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Coordinate Geometry Class 10th Mathematics CBSE Solution

Class 10th Mathematics CBSE Solution
Exercise 7.1
  1. Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii)…
  2. Find the distance between the points (0, 0) and (36, 15). Can you now find the…
  3. Determine if the points (1, 5), (2, 3) and (- 2, - 11) are collinear…
  4. Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles…
  5. In a classroom, 4 friends areseated at the points A, B, C and D as shown in Fig.…
  6. Name the type of quadrilateral formed, if any, by the following points, and give…
  7. Find the point on the x-axis which is equidistant from (2, 5) and (2, 9).…
  8. Find the values of y for which the distance between the points P(2, 3) and Q…
  9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also…
  10. Find a relation between x and y such that the point (x, y) is equidistant from…
Exercise 7.2
  1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3)…
  2. Find the coordinates of the points of trisection of the line segment joining…
  3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD,…
  4. Find the ratio in which the line segment joining the points (- 3, 10) and (6, -…
  5. Find the ratio in which the line segment joining A(1, 5) and B(4, 5) is divided…
  6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken…
  7. Find the coordinates of a point A, where AB is the diameter of a circle whose…
  8. If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P…
  9. Find the coordinates of the points which divide the line segment joining A(- 2,…
  10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (- 1, 4) and (-…
Exercise 7.3
  1. Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, - 4)…
  2. In each of the following find the value of k, for which the points are collinear…
  3. Find the area of the triangle formed by joining the mid-points of the sides of…
  4. Find the area of the quadrilateral whose vertices, taken in order, are (- 4, -…
  5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a…
Exercise 7.4
  1. Determine the ratio in which the line 2x+y-4 = 0 divides the line segment…
  2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are…
  3. Find the centre of a circle passing through the points (6, - 6), (3, - 7) and…
  4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the…
  5. The Class X students of asecondary school inKrishinagarhave been allotteda…
  6. The vertices of a ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to…
  7. Let A (4, 2), B(6, 5) and C (1, 4) be the vertices of ABC (i) The median from A…
  8. ABCD is a rectangle formed by the points A(-1, -1), B(- 1, 4), C(5, 4) and D(5,…
Exercise 7.1
Question 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)


Answer:

(i) We know that distance between the two points is given by:

d = √[(x1 – x2)2 + (y1 – y2)2]


Distance between A(2, 3) and B(4, 1) is:


f =√ [(4 - 2)2 + (1 – 3)2]


= √[4 + 4]= √8


= 2√2


(ii) Distance between A(-5, 7) and B(-1, 3) is:


f = √[(-5 + 1)2 + (7 – 3)2]


= √[16 + 16] = √32


= 4√2


(iii) Distance between A(a, b) and B(-a, -b) is:


f = √[(a - (- a))2 + ( b - (-b))2]

f = √[(a+ a)2 + (b+ b)2]

f = √[2a + 2b] = √(4a2 + 4b2)


f = 2√(a2 + b2)


Question 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


Answer:


To find: Distance between two points
Given: Points A(0, 0), B(36, 15)



For two points A(x1, y1) and B(x2, y2), 

Distance is given by

f = [(x2 - x1)2 + (y2 - y1)2]1/2

Distance between (0, 0) and (36, 15) is:


f = [(36 – 0)2 + (15 – 0)2]1/2


= [1296 + 225]1/2 = (1521)1/2


= 39

Hence Distance between points A and B is 39 units

Yes, we can find the distance between the given towns A and B. Let us take town A at origin point (0, 0)


Hence, town B will be at point (36, 15) with respect to town A


And, as calculated above, the distance between town A and B will be 39 km


Question 3.

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.


Answer:

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.
Let A = (1, 5), B = (2, 3) and C = (- 2,-11)
Case 1)


Since AB + BC ≠ CA
Case 2)
Now,




BA + AC ≠ BC
Case 3)
Now,




As BC+CA≠BA
As three of the cases are not satisfied.
Hence the points are not collinear.


Question 4.

Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle


Answer:



Distance between two points A(x1, y1) and B(x2, y2) is given by

D = √(x2 - x1)2 + (y2 - y1)2

Let us assume that points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively as shown in the figure.

AB = [(5- 6)2 + (-2- 4)2]1/2


= 


= 


BC = [(6- 7)2 + (4 + 2)2]1/2


= 


= 


CA = [(5- 7)2 + (-2+ 2)2]1/2


= 


= 2


Therefore, AB = BC


As two sides are equal in length, therefore, ABC is an isosceles triangle


Question 5.

In a classroom, 4 friends areseated at the points A, B, C and D as shown in Fig. 7.8 Champaand Chameli walk into the classand after observing for a fewminutes Champa asks Chameli,“Don’t you think ABCD is asquare?” Chameli disagrees.Using distance formula, findwhich of them is correct



Answer:

It can be seen that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of 4 friends



distance between two points A(x1, y1) and B(x2, y2) is given by

D = √(x2 - x1)2 + (y2 - y1)2

Hence,


AB = [(3- 6)2 + (4 - 7)2]1/2


= 


= 


BC = [(6 - 9)2 + (7- 4)2]1/2


= 


= 


CD = [(9- 6)2 + (4- 1)2]1/2


= 


= 


AD = [(3 - 6)2 + (4 - 1)2]1/2


= 


= 


Diagonal AC = [(3 - 9)2 + (4 - 4)2]1/2


= 


= 6


Diagonal BD = [(6- 6)2 + (7- 1)2]1/2


= 


= 6


It can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are of the same length


Therefore, ABCD is a square and hence, Champa was correct


Question 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)


Answer:

To Find: Type of quadrilateral formed

(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

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x1, y1) and (x2, y2).

D = √(x2 - x1)2 + (y2 - y1)2

AB = [(-1- 1)2 + (-2- 0)2]1/2


=  = 


= 2√2


BC = √[(1 + 1)2 + (0- 2)2]


=  = 


= 2√2


CD = √[(-1 + 3)2 + (2- 0)2]


=  = 


= 2√2


AD = √[(-1+ 3)2 + (-2- 0)2]


=  = 


= 2√2


Diagonal AC = √[(-1 + 1)2 + (-2 - 2)2]


= 


= 4


Diagonal BD = √[(1 + 3)2 + (0- 0)2]


= 


= 4


It is clear that all sides of this quadrilateral are of the same length and the diagonals are of the same length. 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


The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x1, y1) and (x2, y2).

D = √(x2 - x1)2 + (y2 - y1)2

AB = √[(-3- 3)2 + (5- 1)2]


=  = 


= 2√13


BC = √[(3- 0)2 + (1- 3)2]


= 


= 


CD = √[(0 + 1)2 + (3+ 4)2]


=  = 


= 5√2


AD = √[(-3+ 1)2 + (5+ 4)2]


= 


= 


We can observe that all sides of this quadrilateral are of different lengths.


Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc


(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