Q1. Find the distance between the Following pairs of points:
(i) (2,3),(4,1)
Solution:
Let
By Distance Formula
A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(4 – 2)² + (1 – 3)²
⇒ A B = √(2)² + (-2)²
⇒ A B = √4 + 4
⇒ A B = √8
⇒ A B = 2√2
(ii) (-5,7),(-1,3)
Solution:
Let
A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(-1 + 5)² + (3 – 7)²
⇒ A B = √(4)² + (-4)²
⇒ A B = √16 + 16
⇒ A B = √32
⇒ A B = 4√2
(iii) (a, b),(-a, -b)
Solution:
A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(-a – a)² + (-b – b)²
⇒ A B = √(-2a )² + (-2b)²
⇒ A B = √4a² + 4b²
⇒ A B = √4(a² + b²)
⇒ A B = 2√(a² + b²)
Q.2 Find the distance between the points (0,0) and (36,15).
A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(36 – 0)² + (15 – 0)²
⇒ A B = √(36)² + √(15)²
⇒ A B = √1296+225
⇒ A B = √1521
⇒ A B = √3×3×13×13
⇒ A B = 3×13
⇒ A B = 39 units
Q.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)
Distance A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(2-1)² + (3 – 5)²
⇒ A B = √ (1)² + (-2)²
⇒ A B = √1+4
⇒ A B = √5 units
Distance B C = √(x2 – x1)² + (y2 – y1)²
⇒ B C = √(-2 – 2)² + (-11 – 3)²
⇒ B C = √(-4)² + (-14)²
⇒ B C = √16 + 196
⇒ B C = √212 units
A C = √(x2 – x1)² + (y2 – y1)²
⇒ A C = √(-2 -1)² + (-11 -5)²
⇒ A C = √(-3)² + (-16)²
⇒ A C = √ 9 + 256
⇒ A C = √265
From above we can say that not sum of any two is equal to third therefore these points are not co-linear
Q.4 Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let
Distance A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(6 – 5)² + (4 + 2)²
⇒ A B = √(1)² + (6)²
⇒ A B = √1+36
⇒ A B = √37
Distance B C = √(x2 – x1)² + (y2 – y1)²
⇒ B C = √(7 – 6)² + (-2 – 4)²
⇒ B C = √(1)² + (-6)²
⇒ B C = √1+36
⇒ B C = √37
Distance A C = √(x2 – x1)² + (y2 – y1)²
⇒ A C = √(7 – 5)² + (-2 + 2)²
⇒ A C = √(2)² + (0)²
⇒ A C = √4
⇒ A C = 2units
Hence
A B = B C ≠ A C
Now If two sides are equal then it is isosceles Triangle
∴ Δ ABC is Isosceles Triangle
Q.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.
Solution:
According to figure coordinates of following points are
A = (3,4),
B = (6,7)
C = (9,4)
D = (6,1)
Now we will check for square by distance formula
A (3,4)•——————–•B(6,7)
Distance A B = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A B = √(6 – 3)² + (7 – 4)²
⇒ Distance A B = √(3)² + (3)²
⇒ Distance A B = √9+9
⇒ Distance A B = √18
B (6,7)•——————–•C(9,4)
Distance B C = √(x2 – x1)² + (y2 – y1)²
Distance B C = √(9 – 6)² + (4 -7)²
Distance B C = √(3)² + (-3)²
Distance B C = √(9+9)
Distance B C = √18
C (9,4)•——————–•D(6,1)
Distance C D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance C D = √(6-9)² + (1-4)²
⇒ Distance C D = √(-3)² + (-3)²
⇒ Distance C D = √(9+9)
⇒ Distance C D = √18
A(3,4)•——————–•D(6,1)
Distance A D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A D = √(6 – 3)² + (1 – 4)²
⇒ Distance A D = √(3)² + (-3)²
⇒ Distance A D = √9+9
⇒ Distance A D = √18
A(3,4)•——————–•C(9,4)
Distance A C = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A C = √(9 – 3)² + (4 – 4)²
⇒ Distance A C = √(6)² + (0)²
⇒ Distance A C = √36
⇒ Distance A C = 6
B(6,7)•——————–•D(6,1)
Distance B D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance B D = √(6-6)² + (1-7)²
⇒ Distance B D = √(0)² + (-6)²
⇒ Distance B D = √36
⇒ Distance B D = 6
AB = BC = CD = AD = √18 and Diagonal A C = B D = 6
Hence all sides are equal and diagonals also equal
∴ Chameli was right, it is a square
Q.6 Name the type of quadrilateral formed , if only, by the following points, and give reasons for your answer:
(i) (-1, -2) (1, 0) (-1, 2) (-3, 0)
Let
Distance A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(1 + 1)² + (0 + 2)²
⇒ A B = √(2)² + (2)²
⇒ A B = √4 +4
⇒ A B = √8
⇒ A B = 2√2
Distance B C = √(-1 – 1)² + (2 – 0)²
⇒ B C = √(-2)²+(2)²
⇒ B C = √4 + 4
⇒ B C = √8
⇒ B C = 2√2
Distance C D = √(x2 – x1)² + (y2 – y1)²
⇒ C D = √(-3 + 1)² + (0 – 2)²
⇒ C D = √(-2)² + (-2)²
⇒ C D = √4 + 4
⇒ C D = √8
⇒ C D = 2√2
Distance A D = √(x2 – x1)² + (y2 – y1)²
⇒ A D = √(x2 – x1)² + (y2 – y1)²
⇒ A D = √(-3 + 1)² + (0 + 2)
⇒ A D = √(-2)² + (2)²
⇒ A D = √4+4
⇒ A D = √8
⇒ A D = 2√2
From Above
A B = B C = C D = A D = 2√2
Diagonal A(-1, -2) and C(-1, 2)
Diagonal A C =√(-1+1)² + (2+2)²
⇒ A C = √(0)² + (4)²
⇒ A C = √16
⇒ A C = 4 units
Diagonal B D = √(-3-1)²+(0-0)²
⇒ B D = √(-4)²
⇒ B D = √16
⇒ B D = 4
Diagonal BD = Diagonal AC
Hence All sides are equal and Diagonal Also equal therefore it is the coordinates of Square.
