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) By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
⇒ D = ( 4 − 2 ) 2 + ( 1 − 3 ) 2 = 2 2 + ( − 2 ) 2 = 4 + 4 = 8 = 2 2 . \Rightarrow D = \sqrt{(4 - 2)^2 + (1 - 3)^2} \\[1em] = \sqrt{2^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. ⇒ D = ( 4 − 2 ) 2 + ( 1 − 3 ) 2 = 2 2 + ( − 2 ) 2 = 4 + 4 = 8 = 2 2 .
Hence, distance between (2, 3) and (4, 1) is 2 2 2\sqrt{2} 2 2
(ii) By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
⇒ D = ( 3 − 7 ) 2 + [ − 1 − ( − 5 ) ] 2 = ( − 4 ) 2 + [ − 1 + 5 ] 2 = 16 + 4 2 = 16 + 16 = 32 = 4 2 . \Rightarrow D = \sqrt{(3 - 7)^2 + [-1 - (-5)]^2} \\[1em] = \sqrt{(-4)^2 + [-1 + 5]^2} \\[1em] = \sqrt{16 + 4^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = 4\sqrt{2}. ⇒ D = ( 3 − 7 ) 2 + [ − 1 − ( − 5 ) ] 2 = ( − 4 ) 2 + [ − 1 + 5 ] 2 = 16 + 4 2 = 16 + 16 = 32 = 4 2 .
Hence, distance between (-5, 7) and (-1, 3) is 4 2 4\sqrt{2} 4 2
(iii) By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
⇒ D = ( − b − b ) 2 + ( − a − a ) 2 = ( − 2 b ) 2 + ( − 2 a ) 2 = 4 b 2 + 4 a 2 = 4 ( a 2 + b 2 ) = 2 a 2 + b 2 . \Rightarrow D = \sqrt{(-b - b)^2 + (-a - a)^2} \\[1em] = \sqrt{(-2b)^2 + (-2a)^2} \\[1em] = \sqrt{4b^2 + 4a^2} \\[1em] = \sqrt{4(a^2 + b^2)} \\[1em] = 2\sqrt{a^2 + b^2}. ⇒ D = ( − b − b ) 2 + ( − a − a ) 2 = ( − 2 b ) 2 + ( − 2 a ) 2 = 4 b 2 + 4 a 2 = 4 ( a 2 + b 2 ) = 2 a 2 + b 2 .
Hence, distance between (a, b) and (-a, -b) is 2 a 2 + b 2 2\sqrt{a^2 + b^2} 2 a 2 + b 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
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
⇒ D = ( 15 − 0 ) 2 + ( 36 − 0 ) 2 = ( 15 ) 2 + ( 36 ) 2 = 225 + 1296 = 1521 = 39. \Rightarrow D = \sqrt{(15 - 0)^2 + (36 - 0)^2} \\[1em] = \sqrt{(15)^2 + (36)^2} \\[1em] = \sqrt{225 + 1296} \\[1em] = \sqrt{1521} \\[1em] = 39. ⇒ D = ( 15 − 0 ) 2 + ( 36 − 0 ) 2 = ( 15 ) 2 + ( 36 ) 2 = 225 + 1296 = 1521 = 39.
The positions of towns A & B are given by (0, 0) and (36, 15), hence, as calculated above, the distance between town A and B will be 39 km.
Hence, distance between (0, 0) and (36, 15) is 39 units and distance between towns A and B = 39 km.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Answer
Let the points be A(1, 5), B(2, 3) and C(-2, -11).
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
A B = ( 3 − 5 ) 2 + ( 2 − 1 ) 2 = ( − 2 ) 2 + ( 1 ) 2 = 4 + 1 = 5 . B C = ( − 11 − 3 ) 2 + ( − 2 − 2 ) 2 = ( − 14 ) 2 + ( − 4 ) 2 = 196 + 16 = 212 . A C = ( − 11 − 5 ) 2 + ( − 2 − 1 ) 2 = ( − 16 ) 2 + ( − 3 ) 2 = 256 + 9 = 265 . AB = \sqrt{(3 - 5)^2 + (2 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (1)^2} \\[1em] = \sqrt{4 + 1} \\[1em] = \sqrt{5}. \\[1em] BC = \sqrt{(-11 - 3)^2 + (-2 - 2)^2} \\[1em] = \sqrt{(-14)^2 + (-4)^2} \\[1em] = \sqrt{196 + 16} \\[1em] = \sqrt{212}. \\[1em] AC = \sqrt{(-11 - 5)^2 + (-2 - 1)^2} \\[1em] = \sqrt{(-16)^2 + (-3)^2} \\[1em] = \sqrt{256 + 9} \\[1em] = \sqrt{265}. \\[1em] A B = ( 3 − 5 ) 2 + ( 2 − 1 ) 2 = ( − 2 ) 2 + ( 1 ) 2 = 4 + 1 = 5 . BC = ( − 11 − 3 ) 2 + ( − 2 − 2 ) 2 = ( − 14 ) 2 + ( − 4 ) 2 = 196 + 16 = 212 . A C = ( − 11 − 5 ) 2 + ( − 2 − 1 ) 2 = ( − 16 ) 2 + ( − 3 ) 2 = 256 + 9 = 265 .
Calculating,
⇒ AB + BC = 5 + 212 \sqrt{5} + \sqrt{212} 5 + 212
⇒ BC + AC = 212 + 265 \sqrt{212} + \sqrt{265} 212 + 265
⇒ AB + AC = 5 + 265 \sqrt{5} + \sqrt{265} 5 + 265
Since, AB + AC ≠ BC and BC + AC ≠ AB and AB + BC ≠ AC.
Hence, points (1, 5), (2, 3) and (-2, -11) are not collinear.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Answer
Let vertices be A(5, -2), B(6, 4) and C(7, -2).
