KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry 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)
Solution:
i) If A(x1, y1) = A(2, 3)
B(x2, y2) = B(4, 1), then
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 1

ii) Let P(x1, y1) = P(-5, 7)
Q(x2, y2) = Q(-1, 3)
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 2

iii) A (a, b), B (-a, -b)
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 3

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns /I and B discussed below as following: ‘A town 6 is located 36 km east and 15 km north of the town A’.
Solution:
Part-I
Let the points be A(0, 0) and B(36, 15)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 4
Part-II
We have A(0, 0) and B(36,15) as the positions of two towns.
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Whether the points A (1, 5), B (2, 3) and C (-2, -11) are collinear?
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 6
Here, AB + BC = AC, then it is straight line.
But \(\sqrt{5}+\sqrt{221} \neq \sqrt{261}\)
∴ These points are non-collinear.

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let the points be A(5, -2), 6(6, 4) and C(7, -2).
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 7
We have AB = BC ≠ AC.
∴ ∆ABC is an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points, A, B, C, and D as shown in the following 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 the distance formula, find which of them is correct.
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 8
Solution:
A(3, 4), B (6, 7), C (9, 4), D (6, 1)
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 9
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 10
Four sides of quadrilateral ABCD are equal to each other.
∵ AB = BC = CD = DA = \(3 \sqrt{2}\)
Diagonal AC = Diagonal BD
∴ ABCD is a square.
∴ Champa is correct among the two.

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

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)
Solution:
(i) Let the points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) of a quadrilateral ABCD.
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 11
⇒ AB = BC = CD = AD i.e., all the sides are equal.
Also, AC and BD (the diagonals) are equal.
∴ ABCD is a square.

(ii) Let the points be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 12
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Q6
We see that \(\sqrt{13}+\sqrt{13}=2 \sqrt{13}\)
i. e., AC + BC = AB
⇒ A, B and C are collinear. Thus, ABCD is not a quadrilateral.

(iii) Let the points be A(4, 5), B(7, 6), C(4, 3) and D(1, 2).
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 14
Since, AB = CD, BC = DA i.e., opposite sides of the quadrilateral are equal.
And AC ≠ BD ⇒ Diagonals are unequal.
∴ ABCD is a parallelogram.

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 15
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 16
we have PA = PB
∴ PA2 = PB2
x2 – 4x + 29 = x2 + 4x + 85
-4x – 4x = 85 – 29
-8x = 56
8x = -56
∴ x = \(\frac{-56}{8}\)
∴ x = – 7
∴ Coordinates of P are (x, 0), it means (-7, 0).
∴ Required point (x, 0) = (-7, 0)
∴ x = -7.

Question 8.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:
The given points are P(2, -3) and Q(10, y).
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 17
Squaring both sides, y2 + 6y + 73 = 100
⇒ y2 + 6y – 27 = 0
⇒ y2 – 3y + 9y – 27 = 0
⇒ (y – 3)(y + 9) = 0
Either y – 3 = 0 ⇒ y = 3
or y + 9 = 0 ⇒ y = -9
∴ The required value of y is 3 or -9.

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 9.
If Q (0, 1) is equidistant from P (5, -3) and R(x, 6), find the value of x. Also, find the distances QR and PR.
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 19
Solution:
Point Q is equidistant from P and R, then PQ = QR, x = ?
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 20
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 21

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

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:
Let the points be A(x, y), B(3, 6) and C(-3, 4).
∴ AB = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
And AC = \(\sqrt{[(-3)-x]^{2}+(4-y)^{2}}\)
Since, the point (x, y) is equidistant from (3, 6) and (-3, 4).
∴ AB = AC
⇒ \(\sqrt{(3-x)^{2}+(6-y)^{2}}=\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
Squaring both sides,
(3 – x)2 + (6 – y)2 = (-3 – x)2 + (4 – y)2
⇒ (9 + x2 – 6x) + (36 + y2 – 12y)
⇒ (9 + x2 + 6x) + (16 + y2 – 8y)
⇒ 9 + x2 – 6x + 36 + y2 – 12y – 9 – x2 – 6x – 16 – y2 + 8y = 0
⇒ – 6x – 6x + 36 – 12y – 16 + 8y = 0
⇒ – 12x – 4y + 20 = 0
⇒ -3x – y + 5 = 0
⇒ 3x + y – 5 = 0 which is the required relation between x and y.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter Chapter 7 Coordinate Geometry Ex 7.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter Chapter 7 Coordinate Geometry Exercise 7.1, drop a comment below and we will get back to you at the earliest

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