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Class 10 Maths NCERT Solutions for Coordinate Geometry



In this page we have NCERT book Solutions for Class 10th Maths:Coordinate Geometry for EXERCISE 1 . Hope you like them and do not forget to like , social_share and comment at the end of the page.

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)

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 .

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 triangle

5) In a classroom, 4 friends are seated at the points A, B, C and D as shown below. Champaand Chameli walk into the class and after observing for a few minutes Champa asks Chameli,

“Don’t you think ABCD is a square?” Chameli disagrees.

Using distance formula, find which of them is correct.

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)

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(10, y) is 10 units

9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

10. Find a relation between x and y such that the point (x, y) is equidistant from the point

(3, 6) and (– 3, 4).

 

Solution 1:

Distance between the points AB is given by

Solution 2:

Let P(0,0)  and Q(36,15)  be the given points.

Distance between the points PQ is given b

The position of town A and B can be represented as points P and Q respectively,So distance between town will be 39 km

 

Solution3:

Lets us denote point by P(1,5) , Q(2,3)  and R(-2,-11)

If the points are not collinear, then we should be able to form the triangle

Lets us find the length of PQ, QR and PR  by distance formula

Clearly  None of these is true

PQ+QR=PR

PR+PQ=QR

PQ=QR+PR

Hence  they are not collinear

Solution 4)

Lets us denote point by P(5,-2) , Q(6,4)  and R(7,-2)

Lets us find the length of PQ, QR and PR  by distance formula

Now PQ =QR  ,so it is an isosceles triangle

 

Solution 5)

As per the figure given,The coordinates of the points A,B,C and D  are (3,4), (6,7) ,(9,4)  and (6,1)

Lets us find the length of AB, BC ,CD  and AD by distance formula

So all the sides are equal. But we cannot  still say that it is square as rhombus has all  the sides  equal also.

Now we know that a square has both the diagonal equal also,So lets us calculate the diagonal’s

Hence AC=BD

So it is a square.

So  champa is correct

 

Solution 6)

In these type of problem, we need to find of length of each segment, then check with the properties of different type of quadrilateral

i.e for points A,B,C and D

Line segments are AB,BC,AC,CD,AD  and  BD

We need to find the length of each of these

i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

Here the side AB,BC,CD and AD are equal and diagonal AC and BD are equal. So this is a square

 

ii)  (–3, 5), (3, 1), (0, 3), (–1, – 4)

 

Now here  AC+BC=AB

So  that means ABC are collinear points.

So it is not a quadrilateral infact

iii) (4,5) ,(7,6), (4,3) (1,2)

Now AB=CD  and BC=DA

Now it could be rectangle or parallelogram

But diagonal  AC is not equal diagonal  BD

So It is a parallelogram

 

Solution 7:

Since the point lies on X axis, the point should be of the form (a,0)

Now (a,0) is equidistant from  both the given points

Squaring both the sides

(x-2)2 + (0+5)2 =(x+2)2 + (0-9)2

Solving it  we get

x=-7

Solution 8:

Acoording to the question

PQ=10

Squaring both the sides

 y2 +6y-27=0

(y+9)(y-3)=0

So y=-9 or 3

 

Solution   9:

Now

QP=QR

Squaring both the sides

25+16=x2 +25

x= -4 or 4

So point  R is either (4,6)  or (-4,6)

 

Solution: 10

Let the point P(x,y) is equidistant from the point Q( 3,6)  and R(-3,4)

PQ=PR

Squaring both the sides

x2  -6x+9 +y2-12y +36=x2 +6x+9 +y2 -8y+16

-12x-4y+20=0

Or

3x+y-5=0  ( dividing by -4)


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