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
A B = [ 4 − ( − 2 ) ] 2 + ( 6 − 5 ) 2 = [ 4 + 2 ] 2 + ( 1 ) 2 = 6 2 + 1 = 36 + 1 = 37 . B C = ( − 2 − 4 ) 2 + ( 7 − 6 ) 2 = ( − 6 ) 2 + ( 1 ) 2 = 36 + 1 = 37 . A C = [ − 2 − ( − 2 ) ] 2 + ( 7 − 5 ) 2 = [ − 2 + 2 ] 2 + ( 2 ) 2 = 0 + 4 = 4 = 2. AB = \sqrt{[4 - (-2)]^2 + (6 - 5)^2} \\[1em] = \sqrt{[4 + 2]^2 + (1)^2} \\[1em] = \sqrt{6^2 + 1} \\[1em] = \sqrt{36 + 1} \\[1em] = \sqrt{37}. \\[1em] BC = \sqrt{(-2 - 4)^2 + (7 - 6)^2} \\[1em] = \sqrt{(-6)^2 + (1)^2} \\[1em] = \sqrt{36 + 1} \\[1em] = \sqrt{37}. \\[1em] AC = \sqrt{[-2 - (-2)]^2 + (7 - 5)^2} \\[1em] = \sqrt{[-2 + 2]^2 + (2)^2} \\[1em] = \sqrt{0 + 4} \\[1em] = \sqrt{4} \\[1em] = 2. A B = [ 4 − ( − 2 ) ] 2 + ( 6 − 5 ) 2 = [ 4 + 2 ] 2 + ( 1 ) 2 = 6 2 + 1 = 36 + 1 = 37 . BC = ( − 2 − 4 ) 2 + ( 7 − 6 ) 2 = ( − 6 ) 2 + ( 1 ) 2 = 36 + 1 = 37 . A C = [ − 2 − ( − 2 ) ] 2 + ( 7 − 5 ) 2 = [ − 2 + 2 ] 2 + ( 2 ) 2 = 0 + 4 = 4 = 2.
Since, AB = BC.
Hence, points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
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
From figure,
Co-ordinates of A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1).
By formula,
Distance between two points (D) = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Substituting values we get :
A B = ( 6 − 3 ) 2 + ( 7 − 4 ) 2 = 3 2 + 3 2 = 9 + 9 = 18 = 3 2 units . B C = ( 9 − 6 ) 2 + ( 4 − 7 ) 2 = 3 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 units . C D = ( 6 − 9 ) 2 + ( 1 − 4 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 units A D = ( 6 − 3 ) 2 + ( 1 − 4 ) 2 = 3 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 units . AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}. \\[1em] BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} \\[1em] = \sqrt{3^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}. \\[1em] CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units} \\[1em] AD = \sqrt{(6 - 3)^2 + (1 - 4)^2} \\[1em] = \sqrt{3^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}. A B = ( 6 − 3 ) 2 + ( 7 − 4 ) 2 = 3 2 + 3 2 = 9 + 9 = 18 = 3 2 units . BC = ( 9 − 6 ) 2 + ( 4 − 7 ) 2 = 3 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 units . C D = ( 6 − 9 ) 2 + ( 1 − 4 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 units A D = ( 6 − 3 ) 2 + ( 1 − 4 ) 2 = 3 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 units .
Since, AB = BC = CD = AD.
∴ ABCD is a square.
Hence, Champa is correct.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(-1, -2), (1, 0), (-1, 2), (-3, 0)
Answer
Let coordinates be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).
By formula,
Distance between two points (D) = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Calculating sides, we get :
A B = [ 1 − ( − 1 ) ] 2 + [ 0 − ( − 2 ) ] 2 = [ 1 + 1 ] 2 + [ 0 + 2 ] 2 = 2 2 + 2 2 = 4 + 4 = 8 units B C = ( − 1 − 1 ) 2 + ( 2 − 0 ) 2 = ( − 2 ) 2 + 2 2 = 4 + 4 = 8 units C D = [ − 3 − ( − 1 ) ] 2 + ( 0 − 2 ) 2 = [ − 3 + 1 ] 2 + ( − 2 ) 2 = ( − 2 ) 2 + ( − 2 ) 2 = 4 + 4 = 8 units A D = [ − 3 − ( − 1 ) ] 2 + [ 0 − ( − 2 ) ] 2 = [ − 3 + 1 ] 2 + [ 0 + 2 ] 2 = [ − 2 ] 2 + 2 2 = 4 + 4 = 8 units . AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[1 + 1]^2 + [0 + 2]^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units} \\[1em] BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} \\[1em] = \sqrt{(-2)^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units} \\[1em] CD = \sqrt{[-3 - (-1)]^2 + (0 - 2)^2} \\[1em] = \sqrt{[-3 + 1]^2 + (-2)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units} \\[1em] AD = \sqrt{[-3 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [0 + 2]^2} \\[1em] = \sqrt{[-2]^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}\text{ units}. A B = [ 1 − ( − 1 ) ] 2 + [ 0 − ( − 2 ) ] 2 = [ 1 + 1 ] 2 + [ 0 + 2 ] 2 = 2 2 + 2 2 = 4 + 4 = 8 units BC = ( − 1 − 1 ) 2 + ( 2 − 0 ) 2 = ( − 2 ) 2 + 2 2 = 4 + 4 = 8 units C D = [ − 3 − ( − 1 ) ] 2 + ( 0 − 2 ) 2 = [ − 3 + 1 ] 2 + ( − 2 ) 2 = ( − 2 ) 2 + ( − 2 ) 2 = 4 + 4 = 8 units A D = [ − 3 − ( − 1 ) ] 2 + [ 0 − ( − 2 ) ] 2 = [ − 3 + 1 ] 2 + [ 0 + 2 ] 2 = [ − 2 ] 2 + 2 2 = 4 + 4 = 8 units .
Calculating diagonals, we get :
A C = [ − 1 − ( − 1 ) ] 2 + [ 2 − ( − 2 ) ] 2 = [ − 1 + 1 ] 2 + [ 2 + 2 ] 2 = 0 + 4 2 = 4 units . B D = ( − 3 − 1 ) 2 + ( 0 − 0 ) 2 = ( − 4 ) 2 + 0 = 16 = 4 units . AC = \sqrt{[-1- (-1)]^2 + [2 - (-2)]^2} \\[1em] = \sqrt{[-1 + 1]^2 + [2 + 2]^2} \\[1em] = \sqrt{0 + 4^2} \\[1em] = 4 \text{ units}. \\[1em] BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 0} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. A C = [ − 1 − ( − 1 ) ] 2 + [ 2 − ( − 2 ) ] 2 = [ − 1 + 1 ] 2 + [ 2 + 2 ] 2 = 0 + 4 2 = 4 units . B D = ( − 3 − 1 ) 2 + ( 0 − 0 ) 2 = ( − 4 ) 2 + 0 = 16 = 4 units .
Since, AB = BC = CD = AD and AC = BD.
Hence, the above points form a square.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(-3, 5), (3, 1), (0, 3), (-1, -4)
Answer
Let coordinates be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).
