The exercise solutions provided here are created with a vision to assist the students in their board exam preparation. Exercise solution covers all the questions given in Exercise 7.1 of the NCERT textbook. The NCERT solutions are researched and prepared by our subject experts. Studying this study material will aid you in solving different questions on distance from a point.
Exercise solutions are provided in PDF format for your easy access and download. By studying this solution, you can clear your doubts about distance calculation. By studying using NCERT Class 10 Maths Solutions, you can get well acquainted with all the important formulas. Here, you can also get knowledge of alternative methods to solve distance-from-point problems.
Topics Covered in Exercise 7.1
- Introduction
- Distance formula
How it is helpful
- Provides questions with varied difficulty for your practice
- Has day to day examples for your easy understanding
- Studying using the NCERT Class 10 solution will make you thorough with the concepts of distance calculation from a point
NCERT Solutions for Class 10 Maths Chapter 7- Coordinate Geometry Exercise 7.1
Access other exercise solutions of Class 10 Maths Chapter 7- Coordinate Geometry
Students can access the other exercises of NCERT Class 10 Maths Solutions Chapter 7 using the links below.
Exercise 7.1- 10 Questions (8 practical based questions, 2 reasoning questions)
Exercise 7.2– 10 Questions ( 8 long answer questions, 2 short answer questions)
Exercise 7.3– 5 Questions (3 long answer questions, 2 practical based questions)
Exercise 7.4– 8 Questions (6 long answer questions, 1 practical based question, 1 reasoning question)
Access Answers of Maths NCERT Class 10 Chapter 7 –Coordinate Geometry Exercise 7.1
Exercise 7.1 Page No: 161
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)
Solution:
Distance formula to find the distance between two points (x1, y1) and (x2, y2) is, say d,
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 discussed in Section 7.2.
Solution:
Let us consider, town A at point (0, 0). Therefore, town B will be at point (36, 15).
Distance between points (0, 0) and (36, 15)
In section 7.2, A is (4, 0) and B is (6, 0)
AB2 = (6 – 4)2 – (0 – 0)2 = 4
The distance between town A and B will be 39 km. The distance between the two towns A and B discussed in Section 7.2 is 4 km.
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution: The sum of the lengths of any two line segments is equal to the length of the third line segment then all three points are collinear.
Consider, A = (1, 5) B = (2, 3) and C = (-2, -11)
Find the distance between points; say AB, BC and CA
Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.
4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution:
Since two sides of any isosceles triangle are equal. To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.
Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively.
This implies, whether given points are vertices of an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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:
From figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).
Find distance between points using distance formula, we get
All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.
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)
Solution:
(i) Let the points (- 1, – 2), (1, 0), ( – 1, 2), and ( – 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
Side length = AB = BC = CD = DA = 2√2
Diagonal Measure = AC = BD = 4
Therefore, the given points are the vertices of a square.
(ii) Let the points (- 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
Its also seen that points A, B and C are collinear.
So, the given points can only form 3 sides i.e, a triangle and not a quadrilateral which has 4 sides.
Therefore, the given points cannot form a general quadrilateral.
(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
Opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.
7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
To find a point on x-axis. Therefore, its y-coordinate will be 0. Let the point on x-axis be (x,0).
Consider A = (x, 0); B = (2, – 5) and C = (- 2, 9).
Simplify the above equation,
Remove square root by taking square both the sides, we get
(2 – x)2 + 25 = [-(2 + x)]2 + 81
(2 – x)2 + 25 = (2 + x)2 + 81
x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81
8x = 25 – 81 = -56
x = -7
Therefore, the point is (- 7, 0).
8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
Given: Distance between (2, – 3) and (10, y) is 10.
Using distance formula,
Simplify the above equation and find the value of y.
Squaring both sides,
64 + (y + 3)2 = 100
(y + 3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = −6
y = 6 – 3 = 3 or y = – 6 – 3 = -9
Therefore, y = 3 or -9.
9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
Solution:
Given: Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), which means PQ = QR
Step 1: Find the distance between PQ and QR using distance formula,
Squaring both the sides, to omit square root
41 = x2 + 25
x2 = 16
x = ± 4
x = 4 or x = -4
Coordinates of Point R will be R (4, 6) or R (-4, 6),
If R (4, 6), then QR
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:
Point (x, y) is equidistant from (3, 6) and ( – 3, 4).
Squaring both sides, (x – 3)2+(y – 6)2 = (x + 3)2 +(y – 4)2
x2 + 9 – 6x + y2+ 36 – 12y = x2 + 9 + 6x + y2 +16 – 8y
36 – 16 = 6x + 6x + 12y – 8y
20 = 12x + 4y
3x + y = 5
3x + y – 5 = 0
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