RD SHARMA Solutions for Class 10 Maths Chapter 6 - Co-ordinate Geometry
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.1
Solution 1
Solution 2
Solution 3
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.2
Solution 1 (i)
Solution 1 (ii)
Solution 1 (iii)
Solution 1 (iv)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29(i)
Solution 29(ii)
Solution 29(iii)
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
For an equilateral triangle, the perpendicular bisector of any side passes through the opposite vertex.
Both the points, (0, -3) and (0, 3), lie on the y-axis equidistant from the origin. Hence, the perpendicular bisector joining these two points is the x-axis.
Any point on the x-axis has the coordinates (a, 0).
The distance between (0, -3) and (0, 3) is 6.
Hence, the distance between (a, 0) and (0, 3) should also be 6.
62 = (a - 0)2 + (0 - 3)2
36 = a2 + 9
a2 = 27
Solution 49
Solution 50
Using the distance formula,
AB=
AC=
BC=
PQ=
PR=
QR=
Now,
ΔABC ~ ΔPQR
by the SSS test.
Solution 51
Solution 52
Solution 53
Solution 54
Solution 55
Solution 56
Solution 57
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.3
Solution 1
Solution 2
(i) 
(ii) 
(iii) 
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11(i)
Solution 11(ii)
Solution 12
Solution 13
Solution 14
*Note: (i) Answer given in the book is incorrect.
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Let the point on the x-axis be (a, 0).
Let this point divide the line segment AB in the ratio of r : 1.
Using the section formula for the y-coordinate, we get
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Let P divides the line segment AB is the ratio k: 1.
So, the ratio is 1:1.
Also, 
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
The difference between the x-coordinates of A and B is 6 - 1 = 5
Similarly, the difference between the y-coordinates of A and B is 7 - 2 = 5
Hence, if the line segment joining A(1, 2) and B(6, 7) is divided into 5 equal parts by the points P, Q, R and S, then the coordinates of P, Q, R and S can be found out by increasing the x and the y coordinates of A by 1 successively.
Hence, the coordinates of P are (1 + 1, 2 + 1) = (2, 3)
The coordinates of Q are (2 + 1, 3 + 1) = (3, 4)
The coordinates of R are (3 + 1, 4 + 1) = (4, 5)
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Solution 49
Solution 50 (i)
Solution 50 (ii)
Given: ABCD is a parallelogram
We know that, diagonals of a parallelogram bisect each other.
Therefore, midpoints of diagonals coincide
The midpoints of AC and BD coincide.
Solution 51
Solution 52 (i)
As P and Q trisect AB and P is near to A.
Therefore, P divides AB in the ratio 1:2.
Also, Q divides AB in the ration 2:1.
Solution 52 (ii)
Solution 53
Solution 54
Solution 55
Solution 56
Given points are A(3,-5) and B(-4,8).
P divides AB in the ratio k : 1.
Using the section formula, we have:
Coordinate of point P are {(-4k+3/k+1)(8k-5/k+1)}
Now it is given, that P lies on the line x+y = 0
Therefore,
-4k+3/k+1 + 8k-5/k+1 =0
=> -4k+3+8k-5 =0
=> 4k -2 =0
=> k=2/4
=> k=1/2
Thus, the value of k is 1/2.
Solution 57
Solution 58
Solution 59
Solution 60
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.4
Solution 1(i)
Solution 1(ii)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.5
Solution 1
Solution 1 (iv)
Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is
Therefore, area of triangle with given vertices is
Hence, the area of triangle will be 24 sq. units.
Solution 1 (v)
Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is
Therefore, area of triangle with given vertices is
Hence, the area of triangle will be 53 sq. units.
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13 (i)
Solution 13 (ii)
Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is
Hence, the value of k is 3.
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Let the points (a, a2), (b, b2), (0, 0) represent a triangle. If we can prove that the area of the triangle so formed is not equal to zero, then we can prove that the points (a, a2), (b, b2), (0, 0) are never collinear.
Area of a triangle is given by
Now b≠a≠0.
So,
b - a≠0,
Δ≠0
Thus, points (a, a2), (b, b2), (0, 0) are never collinear.
Solution 22
Area
of a triangle is given by
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Let the points be A, B and C respectively.
If A, B and C are collinear, then the area of ∆ABC is zero.
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Chapter 6 - Co-ordinate Geometry Exercise 6.63
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
So, the correct option is (d).
Solution 8
Solution 9
Solution 10
Chapter 6 - Co-ordinate Geometry Exercise 6.64
Solution 11
Solution 12
Note: The answer does not match the options.
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 25
Solution 24
As the points A, B and C are collinear, then the area formed by these three points is 0.
Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is
Hence, the value of p is -2.
Chapter 6 - Co-ordinate Geometry Exercise 6.65
Solution 26
We know, distance of point (x, y) from x-axis is y.
Hence, distance of point (4, 7) from x-axis is 7.
Hence, correct option is (b).
Solution 27
We know distance of point (x, y) from y-axis is x.
Hence distance of point (4, 7) from y-axis is 4.
Hence, correct option is (a).
Solution 28
Given P is a point on x-axis
Hence P = (x, 0)
Distance from the origin is 3
Hence P = (3, 0)
Given Q is a point on y-axis
So Q is (0, y)
Given that OP = OQ
implies OQ = 3
Distance of Q from the origin is 3
Hence y = 3
implies Q = (0, 3)
Hence, correct option is (a).
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
The Centroid of the triangle is given by
x=-8 and y=3 satisfy(a) 3x + 8y = 0.
Solution 36
Solution 37
Solution 38
Solution 39
Chapter 6 - Co-ordinate Geometry Exercise 6.66
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Chapter 6 - Co-ordinate Geometry Exercise 6.67
Solution 49
Solution 50
Solution 51
As the point (k, 0) divides the line segment AB in the 1:2
Hence, option (d) is correct.
Other Chapters for CBSE Class 10 Mathematics
Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- ProbabilityRD SHARMA Solutions for CBSE Class 10 Subjects
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