By formula,
Distance between two points (D) = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Calculating sides, we get :
A B = [ 3 − ( − 3 ) ] 2 + [ 1 − 5 ] 2 = [ 3 + 3 ] 2 + [ − 4 ] 2 = 6 2 + ( − 4 ) 2 = 36 + 16 = 52 units B C = ( 0 − 3 ) 2 + ( 3 − 1 ) 2 = ( − 3 ) 2 + 2 2 = 9 + 4 = 13 units C D = ( − 1 − 0 ) 2 + ( − 4 − 3 ) 2 = ( − 1 ) 2 + ( − 7 ) 2 = 1 + 49 = 50 units A D = [ − 1 − ( − 3 ) ] 2 + [ − 4 − 5 ] 2 = [ − 1 + 3 ] 2 + [ − 9 ] 2 = [ 2 ] 2 + [ − 9 ] 2 = 4 + 81 = 85 . AB = \sqrt{[3 - (-3)]^2 + [1 - 5]^2} \\[1em] = \sqrt{[3 + 3]^2 + [-4]^2} \\[1em] = \sqrt{6^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units} \\[1em] BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-3)^2 + 2^2} \\[1em] = \sqrt{9 + 4} \\[1em] = \sqrt{13} \text{ units} \\[1em] CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} \\[1em] = \sqrt{(-1)^2 + (-7)^2} \\[1em] = \sqrt{1 + 49} \\[1em] = \sqrt{50} \text{ units} \\[1em] AD = \sqrt{[-1 - (-3)]^2 + [-4 - 5]^2} \\[1em] = \sqrt{[-1 + 3]^2 + [-9]^2} \\[1em] = \sqrt{[2]^2 + [-9]^2} \\[1em] = \sqrt{4 + 81} \\[1em] = \sqrt{85}. A B = [ 3 − ( − 3 ) ] 2 + [ 1 − 5 ] 2 = [ 3 + 3 ] 2 + [ − 4 ] 2 = 6 2 + ( − 4 ) 2 = 36 + 16 = 52 units BC = ( 0 − 3 ) 2 + ( 3 − 1 ) 2 = ( − 3 ) 2 + 2 2 = 9 + 4 = 13 units C D = ( − 1 − 0 ) 2 + ( − 4 − 3 ) 2 = ( − 1 ) 2 + ( − 7 ) 2 = 1 + 49 = 50 units A D = [ − 1 − ( − 3 ) ] 2 + [ − 4 − 5 ] 2 = [ − 1 + 3 ] 2 + [ − 9 ] 2 = [ 2 ] 2 + [ − 9 ] 2 = 4 + 81 = 85 .
Since, AB ≠ BC ≠ CD ≠ AD,
Hence, no special quadrilateral can be formed from the given vertices.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(4, 5), (7, 6), (4, 3), (1, 2)
Answer
Let coordinates be A(4, 5), B(7, 6), C(4, 3) and D(1, 2).
By formula,
Distance between two points (D) = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Calculating sides, we get :
A B = ( 7 − 4 ) 2 + [ 6 − 5 ] 2 = ( 3 ) 2 + ( 1 ) 2 = 9 + 1 = 10 units B C = ( 4 − 7 ) 2 + ( 3 − 6 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 units C D = ( 1 − 4 ) 2 + ( 2 − 3 ) 2 = ( − 3 ) 2 + ( − 1 ) 2 = 9 + 1 = 10 units A D = ( 1 − 4 ) 2 + ( 2 − 5 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 units AB = \sqrt{(7 - 4)^2 + [6 - 5]^2} \\[1em] = \sqrt{(3)^2 + (1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units} \\[1em] BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \text{ units} \\[1em] CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units} \\[1em] AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \text{ units} \\[1em] A B = ( 7 − 4 ) 2 + [ 6 − 5 ] 2 = ( 3 ) 2 + ( 1 ) 2 = 9 + 1 = 10 units BC = ( 4 − 7 ) 2 + ( 3 − 6 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 units C D = ( 1 − 4 ) 2 + ( 2 − 3 ) 2 = ( − 3 ) 2 + ( − 1 ) 2 = 9 + 1 = 10 units A D = ( 1 − 4 ) 2 + ( 2 − 5 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 units
Calculating diagonals, we get :
A C = ( 4 − 4 ) 2 + ( 3 − 5 ) 2 = 0 2 + ( − 2 ) 2 = 0 + 4 = 2 units . B D = ( 1 − 7 ) 2 + ( 2 − 6 ) 2 = ( − 6 ) 2 + ( − 4 ) 2 = 36 + 16 = 52 units . AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} \\[1em] = \sqrt{0^2 + (-2)^2} \\[1em] = \sqrt{0 + 4} \\[1em] = 2 \text{ units}. \\[1em] BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} \\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units}. A C = ( 4 − 4 ) 2 + ( 3 − 5 ) 2 = 0 2 + ( − 2 ) 2 = 0 + 4 = 2 units . B D = ( 1 − 7 ) 2 + ( 2 − 6 ) 2 = ( − 6 ) 2 + ( − 4 ) 2 = 36 + 16 = 52 units .
Since, AB = CD and BC = AD and AC ≠ BD.
Hence, the above points form a parallelogram.
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Answer
We know that,
y-coordinate on x-axis = 0.
Let point on x-axis be (x, 0).
According to question :
Distance between (2, -5) and (x, 0) = Distance between (-2, 9) and (x, 0).
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
⇒ ( x − 2 ) 2 + [ 0 − ( − 5 ) ] 2 = [ x − ( − 2 ) ] 2 + ( 0 − 9 ) 2 ⇒ x 2 + 4 − 4 x + 5 2 = [ x + 2 ] 2 + ( − 9 ) 2 ⇒ x 2 − 4 x + 4 + 25 = x 2 + 4 + 4 x + 81 ⇒ x 2 − 4 x + 29 = x 2 + 4 x + 85 \Rightarrow \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{[x - (-2)]^2 + (0 - 9)^2} \\[1em] \Rightarrow \sqrt{x^2 + 4 - 4x + 5^2} = \sqrt{[x + 2]^2 + (-9)^2} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 4 + 25} = \sqrt{x^2 + 4 + 4x + 81} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 29} = \sqrt{x^2 + 4x + 85} ⇒ ( x − 2 ) 2 + [ 0 − ( − 5 ) ] 2 = [ x − ( − 2 ) ] 2 + ( 0 − 9 ) 2 ⇒ x 2 + 4 − 4 x + 5 2 = [ x + 2 ] 2 + ( − 9 ) 2 ⇒ x 2 − 4 x + 4 + 25 = x 2 + 4 + 4 x + 81 ⇒ x 2 − 4 x + 29 = x 2 + 4 x + 85
Squaring, both sides we get :
⇒ x 2 − 4 x + 29 = x 2 + 4 x + 85 ⇒ x 2 − x 2 + 4 x + 4 x = 29 − 85 ⇒ 8 x = − 56 ⇒ x = − 56 8 ⇒ x = − 7. \Rightarrow x^2 - 4x + 29 = x^2 + 4x + 85 \\[1em] \Rightarrow x^2 - x^2 + 4x + 4x = 29 - 85 \\[1em] \Rightarrow 8x = -56 \\[1em] \Rightarrow x = -\dfrac{56}{8} \\[1em] \Rightarrow x = -7. ⇒ x 2 − 4 x + 29 = x 2 + 4 x + 85 ⇒ x 2 − x 2 + 4 x + 4 x = 29 − 85 ⇒ 8 x = − 56 ⇒ x = − 8 56 ⇒ x = − 7.
