class 10 Maths exercise 7.4 are carefully drafted to assist the student in scoring good marks in the examination. That's why any student can easily score the highest possible marks in the final exam.

• From exercise 7.4 Class 10 Maths we can easily understand the pattern of questions that can be asked in term exams from this chapter and also know the marks weightage of the chapter. From that, they can prepare themselves accordingly for the final examination.

• NCERT solution for Class 10 Maths chapter 7 exercise 7.4 exercises, will be helpful in the further exercise of chapter 10.

Frequently Asked Questions (FAQs)

1. The distance of the point P(4, 9) from the x-axis is _______

The distance from x-axis is equal to its ordinate that is 4 .

2. The distance of the point P(4, 9) from the y-axis is _______

The distance from the x-axis is equal to its ordinate that is 9.

3. State true/false : The area of the triangle is always positive .

The statement is true. The area of the triangle is always positive.

4. Using the area of the triangle formula, how can we find the area of the quadrilateral?

Quadrilaterals can be divided into three triangular areas, each with its own area. The area of the two triangles can then be calculated using the area of the triangle formula. The area of the quadrilateral is then calculated by adding both.

5. Area of the quadrilateral is always _______

The area of the quadrilateral is always positive since the area of the triangle is positive.

6. According to NCERT solutions for Class 10 Maths chapter 7 exercise 7.4 , define collinear points ?

Two or more points are considered to be collinear if they lie on the same line, according to NCERT solutions for Class 10 Maths chapter 7 exercise 7.4.

7. How many questions are there in the NCERT solutions for Class 10 Maths chapter 7 exercise 7.4 and what types of questions are there?

NCERT solutions for Class 10 Maths chapter 7 exercise 7.4  consists of 8 questions in which 6 of them are long answer questions, 1 is a practical based question and remaining 1 is a reasoning question and all the questions are based on topics such as distance formula, section formula, area of the triangle, midpoint and collinearity.

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Questions related to CBSE Class 10th

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Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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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 3.8×107 J 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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