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BSEB Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Book Answers

BSEB Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Book Answers
BSEB Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Book Answers


BSEB Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 books are published by the Bihar Board Publishers. These Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 book solutions pdf online from this page.

Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks Solutions PDF

Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks. These Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Books Solutions

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Class 10th
Subject Maths Chapter 7 Coordinate Geometry Ex 7.4
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BSEB Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

BSEB Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
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).
Solution:
Suppose the line 2x + y – 4 = 0 divides the line segment joining A(2, – 2) and B(3, 7) in the ratio k : 1 at point C. Then, the coordinates of C are (3𝑘+2𝑘+1,7𝑘−2𝑘+1)

But C lies on 2x + y – 4 = 0. Therefore,
2(3𝑘+2𝑘+1) + (7𝑘−2𝑘+1) – 4 = 0
or 6k + 4 + 7k – 2 – 4k – 4 = 0
or 9k – 2 = 0
or 9k = 2 i.e., 29
So, the required ratio is 2 : 9 internally.

Question 2.
Find a relation between x and y, if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
The points A(x, y), B(1, 2) and C(7, 0) will be collinear, if
x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
i.e., 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0
or x + 3y – 7 = 0
which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Solution:
Let P(x, y) be the centre of the circle passing through the points A(6, – 6), B(3, – 7) and C(3, 3).

Then, AP = BP = CP.
Now, AP = BP
So, AP² = BP²
So, (x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
or x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² + 14y + 49 or – 12x + 6x + 12y – 14y + 72 – 58 = 0
or – 6x – 2y + 14 = 0
or 3x +y – 7 = 0 … (1)
Also, BP = CP
So, BP² = CP²
So, (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
or x² – 6x + 9 + y² + 14y + 49 = x² – 6x + 9 + y² – 6y + 9
or – 6x + 6x + 14y + 6y + 58 – 18 = 0
or 20y + 40 = 0
or y = −4020 = – 2 …. (2)
Putting y = – 2 in (1), we get
3x – 2 – 7 = 0
or 3x = 9
or x = 3
Thus, the centre of the circle is (3, – 2).

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let ABCD be a square and let A(- 1, 2) and C(3, 2) be the given opposite vertices.
Let B(x, y) be the unknown vertex.

Then, AB = BC gives AB² = BC²
So, (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
or 2x + 1 = – 6x + 9
or 8x = 8
or x = 1 … (1)
In A ABC, we have : AB² + BC² = AC2. A(“
So, (x + 1)² + (y – 2)² + (x – 3)² + (y – 2)² = (3 + 1)² + (2 – 2)²
or 2x² + 2y² + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
or 2x² + 2y² – 4x – 8y + 2 = 0
or x² + y² – 2x – 4y + 1 = 0 … (2)
Substituting the value of x from (1) in (2), we get
1 + y² – 2 – 4y + 1 = 0
or y² – 4y = 0
or y(y – 4) = 0
i. e., y – 0 or y = 4
Hence, the required vertices of the square are (1, 0) and (1, 4).

Question 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 1 m from each other. There is a triangular grassy lawn in the plot as shown in the 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 ∆ PQR if C is the origin? Also, calculate the area of the triangle in these cases. What do you observe ?
Solution:
(i) Taking A as the origin, AD and AB are the coordinate axes. Clearly, the points P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.
(ii) Taking C as the origin, CB and CD are the coordinate axes. Clearly, the points P, Q and R are given by (12, 2), (13, 6) and (10, 3) respectively.
We know that the area of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by
12[x1(y2 -y3) + x2(y3 -y1) + x3(y1 – y2)].
∴ Area of ∆ PQR in the 1st case
= 12[4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= 12(4 x (-3) + 3 x (- 1) + 6 x 4)
= 12(- 12 – 3 + 24) = 92sq. units
and Area of ∆ PQR in the 2nd case
= 12[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= 12(12 x 3 + 13 x 1 + 10 x (-4))
= 12(36 + 13 – 40) = 92sq. units
Thus, we observe that the areas are the same in both the cases.

Question 6.
The vertices of a ∆ ABC are A(4, 6), B(l, 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 A ADE and compare it with the area of ∆ ABC.
Solution:
Using only Coordinate Geometry

Since D divides AB in the ratio 1:3, therefore coordinates of D are
(1×1+3×41+3,1×5+3×64), i.e., (134, 234)
From 𝐴𝐸𝐴𝐶 = 14 , we find that
𝐴𝐸𝐴𝐶 = 13
Since E divides AC in the ratio 1:3, therefore coordinates of E are

Also, we find the area (∆ ABC) :
Area ∆ ABC = 152 sq. units [already found]
Area (∆ ADE) : Area (∆ ABC)
= 1532 : 152 = 132 : 12 = 1 : 16

Aliter :
We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

∴ DE || BC
Clearly, A ADE ~ A ABC (Being equiangular, so similar)
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Also, from (1), Area (∆ ADE) : Area (∆ ABC) = 1 : 16.

Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(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 you observe?
(Note : The point which is common to all the three medians is called 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 ∆ ABC, find the coordinates of the centroid of the triangle.
Solution:
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of A ABC.
(i) Since AD is the median of ∆ ABC, therefore
D is the mid-point of BC. Its coordinates are

(6+12,5+42), i.e., (72,92)

(ii) Since P divides AD in the ratio 2 : 1, so its coordinates are

(iii) Since BE is the median of ∆ ABC, so E is the mid-point of AC and its coordinates are E(4+12,2+42), i.e., E(52, 3)

Since Q divides BE in the ratio 2 : 1, so its coordinates are

Since CF is the median of ∆ ABC, so F is the mid-point of AB. Therefore, its coordinates are F(4+62,2+52). i.e., F(5, 72)
Since R divides CF in the ratio 2 : 1, so its coordinates are

Since R divides CF in the ratio 2 : 1, so its coordinates are

(iv) We observe that the points P, Q and R coincide, i.e., the medians AD, BE and CF are concurrent at the point (113, 113). This point is known as the centroid of the triangle.

(v) Let A(x1, y1, x2, y2) and C(x3, y3 be the vertices of ∆ ABC whose medians are AD, BE and CF respectively. So, D, E and F are respectively the mid-points of BC, CA and AB.

The coordinates of E are (𝑥1+𝑥32,𝑦1+𝑦32) coordinates of a point dividing BE in the ratio 2 : 1 are

The coordinates of E are (𝑥1+𝑥32,𝑦1+𝑦32). The coordinates of a point dividing BE in the ratio 2 : 1 are

Similarly, the coordinates of a point dividing CF in the ratio 2 : 1 are

AD, BE and CF and divides them in the ratio 2 : 1.
∴ The medians of a triangle are concurrent and the coordinates of the centroid are
(𝑥1+𝑥2+𝑥33,𝑦1+𝑦2+𝑦33)

Question 8.
ABCD is a rectangle formed by joining 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.
Solution:
Various points are marked in the figure shown on the next page.

Since all the sides are equal but the diagonals are not equal, therefore PQRS is a rhombus.


BSEB Textbook Solutions PDF for Class 10th


Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks for Exam Preparations

Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Solutions can be of great help in your Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 exam preparation. The BSEB STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Books State Board syllabus with maximum efficiency.

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