Point = (x, 0) = (-7, 0).
Hence, point (-7, 0) is equidistant from points (2, –5) and (–2, 9).
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
⇒ 10 = [ y − ( − 3 ) ] 2 + ( 10 − 2 ) 2 ⇒ 10 = [ y + 3 ] 2 + 8 2 ⇒ 10 = y 2 + 9 + 6 y + 64 ⇒ 10 = y 2 + 6 y + 73 ⇒ 1 0 2 = ( y 2 + 6 y + 73 ) 2 ⇒ 100 = y 2 + 6 y + 73 ⇒ y 2 + 6 y + 73 − 100 = 0 ⇒ y 2 + 6 y − 27 = 0 ⇒ y 2 + 9 y − 3 y − 27 = 0 ⇒ y ( y + 9 ) − 3 ( y + 9 ) = 0 ⇒ ( y − 3 ) ( y + 9 ) = 0 ⇒ y − 3 = 0 or y + 9 = 0 ⇒ y = 3 or y = − 9. \Rightarrow 10 = \sqrt{[y - (-3)]^2 + (10 - 2)^2} \\[1em] \Rightarrow 10 = \sqrt{[y + 3]^2 + 8^2} \\[1em] \Rightarrow 10 = \sqrt{y^2 + 9 + 6y + 64} \\[1em] \Rightarrow 10 = \sqrt{y^2 + 6y + 73} \\[1em] \Rightarrow 10^2 = \Big(\sqrt{y^2 + 6y + 73}\Big)^2 \\[1em] \Rightarrow 100 = y^2 + 6y + 73 \\[1em] \Rightarrow y^2 + 6y + 73 - 100 = 0 \\[1em] \Rightarrow y^2 + 6y - 27 = 0 \\[1em] \Rightarrow y^2 + 9y - 3y - 27 = 0 \\[1em] \Rightarrow y(y + 9) - 3(y + 9) = 0 \\[1em] \Rightarrow (y - 3)(y + 9) = 0 \\[1em] \Rightarrow y - 3 = 0 \text{ or } y + 9 = 0 \\[1em] \Rightarrow y = 3 \text{ or } y = -9. ⇒ 10 = [ y − ( − 3 ) ] 2 + ( 10 − 2 ) 2 ⇒ 10 = [ y + 3 ] 2 + 8 2 ⇒ 10 = y 2 + 9 + 6 y + 64 ⇒ 10 = y 2 + 6 y + 73 ⇒ 1 0 2 = ( y 2 + 6 y + 73 ) 2 ⇒ 100 = y 2 + 6 y + 73 ⇒ y 2 + 6 y + 73 − 100 = 0 ⇒ y 2 + 6 y − 27 = 0 ⇒ y 2 + 9 y − 3 y − 27 = 0 ⇒ y ( y + 9 ) − 3 ( y + 9 ) = 0 ⇒ ( y − 3 ) ( y + 9 ) = 0 ⇒ y − 3 = 0 or y + 9 = 0 ⇒ y = 3 or y = − 9.
Hence, y = 3 or -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
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Given,
Q(0, 1) is equidistant from P(5, –3) and R(x, 6).
∴ PQ = QR
Substituting values we get :
⇒ ( − 3 − 1 ) 2 + ( 5 − 0 ) 2 = ( 6 − 1 ) 2 + ( x − 0 ) 2 ⇒ ( − 4 ) 2 + 5 2 = 5 2 + x 2 ⇒ 16 + 25 = x 2 + 25 ⇒ x 2 + 25 = 41 \Rightarrow \sqrt{(-3 - 1)^2 + (5 - 0)^2} = \sqrt{(6 - 1)^2 + (x - 0)^2} \\[1em] \Rightarrow \sqrt{(-4)^2 + 5^2} = \sqrt{5^2 + x^2} \\[1em] \Rightarrow \sqrt{16 + 25} = \sqrt{x^2 + 25} \\[1em] \Rightarrow \sqrt{x^2 + 25} = \sqrt{41} ⇒ ( − 3 − 1 ) 2 + ( 5 − 0 ) 2 = ( 6 − 1 ) 2 + ( x − 0 ) 2 ⇒ ( − 4 ) 2 + 5 2 = 5 2 + x 2 ⇒ 16 + 25 = x 2 + 25 ⇒ x 2 + 25 = 41
Squaring both sides we get :
⇒ x 2 + 25 = 41 ⇒ x 2 = 41 − 25 ⇒ x 2 = 16 ⇒ x = 16 ⇒ x = ± 4. \Rightarrow x^2 + 25 = 41 \\[1em] \Rightarrow x^2 = 41 - 25 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x = \sqrt{16} \\[1em] \Rightarrow x = \pm 4. ⇒ x 2 + 25 = 41 ⇒ x 2 = 41 − 25 ⇒ x 2 = 16 ⇒ x = 16 ⇒ x = ± 4.
When x = 4,
P = (5, -3), Q = (0, 1) and R = (4, 6).
Q R = ( 6 − 1 ) 2 + ( 4 − 0 ) 2 = 5 2 + 4 2 = 25 + 16 = 41 . P R = [ 6 − ( − 3 ) ] 2 + ( 4 − 5 ) 2 = [ 6 + 3 ] 2 + ( − 1 ) 2 = 9 2 + 1 = 81 + 1 = 82 . QR = \sqrt{(6 - 1)^2 + (4 - 0)^2} \\[1em] = \sqrt{5^2 + 4^2} \\[1em] = \sqrt{25 + 16} \\[1em] = \sqrt{41}. \\[1em] PR = \sqrt{[6 - (-3)]^2 + (4 - 5)^2} \\[1em] = \sqrt{[6 + 3]^2 + (-1)^2} \\[1em] = \sqrt{9^2 + 1} \\[1em] = \sqrt{81 + 1} \\[1em] = \sqrt{82}. QR = ( 6 − 1 ) 2 + ( 4 − 0 ) 2 = 5 2 + 4 2 = 25 + 16 = 41 . PR = [ 6 − ( − 3 ) ] 2 + ( 4 − 5 ) 2 = [ 6 + 3 ] 2 + ( − 1 ) 2 = 9 2 + 1 = 81 + 1 = 82 .
