NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Table of Contents
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
(i) Let A (2, 3) and B (4, 1)
\(\quad\) AB = \(\sqrt{(4 – 2)^{2} + (1 – 3)^{2}}\)
\(\quad \quad\) = \(\sqrt{(2)^{2} + (-2)^{2}}\)
\(\quad \quad\) = \(\sqrt{4 + 4}\)
\(\quad \quad\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
(ii) Let A (-5, 7) and B (-1, 3)
\(\quad\) AB = \(\sqrt{(-1 – (-5))^{2} + (3 – 7)^{2}}\)
\(\quad \quad\) = \(\sqrt{(-1 + 5)^{2} + (-4)^{2}}\)
\(\quad \quad\) = \(\sqrt{16 + 16}\)
\(\quad \quad\) = \(\sqrt{32}\) = \(4\sqrt{2}\)
(iii) Let A (a, b) and B (-a, -b)
\(\quad\) AB = \(\sqrt{(-a – a)^{2} + (-b – b)^{2}}\)
\(\quad \quad\) = \(\sqrt{(-2a)^{2} + (-2b)^{2}}\)
\(\quad \quad\) = \(\sqrt{4a^{2} + 4b^{2}}\)
\(\quad \quad\) = \(\sqrt{4(a^2 + b^2)}\) = 2\(\sqrt{a^2 + b^2}\)
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
Let A(0, 0) and B (36, 15)
\(\quad\) AB = \(\sqrt{(36 – 0)^{2} + (15 – 0)^{2}}\)
\(\quad \quad\) = \(\sqrt{(36)^{2} + (15)^{2}}\)
\(\quad \quad\) = \(\sqrt{1296 + 225}\)
\(\quad \quad\) = \(\sqrt{1521}\) = 39
Question 3:
Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Answer
Let A (1, 5), B (2, 3) and C (-2, -11)
\(\quad\) AB = \(\sqrt{(2 – 1)^{2} + (3 – 5)^{2}}\) = \(\sqrt{(1)^{2} + (-2)^{2}}\) = \(\sqrt{1 + 4}\) = \(\sqrt{5}\)
\(\quad\) BC = \(\sqrt{(-2 – 2)^{2} + (-11 – 3)^{2}}\) = \(\sqrt{(-4)^{2} + (-14)^{2}}\) = \(\sqrt{16 + 196}\) = \(\sqrt{212}\)
\(\quad\) CA = \(\sqrt{(-2 – 1)^{2} + (-11 – 5)^{2}}\) = \(\sqrt{(-3)^{2} + (-16)^{2}}\) = \(\sqrt{9+ 256}\) = \(\sqrt{265}\)
Since AB + BC ≠ CA
Therefore, 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
Let A (5, -2), B (6, 4) and C (7, -2)
\(\quad\) AB = \(\sqrt{(6 – 5)^{2} + {4 – (-2)}^{2}}\) = \(\sqrt{(1)^{2} + (6)^{2}}\) = \(\sqrt{1 + 36}\) = \(\sqrt{37}\)
\(\quad\) BC = \(\sqrt{(7 – 6)^{2} + (-2 – 4)^{2}}\) = \(\sqrt{(1)^{2} + (-6)^{2}}\) = \(\sqrt{1 + 36}\) = \(\sqrt{37}\)
\(\quad\) CA = \(\sqrt{(7 – 5)^{2} + {-2 – (-2)}^{2}}\) = \(\sqrt{(2)^{2} + (0)^{2}}\) = \(\sqrt{4+ 0}\) = 2
Here, AB = BC
Hence, the 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
Let A (3, 4), B (6, 7), C (9, 4) and D (6, 1)
\(\quad\) AB = \(\sqrt{(6 – 3)^{2} + {7 – 4}^{2}}\) = \(\sqrt{(3)^{2} + (3)^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
\(\quad\) BC = \(\sqrt{(9 – 6)^{2} + (4 – 7)^{2}}\) = \(\sqrt{(3)^{2} + (-3)^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
\(\quad\) CD = \(\sqrt{(6 – 9)^{2} + {1 – 4}^{2}}\) = \(\sqrt{(-3)^{2} + (-3)^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
\(\quad\) DA = \(\sqrt{(6 – 3)^{2} + {1 – 4}^{2}}\) = \(\sqrt{(3)^{2} + (-3)^{2}}\) = \(\sqrt{9+ 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
And,
Diagonal AC = \(\sqrt{(9 – 3)^{2} + {4 – 4}^{2}}\) = \(\sqrt{(6)^{2} + (0)^{2}}\) = \(\sqrt{36+ 0}\) = \(\sqrt{36}\) = 6
