In this post, you will find the NCERT solutions for class 10 maths ex 7.1. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 7.1
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)
Solution:
(i) Using the distance
formula to find the distance between points (2, 3) and (4, 1), we have
d = 
(ii) Using the distance formula to
find the distance between points (–5, 7) and (–1, 3), we have
d = 
d = 
2. Find the distance between the points (0, 0)
and (36, 15). Also, find the distance between towns A and B if the town B is
located at 36 km east and 15 km north of the town A.
Solution: Using the distance formula to find the
distance between points (0, 0) and (36, 15), we have
d = 
Town B is located
at 36 km east and 15 km north of town A. So, the location of town A and B can
be shown as:
To
find the distance between them, we use distance formula:
d = 
Thus,
d = 39 km
3. Determine if the points (1, 5), (2, 3) and
(–2, –11) are collinear.
Solution: Let A(1, 5), B(2, 3) and C(–2, –11)
are three points.
Let’s
use the distance formula to find the distances AB, BC and CA.
AB
= 
BC
= 
CA
= 
Therefore,
the points A(1, 5), B(2, 3) and C(–2, –11) are not collinear.
4. Check whether (5, –2), (6, 4) and (7, –2)
are the vertices of an isosceles triangle.
Solution:
Let A(5, –2),
B(6, 4) and C(7, –2) are three points.
Let’s
use the distance formula to find the distances AB, BC and CA.
AB
= 
BC
= 
CA
= 
Therefore,
A(5, –2), B(6, 4) and C(7, –2) are vertices of an isosceles triangle.
5. In a classroom, 4 friends are seated at the
points A, B, C and D as shown in the figure. Champa and 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.
Solution:
We have A(3, 4),
B(6, 7), C(9, 4) and D(6, 1) are the four points where four friends are seated.
Let’s use the distance
formula to find the distances AB, BC, CD and DA.
AB
= 
BC
= 
CD
= 
DA
= 
Now,
we will check the length of its diagonals.
Let’s use the distance
formula to find the distances AC and BD.
AC
= 
BD
= 
Since
all the four sides and two diagonals of quadrilateral ABCD are equal.
We
can definitely say that ABCD is a square.
Therefore,
Champa 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)
Solution: (i) Let A(–1, –2), B(1, 0), C(–1, 2)
and D(–3, 0) are the three points.
Let’s
use the distance formula to find the distances AB, BC, CD and DA.
AB
= 
BC
= 
CD
= 
DA
= 
Now,
we will check the lengths of diagonals.
Let’s use the distance
formula to find the distances AC and BD.
AC
= 
BD
= 
Since
all the four sides and two diagonals of quadrilateral ABCD are equal.
Therefore, the quadrilateral
ABCD is a square.
(ii) Let A(–3, 5), B(3, 1), C(0, 3) and D(–1, –4) are the four points.
Let’s
use the distance formula to find the distances AB, BC, CD and DA.
AB
= 
BC
= 
CD
= 
DA
= 
Therefore,
we cannot give any name to the quadrilateral ABCD.
(iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1,
2) are the four points.
Let’s
use the distance formula to find the distances AB, BC, CD and DA.
AB
= 
BC
= 
CD
= 
DA
= 
Now,
we will check the lengths of diagonals.
Let’s use the distance
formula to find the distances AC and BD.
AC
= 
BD
= 
Here
the diagonals of the quadrilateral ABCD are not equal.
In
quadrilateral ABCD, the opposite sides are equal but the diagonals are not
equal.
Therefore,
the quadrilateral is a parallelogram.
7. Find the point on the x-axis
which is equidistant from (2, –5) and (–2, 9).
Solution: Let the point be (x, 0) on x-axis which is
equidistant from (2, –5) and (–2, 9).
Using
distance formula according to the given conditions, we have
The
distance between (x, 0) and (2, –5) =
The distance between (x, 0) and (–2,
9)
⇒ 
⇒ 
⇒ x2 + 4 − 4x +
25 = x2 + 4 + 4x +81
⇒ −4x
+ 29 = 4x + 85
⇒ 8x
= −56
⇒ x
= −7
Therefore, the
point on the x-axis which is
equidistant from (2, –5) and (–2, 9) is (–7, 0).
8. Find the values of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units.
Solution: Using the distance formula, we have
⇒ 
⇒ 
⇒ 
100
= 73 + y2 + 6y
⇒ y2 + 6y – 27 =
0
Let’s
solve this quadratic equation by factorization.
⇒ y2 + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y
+ 9) (y − 3) = 0
⇒ y
= 3, −9
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.
Solution:
It is given that Q(0, 1) is equidistant from P(5, –3) and R(x,
6).
Since
PQ = RQ
Using
the distance formula, we get
⇒ 
⇒ 
⇒ 
⇒ 25 + 16 = x2 + 25
⇒ x2 = 16
⇒ x
= 4, −4
Using
the distance formula to find QR, we have
Using
the value of x = 4, QR =
Using
the value of x = –4, QR =
Using
the distance formula to find PR, we have
Using
the value of x = 4, PR =
Using
the value of x = –4, PR =
Therefore,
x = 4, –4
QR
=√41 and PR = √82, 9√2
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:
According to
question, the point (x, y) is equidistant from the point (3, 6)
and (–3, 4).
Using
the distance formula, we have
⇒ 
⇒ 
⇒ x2 + 9 – 6x + y2 + 36 –12y = x2
+ 9 + 6x + y2 + 16 – 8y
⇒ −6x
− 12y + 45 = 6x − 8y + 25
⇒ 12x
+ 4y = 20
⇒ 3x
+ y = 5
This is the required relation between x and y.