When x = -4,
P = (5, -3), Q = (0, 1) and R = (-4, 6).
Q R = ( 6 − 1 ) 2 + ( − 4 − 0 ) 2 = 5 2 + ( − 4 ) 2 = 25 + 16 = 41 . P R = [ 6 − ( − 3 ) ] 2 + ( − 4 − 5 ) 2 = [ 6 + 3 ] 2 + ( − 9 ) 2 = 9 2 + ( − 9 ) 2 = 81 + 81 = 162 = 9 2 . QR = \sqrt{(6 - 1)^2 + (-4 - 0)^2} \\[1em] = \sqrt{5^2 + (-4)^2} \\[1em] = \sqrt{25 + 16} \\[1em] = \sqrt{41}. \\[1em] PR = \sqrt{[6 - (-3)]^2 + (-4 - 5)^2} \\[1em] = \sqrt{[6 + 3]^2 + (-9)^2} \\[1em] = \sqrt{9^2 + (-9)^2} \\[1em] = \sqrt{81 + 81} \\[1em] = \sqrt{162} \\[1em] = 9\sqrt{2}. QR = ( 6 − 1 ) 2 + ( − 4 − 0 ) 2 = 5 2 + ( − 4 ) 2 = 25 + 16 = 41 . PR = [ 6 − ( − 3 ) ] 2 + ( − 4 − 5 ) 2 = [ 6 + 3 ] 2 + ( − 9 ) 2 = 9 2 + ( − 9 ) 2 = 81 + 81 = 162 = 9 2 .
Hence, x = ± 4 , Q R = 41 , P R = 82 , 9 2 . \pm 4, QR = \sqrt{41}, PR = \sqrt{82}, 9\sqrt{2}. ± 4 , QR = 41 , PR = 82 , 9 2 .
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Answer
By formula,
Distance between two points (D) = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Given,
Point (x, y) is equidistant from the point (3, 6) and (-3, 4).
⇒ ( y − 6 ) 2 + ( x − 3 ) 2 = ( y − 4 ) 2 + [ x − ( − 3 ) 2 ] ⇒ y 2 + 36 − 12 y + x 2 + 9 − 6 x = y 2 + 16 − 8 y + [ x + 3 ] 2 ⇒ y 2 + x 2 − 6 x − 12 y + 45 = y 2 + 16 − 8 y + x 2 + 9 + 6 x ⇒ x 2 + y 2 − 12 y − 6 x + 45 = x 2 + y 2 + 6 x − 8 y + 25 \Rightarrow \sqrt{(y - 6)^2 + (x - 3)^2} = \sqrt{(y - 4)^2 + [x - (-3)^2]} \\[1em] \Rightarrow \sqrt{y^2 + 36 - 12y + x^2 + 9 - 6x} = \sqrt{y^2 + 16 - 8y + [x + 3]^2} \\[1em] \Rightarrow \sqrt{y^2 + x^2 - 6x - 12y + 45} = \sqrt{y^2 + 16 - 8y + x^2 + 9 + 6x} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 12y - 6x + 45} = \sqrt{x^2 + y^2 + 6x - 8y + 25} ⇒ ( y − 6 ) 2 + ( x − 3 ) 2 = ( y − 4 ) 2 + [ x − ( − 3 ) 2 ] ⇒ y 2 + 36 − 12 y + x 2 + 9 − 6 x = y 2 + 16 − 8 y + [ x + 3 ] 2 ⇒ y 2 + x 2 − 6 x − 12 y + 45 = y 2 + 16 − 8 y + x 2 + 9 + 6 x ⇒ x 2 + y 2 − 12 y − 6 x + 45 = x 2 + y 2 + 6 x − 8 y + 25
Squaring both sides we get :
⇒ x 2 + y 2 − 12 y − 6 x + 45 = x 2 + y 2 + 6 x − 8 y + 25 ⇒ x 2 − x 2 + y 2 − y 2 + 6 x + 6 x − 8 y + 12 y = 45 − 25 ⇒ 12 x + 4 y = 20 ⇒ 4 ( 3 x + y ) = 20 ⇒ 3 x + y = 5 ⇒ 3 x + y − 5 = 0. \Rightarrow x^2 + y^2 - 12y - 6x + 45 = x^2 + y^2 + 6x - 8y + 25 \\[1em] \Rightarrow x^2 - x^2 + y^2 - y^2 + 6x + 6x - 8y + 12y = 45 - 25 \\[1em] \Rightarrow 12x + 4y = 20 \\[1em] \Rightarrow 4(3x + y) = 20 \\[1em] \Rightarrow 3x + y = 5 \\[1em] \Rightarrow 3x + y - 5 = 0. ⇒ x 2 + y 2 − 12 y − 6 x + 45 = x 2 + y 2 + 6 x − 8 y + 25 ⇒ x 2 − x 2 + y 2 − y 2 + 6 x + 6 x − 8 y + 12 y = 45 − 25 ⇒ 12 x + 4 y = 20 ⇒ 4 ( 3 x + y ) = 20 ⇒ 3 x + y = 5 ⇒ 3 x + y − 5 = 0.
Hence, 3x + y - 5 = 0.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Answer
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Let (x, y) divide the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Substituting values we get :
⇒ ( x , y ) = ( 2 × 4 + 3 × − 1 2 + 3 , 2 × − 3 + 3 × 7 2 + 3 ) = ( 8 + ( − 3 ) 5 , − 6 + 21 5 ) = ( 5 5 , 15 5 ) = ( 1 , 3 ) . \Rightarrow (x, y) = \Big(\dfrac{2 \times 4 + 3 \times -1}{2 + 3}, \dfrac{2 \times -3 + 3 \times 7}{2 + 3}\Big) \\[1em] = \Big(\dfrac{8 + (-3)}{5}, \dfrac{-6 + 21}{5}\Big) \\[1em] = \Big(\dfrac{5}{5}, \dfrac{15}{5}\Big) \\[1em] = (1, 3). ⇒ ( x , y ) = ( 2 + 3 2 × 4 + 3 × − 1 , 2 + 3 2 × − 3 + 3 × 7 ) = ( 5 8 + ( − 3 ) , 5 − 6 + 21 ) = ( 5 5 , 5 15 ) = ( 1 , 3 ) .
Hence, point (1, 3) divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Answer
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Let A and B be the points of trisection.
