NCERT Solutions for Class 10 Maths Chapter 7:Coordinate Geometry Exercise 7.1
Coordinate Geometry Exercise 7.1
In this page we have NCERT Solutions for Class 10 Maths Chapter 7:Coordinate Geometry for
Exercise 7.1 on pages 161 and 162. Hope you like them and do not forget to like , social_share
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Question 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) Question 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 . Question 3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear. Question 4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle Question 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. Question 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) Question 7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). Question 8. . Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units Question 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. Question 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
y^{2} +6y-27=0
(y+9)(y-3)=0
So y=-9 or 3 Solution 9:
Now
QP=QR
Squaring both the sides
25+16=x^{2} +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
x^{2} -6x+9 +y^{2}-12y +36=x^{2} +6x+9 +y^{2} -8y+16
-12x-4y+20=0
Or
3x+y-5=0 ( dividing by -4)
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