NCERT Solutions class 10 maths chapter 7
Coordinate Geometry
NCERT Solutions class 10 maths chapter 7 Exercise 7.1
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)
Answer
Distance between two points (x
1, y
1) and (x
2, y
2) is
Distance
.
.
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 discussed in Section 7.2.
Answer
Distance between points (0, 0) and (36, 15)
=
√(36-0)2 +
(15-0)2
=
√362 + 152
=
√ 1296 + 225
=
√ 1521
=39
The distance between town A and B will be 39 km.
Question 3
. Determine if the points (1, 5), (2, 3) and (-2, -11) are
collinear.
Answer 3
In collinear AB + BC = CA
A = (1, 5) B = (2, 3) and C = (-2, -11)
.
Hrer AB + BC ≠ CA
So the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.
Question 4
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an
isosceles triangle.
Answer
Since two sides of any isosceles triangle are equal.
Let the three point of triangle are A(5, – 2), B(6, 4), and C(7, – 2)
.
Here AB = BC
here Two sides of isosceles triangle are equal .Given points are vertices of an isosceles triangle.
Question 5
In a classroom, 4 friends are seated at the points A, B, C and D as
shown in Fig. 7.8. 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.
.
Answer
As shown figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).
Using Distance formula
.
AB = BC = CD = DA = 3√2
. ABCD is a square because all sides are of equal length so Champa was 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)
Answer
(i) Let the points A(- 1, – 2), B(1, 0),C( – 1, 2), and D( – 3, 0)
Side length = AB = BC = CD = DA = 2√2
Diagonal = AC = BD = 4
Therefore, the given points are the vertices of a square.
(ii) Let the points (- 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D
of the given quadrilateral respectively.
All sides of this quadrilateral are of different then it is a general quadrilateral.
(iii) Let the points A(4, 5), B(7, 6), C(4, 3), and D(1, 2)
NCERT Solutions for Class 10 Chapter 7-9
Opposite sides of this quadrilateral are of the same length. and the diagonals are of different lengths.
So the given points are the vertices of a parallelogram.
Question 7
Find the point on the x-axis which is equidistant from (2, – 5) and
(- 2, 9).
Answer
Let the point on x-axis be (x,0).
A(x, 0); B(2, – 5) and C(- 2, 9)
Therefore, the point is (- 7, 0).
Question 8
Find the values of y for which the distance between the points P (2,
– 3) and Q (10, y) is 10 units.
Answer
Distance between (2, – 3) and (10, y) is 10.
Squaring both sides,
64 +(y+3)
2 = 100
(y+3)
2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = −6
y = 3 or -9.
Question 9
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the
values of x. Also find the distance QR and PR.
Answer
Given: Q (0, 1) is equidistant from P (5, – 3) and R (x, 6)
PQ = QR
So
Squaring both the sides we,get
41 = x
2+25
xx
2 = 16
x = ± 4
x = 4 or x = -4
Coordinates of Point R(4, 6) OR
If R (4, 6), then QR
If R(-4, 6),then QR
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).
Answer
Point Q(x, y) is equidistant from P(3, 6) and R( – 3, 4).
We know in equidistant