So, A divides line segment joining (4, –1) and (–2, –3) in the ratio 1 : 2.
Let co-ordinates of A be (a, b).
⇒ ( a , b ) = ( 1 × − 2 + 2 × 4 1 + 2 , 1 × − 3 + 2 × − 1 1 + 2 ) = ( − 2 + 8 3 , − 3 − 2 3 ) = ( 6 3 , − 5 3 ) = ( 2 , − 5 3 ) . \Rightarrow (a, b) = \Big(\dfrac{1 \times -2 + 2 \times 4}{1 + 2}, \dfrac{1 \times -3 + 2 \times -1}{1 + 2}\Big) \\[1em] = \Big(\dfrac{-2 + 8}{3}, \dfrac{-3 - 2}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{-5}{3}\Big) \\[1em] = \Big(2, -\dfrac{5}{3}\Big). ⇒ ( a , b ) = ( 1 + 2 1 × − 2 + 2 × 4 , 1 + 2 1 × − 3 + 2 × − 1 ) = ( 3 − 2 + 8 , 3 − 3 − 2 ) = ( 3 6 , 3 − 5 ) = ( 2 , − 3 5 ) .
B divides line segment joining (4, –1) and (–2, –3) in the ratio 2 : 1.
Let co-ordinates of B be (c, d).
⇒ ( c , d ) = ( 2 × − 2 + 1 × 4 2 + 1 , 2 × − 3 + 1 × − 1 2 + 1 ) = ( − 4 + 4 3 , − 6 − 1 3 ) = ( 0 3 , − 7 3 ) = ( 0 , − 7 3 ) . \Rightarrow (c, d) = \Big(\dfrac{2 \times -2 + 1 \times 4}{2 + 1}, \dfrac{2 \times -3 + 1 \times -1}{2 + 1}\Big) \\[1em] = \Big(\dfrac{-4 + 4}{3}, \dfrac{-6 - 1}{3}\Big) \\[1em] = \Big(\dfrac{0}{3}, \dfrac{-7}{3}\Big) \\[1em] = \Big(0, -\dfrac{7}{3}\Big). ⇒ ( c , d ) = ( 2 + 1 2 × − 2 + 1 × 4 , 2 + 1 2 × − 3 + 1 × − 1 ) = ( 3 − 4 + 4 , 3 − 6 − 1 ) = ( 3 0 , 3 − 7 ) = ( 0 , − 3 7 ) .
Hence, points of intersection are ( 2 , − 5 3 ) , ( 0 , − 7 3 ) \Big(2, -\dfrac{5}{3}\Big), \Big(0, -\dfrac{7}{3}\Big) ( 2 , − 3 5 ) , ( 0 , − 3 7 )
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 1 4 \dfrac{1}{4} 4 1 1 5 \dfrac{1}{5} 5 1
Answer
Given,
100 flower pots have been placed at a distance of 1 m from each other along AD.
AD = 100 × 1 = 100 m.
It can be observed that Niharika posted the green flag at 1 4 \dfrac{1}{4} 4 1 1 4 × 100 = 25 \dfrac{1}{4} \times 100 = 25 4 1 × 100 = 25
Similarly, Preet posted red flag at 1 5 \dfrac{1}{5} 5 1 1 5 × 100 = 20 \dfrac{1}{5} \times 100 = 20 5 1 × 100 = 20
By formula,
Distance between two points = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2
Substituting values we get :
G R = ( 20 − 25 ) 2 + ( 8 − 2 ) 2 = ( − 5 ) 2 + ( 6 ) 2 = 25 + 36 = 61 m . GR = \sqrt{(20 - 25)^2 + (8 - 2)^2} \\[1em] = \sqrt{(-5)^2 + (6)^2} \\[1em] = \sqrt{25 + 36} \\[1em] = \sqrt{61} \text{ m}. GR = ( 20 − 25 ) 2 + ( 8 − 2 ) 2 = ( − 5 ) 2 + ( 6 ) 2 = 25 + 36 = 61 m .
The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y).
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
A ( x , y ) = ( 2 + 8 2 , 20 + 25 2 ) = ( 10 2 , 45 2 ) = ( 5 , 22.5 ) A (x, y) = \Big(\dfrac{2 + 8}{2}, \dfrac{20 + 25}{2}\Big) \\[1em] = \Big(\dfrac{10}{2}, \dfrac{45}{2}\Big) \\[1em] = (5, 22.5) A ( x , y ) = ( 2 2 + 8 , 2 20 + 25 ) = ( 2 10 , 2 45 ) = ( 5 , 22.5 )
From figure,
The x-coordinate represets the no. of line and y-coordinate represents the vertical distance.
Hence, Rashmi should post her blue flag at a distance of 22.5m on 5th line.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer
Let ratio be k : 1.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get :
⇒ ( − 1 , 6 ) = ( k × 6 + 1 × − 3 k + 1 , k × − 8 + 1 × 10 k + 1 ) ⇒ ( − 1 , 6 ) = ( 6 k − 3 k + 1 , − 8 k + 10 k + 1 ) ⇒ − 1 = 6 k − 3 k + 1 and 6 = − 8 k + 10 k + 1 ⇒ − k − 1 = 6 k − 3 and 6 k + 6 = − 8 k + 10 ⇒ 6 k + k = − 1 + 3 and 6 k + 8 k = 10 − 6 ⇒ 7 k = 2 and 14 k = 4 ⇒ k = 2 7 and k = 4 14 = 2 7 . \Rightarrow (-1, 6) = \Big(\dfrac{k \times 6 + 1 \times -3}{k + 1}, \dfrac{k \times -8 + 1 \times 10}{k + 1}\Big) \\[1em] \Rightarrow (-1, 6) = \Big(\dfrac{6k - 3}{k + 1}, \dfrac{-8k + 10}{k + 1}\Big) \\[1em] \Rightarrow -1 = \dfrac{6k - 3}{k + 1} \text{ and } 6 = \dfrac{-8k + 10}{k + 1} \\[1em] \Rightarrow -k - 1 = 6k - 3 \text{ and } 6k + 6 = -8k + 10 \\[1em] \Rightarrow 6k + k = -1 + 3 \text{ and } 6k + 8k = 10 - 6 \\[1em] \Rightarrow 7k = 2 \text{ and } 14k = 4 \\[1em] \Rightarrow k = \dfrac{2}{7} \text{ and } k = \dfrac{4}{14} = \dfrac{2}{7}. ⇒ ( − 1 , 6 ) = ( k + 1 k × 6 + 1 × − 3 , k + 1 k × − 8 + 1 × 10 ) ⇒ ( − 1 , 6 ) = ( k + 1 6 k − 3 , k + 1 − 8 k + 10 ) ⇒ − 1 = k + 1 6 k − 3 and 6 = k + 1 − 8 k + 10 ⇒ − k − 1 = 6 k − 3 and 6 k + 6 = − 8 k + 10 ⇒ 6 k + k = − 1 + 3 and 6 k + 8 k = 10 − 6 ⇒ 7 k = 2 and 14 k = 4 ⇒ k = 7 2 and k = 14 4 = 7 2 .