(ii) (-3, 5) (3, 1) (0, 3) (-1, -4)
According to figure
Distance A B = √(x2 – x1)² + (y2 – y1)²
⇒ A B = √(3 + 3)² + (1 – 5)²
⇒ A B = √(6)² + (- 4)²
⇒ A B = √(36 + 16)²
⇒ A B = √(52)²
⇒ A B = 2√13units
Distance B C = √(x2 – x1)² + (y2 – y1)²
⇒ Distance B C = √(0 – 3)² + (3 – 1)²
⇒ Distance B C = √(-3)² + (2)²
⇒ Distance B C = √(9+4)
⇒ Distance B C = √13units
Distance C D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance C D = √(-1 – 0)² + (3+4)²
⇒ Distance CD = √(-1 )² + (7)²
⇒ Distance CD = √1 + 49
⇒ Distance CD = √50
⇒ Distance CD = 5√2units
Distance A D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A D = √(-1 + 3)² + (-4 – 5)²
⇒ Distance A D = √(2)² + (-9)²
⇒ Distance A D = √4 + 81
⇒ Distance A D = √85 units
Distance A C = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A C = √(0+3)² + (3-5)²
⇒ Distance A C = √(3)² + (-2)²
⇒ Distance A C = √9 + 4
⇒ Distance A C = √13
Here
B C + A C = A B
√13 + √13 = 2√13
ABC are colinear
Hence Given Co-ordinates are not form a quadrilateral
(iii) (4, 5) (7, 6) (4, 3) (1, 2)
Let A (4, 5) , B (7, 6) , C (4, 3) and D(1, 2)
Distance A B = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A B = √(7 – 4)² + (6 – 5)²
⇒ Distance A B = √(3)² + (1)²
⇒ Distance A B = √9+1
⇒ Distance A B = √10 units
Distance B C = √(x2 – x1)² + (y2 – y1)²
⇒ Distance B C = √(4 – 7)² + (3 – 6)²
⇒ Distance B C = √(-3)² + (-3)²
⇒ Distance B C = √(9 + 9)
⇒ Distance B C = √18
⇒ Distance B C = 3√2 units
Distance C D =√(x2 – x1)² + (y2 – y1)²
⇒ Distance C D = √(1 – 4)² + (2 – 3)²
⇒ Distance C D = √(-3)² + (-1)²
⇒ Distance C D = √(9+1)
⇒ Distance C D = √10
Distance A D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A D = √(1 – 4)² + (2 – 5)
⇒ Distance A D = √(-3)² + (-3)²
⇒ Distance A D = √(9+9)
⇒ Distance A D = √19
⇒ Distance A D = 3√2units
Diagonal A (4,5) and C(4,3)
Distance A C = √(x2 – x1)² + (y2 – y1)²
⇒ Distance A C = √(4 – 4)² + (3 – 5)²
⇒ Distance A C = √(0)² + (-2)²
⇒ Distance A C = √4
⇒ Distance A C = 2 units
B(7,6) •——————• D(1,2)
Distance B D = √(x2 – x1)² + (y2 – y1)²
⇒ Distance B D = √(1 – 7)² + (2 – 6)²
⇒ Distance B D = √(-6)² + (- 4)²
⇒ Distance B D = √(36+16)
⇒ Distance B D = √(52)
⇒ Distance B D = √2×2×13
⇒ Distance B D = 2√13units
Now
A B = C D = √10 and B C = A D = 3√2 but Diagonal A C ≠ B D
Hence ABCD is a parallelogram.