Diagonal BD = \(\sqrt{(6 – 6)^{2} + {1 – 7}^{2}}\) = \(\sqrt{(0)^{2} + (-6)^{2}}\) = \(\sqrt{0+ 36}\) = \(\sqrt{36}\) = 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)
Answer
(i) Let A(– 1, – 2), B (1, 0), C (– 1, 2), D(– 3, 0)
\(\quad\) AB = \(\sqrt{(1 – (-1))^{2} + {0 – (-2)}^{2}}\) = \(\sqrt{(2)^{2} + (2)^{2}}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
\(\quad\) BC = \(\sqrt{(-1 – 1)^{2} + (2 – 0)^{2}}\) = \(\sqrt{(-2)^{2} + (2)^{2}}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
\(\quad\) CD = \(\sqrt{(-3 – (-1))^{2} + {0 – 2}^{2}}\) = \(\sqrt{(-2)^{2} + (-2)^{2}}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
\(\quad\) DA = \(\sqrt{(-3 – (-1))^{2} + {0 – (-2)}^{2}}\) = \(\sqrt{(-2)^{2} + (2)^{2}}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
And,
Diagonal AC = \(\sqrt{(-1 – (-1))^{2} + {2 – (-2)}^{2}}\) = \(\sqrt{(0)^{2} + (4)^{2}}\) = \(\sqrt{0 + 16}\) = \(\sqrt{16}\) = 4
Diagonal BD = \(\sqrt{(-3 – 1)^{2} + {0 – 0}^{2}}\) = \(\sqrt{(-4)^{2} + (0)^{2}}\) = \(\sqrt{16 + 0}\) = \(\sqrt{16}\) = 4
Here, AB = BC = CD = DA = 2√2
Diagonal AC = BD = 4
Therefore, the given points are the vertices of a square.
(ii) Let A (–3, 5), B (3, 1), C (0, 3), D (–1, – 4)
\(\quad\) AB = \(\sqrt{(3 – (-3))^{2} + {1 – 5}^{2}}\) = \(\sqrt{(6)^{2} + (-4)^{2}}\) = \(\sqrt{36 + 16}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\)
\(\quad\) BC = \(\sqrt{(0 – 3)^{2} + (3 – 1)^{2}}\) = \(\sqrt{(-3)^{2} + (2)^{2}}\) = \(\sqrt{9 + 4}\) = \(\sqrt{13}\)
\(\quad\) CD = \(\sqrt{(-1 – 0)^{2} + {-4 – 3}^{2}}\) = \(\sqrt{(-1)^{2} + (-7)^{2}}\) = \(\sqrt{1 + 49}\) = \(\sqrt{50}\) = 5\(\sqrt{2}\)
\(\quad\) DA = \(\sqrt{(-1 – (-3))^{2} + {-4 – 5}^{2}}\) = \(\sqrt{(2)^{2} + (-9)^{2}}\) = \(\sqrt{4 + 81}\) = \(\sqrt{85}\)
Here, AB \(\neq\) BC \(\neq\) CD \(\neq\) DA
Therefore, the given points cannot form a general quadrilateral
(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)
\(\quad\) AB = \(\sqrt{(7 – 4)^{2} + (6 – 5)^{2}}\) = \(\sqrt{(3)^{2} + (1)^{2}}\) = \(\sqrt{9 + 1}\) = \(\sqrt{10}\)
\(\quad\) BC = \(\sqrt{(4 – 7)^{2} + (3 – 6)^{2}}\) = \(\sqrt{(-3)^{2} + (-3)^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
\(\quad\) CD = \(\sqrt{(1 – 4)^{2} + {2 – 3}^{2}}\) = \(\sqrt{(-3)^{2} + (-1)^{2}}\) = \(\sqrt{9 + 1}\) = \(\sqrt{10}\)
\(\quad\) DA = \(\sqrt{(1 – 4)^{2} + {2 – 5}^{2}}\) = \(\sqrt{(-3)^{2} + (-3)^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
And,
Diagonal AC = \(\sqrt{(4 – 4)^{2} + {3 – 5}^{2}}\) = \(\sqrt{(0)^{2} + (-2)^{2}}\) = \(\sqrt{0 + 4}\) = \(\sqrt{4}\) = 2
Diagonal BD = \(\sqrt{(1 – 7)^{2} + {2 – 6}^{2}}\) = \(\sqrt{(-6)^{2} + (-4)^{2}}\) = \(\sqrt{36 + 16}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\)
Here, AB = CD and BC = DA
But, Diagonal AC \(\neq\) BD
Therefore, 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
x-axis \(\Rightarrow\) y = 0.