∴ k : 1 = 2 7 : 1 \dfrac{2}{7} : 1 7 2 : 1
Hence, ratio = 2 : 7.
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.
Answer
Let point on x-axis be (x, 0) as y-coordinate = 0 on x-axis.
Let ratio be k : 1.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting for y-coordinate we get,
⇒ 0 = k × 5 + 1 × − 5 k + 1 ⇒ 0 = 5 k − 5 k + 1 ⇒ 5 k − 5 = 0 ⇒ 5 k = 5 ⇒ k = 5 5 = 1. \Rightarrow 0 = \dfrac{k \times 5 + 1 \times -5}{k + 1} \\[1em] \Rightarrow 0 = \dfrac{5k - 5}{k + 1} \\[1em] \Rightarrow 5k - 5 = 0 \\[1em] \Rightarrow 5k = 5 \\[1em] \Rightarrow k = \dfrac{5}{5} = 1. ⇒ 0 = k + 1 k × 5 + 1 × − 5 ⇒ 0 = k + 1 5 k − 5 ⇒ 5 k − 5 = 0 ⇒ 5 k = 5 ⇒ k = 5 5 = 1.
k : 1 = 1 : 1.
Substituting value for x-coordinate, we get :
⇒ x = 1 × − 4 + 1 × 1 1 + 1 ⇒ x = − 4 + 1 2 ⇒ x = − 3 2 . \Rightarrow x = \dfrac{1 \times -4 + 1 \times 1}{1 + 1} \\[1em] \Rightarrow x = \dfrac{-4 + 1}{2} \\[1em] \Rightarrow x = -\dfrac{3}{2}. ⇒ x = 1 + 1 1 × − 4 + 1 × 1 ⇒ x = 2 − 4 + 1 ⇒ x = − 2 3 .
Point = (x, 0) = ( − 3 2 , 0 ) \Big(-\dfrac{3}{2}, 0\Big) ( − 2 3 , 0 )
Hence, ratio = 1 : 1 and point of division = ( − 3 2 , 0 ) \Big(-\dfrac{3}{2}, 0\Big) ( − 2 3 , 0 )
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram..
We know that,
Diagonals of parallelogram bisect each other.
So, mid-point will be same let O.
i.e., Mid-point of AC = Mid-point of BD.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( 1 + x 2 , 2 + 6 2 ) = ( 4 + 3 2 , y + 5 2 ) ⇒ ( 1 + x 2 , 8 2 ) = ( 7 2 , y + 5 2 ) ⇒ 1 + x 2 = 7 2 and 8 2 = y + 5 2 ⇒ 1 + x = 7 and 8 = y + 5 ⇒ x = 7 − 1 and y = 8 − 5 ⇒ x = 6 and y = 3. \Rightarrow \Big(\dfrac{1 + x}{2}, \dfrac{2 + 6}{2}\Big) = \Big(\dfrac{4 + 3}{2}, \dfrac{y + 5}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{1 + x}{2}, \dfrac{8}{2}\Big) = \Big(\dfrac{7}{2}, \dfrac{y + 5}{2}\Big) \\[1em] \Rightarrow \dfrac{1 + x}{2} = \dfrac{7}{2} \text{ and } \dfrac{8}{2} = \dfrac{y + 5}{2} \\[1em] \Rightarrow 1 + x = 7 \text{ and } 8 = y + 5 \\[1em] \Rightarrow x = 7 - 1 \text{ and } y = 8 - 5 \\[1em] \Rightarrow x = 6 \text{ and } y = 3. ⇒ ( 2 1 + x , 2 2 + 6 ) = ( 2 4 + 3 , 2 y + 5 ) ⇒ ( 2 1 + x , 2 8 ) = ( 2 7 , 2 y + 5 ) ⇒ 2 1 + x = 2 7 and 2 8 = 2 y + 5 ⇒ 1 + x = 7 and 8 = y + 5 ⇒ x = 7 − 1 and y = 8 − 5 ⇒ x = 6 and y = 3.
Hence, x = 6 and y = 3.
Find the coordinates of a point A, where AB is the diameter of a circle whose center is (2, -3) and B is (1, 4).
Answer
Let coordinates of A be (x, y) and C be the center (2, -3).
Given,
B = (1, 4).
We know that,
Center is the mid-point of diameter.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( 2 , − 3 ) = ( x + 1 2 , y + 4 2 ) ⇒ 2 = x + 1 2 , − 3 = y + 4 2 ⇒ x + 1 = 4 , y + 4 = − 6 ⇒ x = 4 − 1 = 3 , y = − 6 − 4 = − 10. \Rightarrow (2, -3) = \Big(\dfrac{x + 1}{2}, \dfrac{y + 4}{2}\Big) \\[1em] \Rightarrow 2 = \dfrac{x + 1}{2}, -3 = \dfrac{y + 4}{2} \\[1em] \Rightarrow x + 1 = 4, y + 4 = -6 \\[1em] \Rightarrow x = 4 - 1 = 3, y = -6 - 4 = -10. ⇒ ( 2 , − 3 ) = ( 2 x + 1 , 2 y + 4 ) ⇒ 2 = 2 x + 1 , − 3 = 2 y + 4 ⇒ x + 1 = 4 , y + 4 = − 6 ⇒ x = 4 − 1 = 3 , y = − 6 − 4 = − 10.
A = (x, y) = (3, -10).
Hence, co-ordinates of A = (3, -10).
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3 7 \dfrac{3}{7} 7 3
Answer
Given,
AP = 3 7 \dfrac{3}{7} 7 3
From figure,
AB = AP + PB
PB = AB - AP = A B − 3 7 A B = 7 A B − 3 A B 7 = 4 7 AB - \dfrac{3}{7}AB = \dfrac{7AB - 3AB}{7} = \dfrac{4}{7} A B − 7 3 A B = 7 7 A B − 3 A B = 7 4
AP : PB = 3 7 A B : 4 7 A B \dfrac{3}{7}AB : \dfrac{4}{7}AB 7 3 A B : 7 4 A B
∴ P divides line segment AB in the ratio 3 : 4.
