- Flash Back from Class IX notes
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- Distance formula
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- Section Formula
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- Area of triangle
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- How to Solve the line segment bisection ,trisection and four-section problem's
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- How to Prove three points are collinear
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- How to solve general Problems of Area in Coordinate geometry

- Coordinate Geometry Problem and Solutions
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- Coordinate Geometry Short questions
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- Coordinate Geometry 3 Marks Questions
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- Coordinate Geometry 5 Marks questions

In this page we have *NCERT Solutions for Class 10 Maths Coordinate Geometry* for
EXERCISE 2 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

Let P (x, y) be the required point. Using the section formula, we obtain.

So

Therefore, the point is (1, 3).

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining

The given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2. And Q divides PQ in 2:1

Using the section formula, we obtain.

=2

=-5/3

=0

=-7/3

It can be observed that Niharika posted the green flag at ¼ of the distance AD i.e., (1/4) ×100=25 m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).

Similarly, Preeti posted red flag at 1/5 of the distance AD i.e., (1/5) ×100=20 m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).

Distance between these flags by using distance formula = GR

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y). Now from mid-point section formula

Therefore, Rashmi should post her blue flag at 22.5m on 5th line

Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by

point (−1, 6) be k: 1.

Then from section formula

Solving this

k=2/7

Therefore, the required ratio is 2:7

Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by

x-axis be k:1

Using the section formula, we obtain.

Now we know that point lies on x-axis so y=0

Or k=1

Then

So coordinates are (-3/2,0)

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are

(1+x)/2, (2+6)/2

or (1+x)/2 ,4

If O is the mid-point of BD, then the coordinates of O are

(3+4)/2, (5+y)/2

or 7/2, (5+y)/2

Since both the coordinates are of the same point O,

So

(1+x)/2= 7/2

or x=6

(5+y)/2= 4

or y=3

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, −3), which is the center of the circle.

So

(2, -3) = [(x+1)/2, (y+4)/2]

Or

(x+1)/2= 2

x=3

(y+4)/2=-3

y=-10

Therefore, coordinates are (3, -10)

The coordinates of point A and B are (−2, −2) and (2, −4) respectively.

Since AP = (3/7) AB

Therefore, AP: PB = 3:4

So P divides the line into the ration 3:4

Let P, Q, R are the point dividing line AB into four equal parts ,it can be observed that points P, Q, Rare dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

Now coordinates of Point P will be given

= (-1,7/2)

Now coordinates of Point Q will be given

= (0,5)

Now coordinates of Point R will be given

= (1,13/2)

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.

Now we know that Area of rhombus is given by

A= (1/2) d

Where d

So here we just need to find the diagonals and then area can be calculated easily

So Area will be

A= (1/2) ×6√2×4√2=24

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