Let point P (x, 0) and A (2, – 5), B (-2, 9).
\(\;\) PA = PB (Given)
\(\Rightarrow\) \(\sqrt{(x – 2)^2 + (0 – (-5)^2)}\) = \(\sqrt{(x – (-2))^2 + (0 – 9^2)}\)
\(\Rightarrow\) \(\sqrt{(x – 2)^2 + ( (-5)^2)}\) = \(\sqrt{(x + 2)^2 + ( – 9)^2}\)
Squaring on both sides,
\(\Rightarrow\) \(x^2\) + 4 – 4x + 25 = \(x^2\) + 4 + 4x + 81
\(\Rightarrow\) \(x^2\) + 4 – 4x + 25 – \(x^2\) – 4 – 4x – 81 = 0
\(\Rightarrow\) – 8x – 56 = 0
\(\Rightarrow\) – 8x = 56 \(\Rightarrow\) x = \(\frac{56}{-8}\) = – 7
Therefore, 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
P (2, -3) and Q (10, y)
\(\;\) PQ = 10 units (Given)
\(\Rightarrow\) \(\sqrt{(10 – 2)^2 + (y – (-3))^{2}}\) = 10
\(\Rightarrow\) \(\sqrt{(8)^2 + (y + 3)^2}\) = 10
\(\quad\) Squaring on both sides,
\(\Rightarrow\) \(8^2\) + \((y + 3)^2\) = 100 \(\Rightarrow\) 64 + y2 + 9 + 6y = 100
\(\Rightarrow\) y2 + 6 y + 73 – 100 = 0 \(\Rightarrow\) y2 + 6 y – 27 = 0
\(\Rightarrow\) (y + 9)(y – 3) = 0 \(\Rightarrow\) y = -9, 3
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.
Answer
\(\;\) QP = QR (Given)
\(\Rightarrow\) \(\sqrt{(5 – 0)^2 + (-3 – 1)^2}\) = \(\sqrt{(x – 0)^2 + (6 – 1)^2}\)
\(\Rightarrow\) \(\sqrt{(5)^2 + (-4)^2}\) = \(\sqrt{x^2 + (5)^2}\)
\(\Rightarrow\) 25 + 16 = x2 + 25
\(\Rightarrow\) 16 = x2
\(\Rightarrow\) x = \(\pm\)4
Points R (4, 6) or (-4, 6)
If R (4, 6), then QR
\(\quad\) QR = \(\sqrt{(0 – 4)^{2} + (1 – 6)^{2}}\) = \(\sqrt{(-4)^{2} + (-5)^{2}}\) = \(\sqrt{16 + 25}\) = \(\sqrt{41}\)
and, PR = \(\sqrt{(5 – 4)^{2} + (-3 – 6)^{2}}\) = \(\sqrt{(1)^{2} + (-9)^{2}}\) = \(\sqrt{1 + 81}\) = \(\sqrt{81}\)
If R (-4, 6), then QR
\(\quad\) QR = \(\sqrt{(0 + 4)^{2} + (1 – 6)^{2}}\) = \(\sqrt{(4)^{2} + (-5)^{2}}\) = \(\sqrt{16 + 25}\) = \(\sqrt{41}\)
and, PR = \(\sqrt{(5 + 4)^{2} + (-3 – 6)^{2}}\) = \(\sqrt{(9)^{2} + (-9)^{2}}\) = \(\sqrt{81 + 81}\) = \(\sqrt{162}\) = 9\(\sqrt{2}\)
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
Let P (x, y) and A (3, 6) and B (-3 , 4)
\(\;\) PA = PB [Given]
\(\Rightarrow\) \(\sqrt{(x – 3)^2 + (y – 6)^2}\) = \(\sqrt{(x + 3)^2 + (y – 4)^2}\)
\(quad\) Squaring on both sides,
\(\Rightarrow\) x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 16 – 8y
\(\Rightarrow\) x2 + 9 – 6x + y2 + 36 – 12y – x2 – 9 – 6x – y2 – 16 + 8y = 0
\(\Rightarrow\) -12x – 4y + 20 = 0
\(\Rightarrow\) -4 (3x + y – 5) = 0
\(\Rightarrow\) 3x + y – 5 = 0
Exercise 7.2
Question 1:
Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Question 2:
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Question 3:
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs \(\frac{1}{4}\)th the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac{1}{5}\)th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Question 4:
Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
Question 5:
Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Question 6:
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Question 7:
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Question 8:
If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the line segment AB.
Question 9:
Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Question 10:
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.
[Hint : Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals)]