Let co-ordinates of P be (x, y).
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get,
⇒ ( x , y ) = ( 3 × 2 + 4 × − 2 3 + 4 , 3 × − 4 + 4 × − 2 3 + 4 ) ⇒ ( x , y ) = ( 6 − 8 7 , − 12 − 8 7 ) ⇒ ( x , y ) = ( − 2 7 , − 20 7 ) . \Rightarrow (x, y) = \Big(\dfrac{3 \times 2 + 4 \times -2}{3 + 4}, \dfrac{3 \times -4 + 4 \times -2}{3 + 4}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{6 - 8}{7}, \dfrac{-12 - 8}{7}\Big) \\[1em] \Rightarrow (x, y) = \Big(-\dfrac{2}{7}, -\dfrac{20}{7}\Big). ⇒ ( x , y ) = ( 3 + 4 3 × 2 + 4 × − 2 , 3 + 4 3 × − 4 + 4 × − 2 ) ⇒ ( x , y ) = ( 7 6 − 8 , 7 − 12 − 8 ) ⇒ ( x , y ) = ( − 7 2 , − 7 20 ) .
Hence, P = ( − 2 7 , − 20 7 ) . \Big(-\dfrac{2}{7}, -\dfrac{20}{7}\Big). ( − 7 2 , − 7 20 ) .
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer
Let C, D and E divide the line segment AB into four equal parts.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Let co-ordinates of C be (a, b) and from figure we see that C divides AB in the ratio 1 : 3.
Substituting values we get :
⇒ ( a , b ) = ( 1 × 2 + 3 × − 2 1 + 3 , 1 × 8 + 3 × 2 1 + 3 ) = ( 2 − 6 4 , 8 + 6 4 ) = ( − 4 4 , 14 4 ) = ( − 1 , 7 2 ) . \Rightarrow (a, b) = \Big(\dfrac{1 \times 2 + 3 \times -2}{1 + 3}, \dfrac{1 \times 8 + 3 \times 2}{1 + 3}\Big) \\[1em] = \Big(\dfrac{2 - 6}{4}, \dfrac{8 + 6}{4}\Big) \\[1em] = \Big(-\dfrac{4}{4}, \dfrac{14}{4}\Big) \\[1em] = \Big(-1, \dfrac{7}{2}\Big). ⇒ ( a , b ) = ( 1 + 3 1 × 2 + 3 × − 2 , 1 + 3 1 × 8 + 3 × 2 ) = ( 4 2 − 6 , 4 8 + 6 ) = ( − 4 4 , 4 14 ) = ( − 1 , 2 7 ) .
From figure,
D is the mid-point of AB. Let co-ordinates of D be (g, h).
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( g , h ) = ( − 2 + 2 2 , 2 + 8 2 ) = ( 0 2 , 10 2 ) = ( 0 , 5 ) . \Rightarrow (g, h) = \Big(\dfrac{-2 + 2}{2}, \dfrac{2 + 8}{2}\Big) \\[1em] = \Big(\dfrac{0}{2}, \dfrac{10}{2}\Big) \\[1em] = (0, 5). ⇒ ( g , h ) = ( 2 − 2 + 2 , 2 2 + 8 ) = ( 2 0 , 2 10 ) = ( 0 , 5 ) .
Let co-ordinates of E be (p, q) and from figure we see that E divides AB in the ratio 3 : 1.
Substituting values we get :
⇒ ( p , q ) = ( 3 × 2 + 1 × − 2 1 + 3 , 3 × 8 + 1 × 2 1 + 3 ) = ( 6 − 2 4 , 24 + 2 4 ) = ( 4 4 , 26 4 ) = ( 1 , 13 2 ) . \Rightarrow (p, q) = \Big(\dfrac{3 \times 2 + 1 \times -2}{1 + 3}, \dfrac{3 \times 8 + 1 \times 2}{1 + 3}\Big) \\[1em] = \Big(\dfrac{6 - 2}{4}, \dfrac{24 + 2}{4}\Big) \\[1em] = \Big(\dfrac{4}{4}, \dfrac{26}{4}\Big) \\[1em] = \Big(1, \dfrac{13}{2}\Big). ⇒ ( p , q ) = ( 1 + 3 3 × 2 + 1 × − 2 , 1 + 3 3 × 8 + 1 × 2 ) = ( 4 6 − 2 , 4 24 + 2 ) = ( 4 4 , 4 26 ) = ( 1 , 2 13 ) .
Hence, co-ordinates of required points are ( − 1 , 7 2 ) , ( 0 , 5 ) , ( 1 , 13 2 ) \Big(-1, \dfrac{7}{2}\Big), (0, 5), \Big(1, \dfrac{13}{2}\Big) ( − 1 , 2 7 ) , ( 0 , 5 ) , ( 1 , 2 13 )
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
Answer
Let co-ordinates of rhombus be A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1).
By distance formula,
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Substituting values we get :
⇒ A C = ( − 1 − 3 ) 2 + ( 4 − 0 ) 2 = ( − 4 ) 2 + 4 2 = 16 + 16 = 32 = 4 2 units . ⇒ B D = ( − 2 − 4 ) 2 + ( − 1 − 5 ) 2 = ( − 6 ) 2 + ( − 6 ) 2 = 36 + 36 = 72 = 6 2 units . \Rightarrow AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 4^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = 4\sqrt{2} \text{ units}. \\[1em] \Rightarrow BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} \\[1em] = \sqrt{(-6)^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2} \text{ units}. ⇒ A C = ( − 1 − 3 ) 2 + ( 4 − 0 ) 2 = ( − 4 ) 2 + 4 2 = 16 + 16 = 32 = 4 2 units . ⇒ B D = ( − 2 − 4 ) 2 + ( − 1 − 5 ) 2 = ( − 6 ) 2 + ( − 6 ) 2 = 36 + 36 = 72 = 6 2 units .
By formula,
Area of rhombus (A) = 1 2 × \dfrac{1}{2} \times 2 1 ×
Substituting values we get :
A = 1 2 × A C × B D = 1 2 × 4 2 × 6 2 = 1 2 × 24 × 2 = 1 2 × 48 = 24 sq. units . A = \dfrac{1}{2} \times AC \times BD \\[1em] = \dfrac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} \\[1em] = \dfrac{1}{2} \times 24 \times 2 \\[1em] = \dfrac{1}{2} \times 48 \\[1em] = 24 \text{ sq. units}. A = 2 1 × A C × B D = 2 1 × 4 2 × 6 2 = 2 1 × 24 × 2 = 2 1 × 48 = 24 sq. units .
Hence, area of rhombus = 24 sq. units.
