RD SHARMA Solutions for Class 10 Maths Chapter 6 - Co-ordinate Geometry
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.1
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.2
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
Using the distance formula,
AB=
AC=
BC=
PQ=
PR=
QR=
Now,
ΔABC ~ ΔPQR
by the SSS test.
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.3
(i)
(ii)
(iii)

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
Let P divides the line segment AB is the ratio k: 1.
So, the ratio is 1:1.
Also,
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)


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.
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.
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.
Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.4


Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.5

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.
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.


Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is
Hence, the value of k is 3.
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.
Area
of a triangle is given by
Let the points be A, B and C respectively.
If A, B and C are collinear, then the area of ∆ABC is zero.
Chapter 6 - Co-ordinate Geometry Exercise 6.63
So, the correct option is (d).
Chapter 6 - Co-ordinate Geometry Exercise 6.64
Note: The answer does not match the options.
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
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).
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).
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).
The Centroid of the triangle is given by
x=-8 and y=3 satisfy(a) 3x + 8y = 0.
Chapter 6 - Co-ordinate Geometry Exercise 6.66
Chapter 6 - Co-ordinate Geometry Exercise 6.67
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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