NCERT Solutions for Exercise 7.4 Class 10 Maths Chapter 7

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.

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Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1
More About NCERT Solutions for Class 10 Maths Exercise 7.4:
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:

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

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, 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:

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

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.

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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sadhu ashram ramgart road aligarh

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sai raja 01 September,2024

show the previous year exams ?

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

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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

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) $20,000 \, \, J - 50,000 \, \, J$

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

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 .

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

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.

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

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

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

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

Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1

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