Q.7 Find the point in the x- axis which is equidistant from (2,-5) and (-2,9).
Solution:
Let the coordinate in X- axis be (x,0), and A (2,-5) and B(-2, 9)
Distance P A = Distance P B
√(x2 – x1)² + (y2 – y1)² = √(x2 – x1)² + (y2 – y1)²
⇒ √(2 – x)² + (-5 – 0)² = √(-2 – x)² + (9 – 0)²
⇒ (2 – x)² + (-5 – 0)² = (-2 – x)² + (9 – 0)²
⇒ (2)² – 2×2×x + x² + (-5)² = (-2)² -2×-2×x + x²
⇒ 4 – 4x + x² + 25 = 4 + 4x + x²
⇒ 4 – 4x + x² + 25 – 4 – 4x – x² = 0
⇒ -8x + 25 = 0
⇒ 8x = – 25
⇒ x = -25/8
⇒ x = -7
Hence the coordinate on x-axis = (-7,0)
Q.8 Find the values of y for which the distance between the points P (2,-3) and Q (10,y) is 10 units.
Solution:
By Distance Formula
Distance P Q = 10
√(x2 – x1)² + (y2 – y1)² = Distance P Q
⇒ √(10 – 2)² + (y +3)² = 10
squaring both side
⇒ (8)² + (y +3)² = 10²
⇒ 64 + y² + 2×y×3 + 3² = 100
⇒ 64 + y² + 6y + 9 = 100
⇒ 73 + y² + 6y = 100
⇒ 73 + y² + 6y – 100 = 0
⇒ y² + 6y – 27 = 0
⇒ y² + (9-3)y – 27 = 0
⇒ y² + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3(y – 9) = 0
⇒ (y + 9)(y – 3) = 0
y + 9 = 0 ; y – 3 = 0
y = -9 ; y = 3
y = -9,3
Q.9 If Q (0, 1) is equidistance from P(5, -3) and R(x, 6).Find the value of x also find the distances Q R and P R.
Solution:
Let
According to Question
Q P = Q R
⇒ √(x2 – x1)² + (y2 – y1)² = √(x2 – x1)² + (y2 – y1)²
⇒ √(5 – 0)² + (-3 – 1)² = √(x – 0)² + (6 – 1)²
⇒ √(5) ² + (-4)² = √(x)² + (5)²
⇒ √(25+16) = √(x²+25)
squaring both side
⇒ 41 = x² + 25
⇒ 41 – 25 = x²
⇒ 16 = x²
⇒ ±4 = x
Hence x = 4,-4
Q (0, 1) •——————• R(5,-3)
Distance Q P = √(x2 – x1)² + (y2 – y1)²
⇒ Distance Q P = √(5 – 0)² + (-3 – 1)²
⇒ Distance Q P = √(5) ² + (-4)²
⇒ Distance Q P = √(25+16)
⇒ Distance Q P = √(41) units
If x = -4
Q (0,1) ⋅———————–⋅ R (-4,6)
Q R = √(x2 – x1)² + (y2 – y1)²
⇒ Q R = √(-4 – 0)² + (6 – 1)²
⇒ Q R = √(-4 )² + (5)²
⇒ Q R = √(16+25)
⇒ Q R = √41
P(5, -3) ⋅———————–⋅R(-4, 6)
P R = √(x2 – x1)² + (y2 – y1)²
⇒ P R = √(-4 – 5)² + (6 + 3 )²
⇒ P R = √(-1)² + (9)²
⇒ P R = √1 + 81
⇒ P R = √82
If x = 4 then
P(5, -3) ⋅———————–⋅ Q(4,6)
P Q = √(4-5)² + (6+3)
⇒ P Q = √(-1)² + (9)²
⇒ P Q = √(1+81)
⇒ P Q = √82
Q.10 Find a relationship between x and y such that the point (x, y) is equidistant from the point (3,6) and (-3,4).
Given:
Let P(x, y) Equidistance from Q(3,6) and R(-3,4)
Distance P Q = Distance P R
⇒ √(x2 – x1)² + (y2 – y1)² = √(x2 – x1)² + (y2 – y1)²
⇒ √(3 – x)² + (6 – y)² = √(-3 – x)² + (4 – y)²
Squaring both side
⇒ (3 – x)² + (6 – y)² = (-3 – x)² + (4 – y)²
⇒ 3² – 2×3×x + x² + 6² – 2×6×y + y² = (-3)² – 2×-3×x + x² + 4² – 2×4×y +y²
⇒ 9 – 6x + x² + 36 – 12y + y² = 9 + 6x + x² + 16 – 8y + y²
⇒ 9 – 6x + x² + 36 – 12y + y² -9 -6x – x² – 16 + 8y – y² = 0
⇒ -12x – 4y + 20 = 0
⇒ -4(3x + y + 5) = 0
⇒ 3x + y + 5 = 0
⇒ 3x + y = -5