# 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.

6^{2 }= (a - 0)^{2 }+ (0 - 3)^{2}

36 = a^{2 }+ 9

a^{2 }= 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 (x_{1}, y_{1}), (x_{2}, y_{2})
and (x_{3}, y_{3}) is

Therefore, area of triangle with given vertices is

Hence, the area of triangle will be 24 sq. units.

Area of triangle
with vertices (x_{1}, y_{1}), (x_{2}, y_{2})
and (x_{3}, y_{3}) is

Therefore, area of triangle with given vertices is

Hence, the area of triangle will be 53 sq. units.

Area of triangle
with vertices (x_{1}, y_{1}), (x_{2}, y_{2})
and (x_{3}, y_{3}) is

Hence, the value of k is 3.

Let the points (a, a^{2}), (b, b^{2}), (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, a^{2}), (b, b^{2}), (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, a^{2}), (b, b^{2}), (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 (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) 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- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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