Exercise 7.1 Solutions
1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
Using the distance formula:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(4 - 2)² + (1 - 3)²]
= √[2² + (-2)²]
= √[4 + 4]
= √8
= 2√2 units
(ii) (-5, 7), (-1, 3)
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(-1 - (-5))² + (3 - 7)²]
= √[4² + (-4)²]
= √[16 + 16]
= √32
= 4√2 units
(iii) (a, b), (-a, -b)
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(-a - a)² + (-b - b)²]
= √[(-2a)² + (-2b)²]
= √[4a² + 4b²]
= 2√(a² + b²) units
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?
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(36 - 0)² + (15 - 0)²]
= √[36² + 15²]
= √[1296 + 225]
= √1521
= 39 units
Yes, this is the same distance between the two towns A and B discussed in Section 7.2, which was 39 km.
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Let A = (1, 5), B = (2, 3), C = (-2, -11)
AB = √[(2 - 1)² + (3 - 5)²] = √[1 + 4] = √5
BC = √[(-2 - 2)² + (-11 - 3)²] = √[16 + 196] = √212 = 2√53
AC = √[(-2 - 1)² + (-11 - 5)²] = √[9 + 256] = √265
Since AB + BC ≠ AC, and no two distances sum to the third, the points are not collinear.
4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Let A = (5, -2), B = (6, 4), C = (7, -2)
AB = √[(6 - 5)² + (4 - (-2))²] = √[1 + 36] = √37
BC = √[(7 - 6)² + (-2 - 4)²] = √[1 + 36] = √37
AC = √[(7 - 5)² + (-2 - (-2))²] = √[4 + 0] = 2
Since AB = BC = √37, the triangle has two equal sides, so it is an isosceles triangle.
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.
From the figure, the points are approximately:
A(3, 4), B(6, 7), C(9, 4), D(6, 1)
AB = √[(6 - 3)² + (7 - 4)²] = √[9 + 9] = √18 = 3√2
BC = √[(9 - 6)² + (4 - 7)²] = √[9 + 9] = √18 = 3√2
CD = √[(6 - 9)² + (1 - 4)²] = √[9 + 9] = √18 = 3√2
DA = √[(3 - 6)² + (4 - 1)²] = √[9 + 9] = √18 = 3√2
AC = √[(9 - 3)² + (4 - 4)²] = √[36 + 0] = 6
BD = √[(6 - 6)² + (1 - 7)²] = √[0 + 36] = 6
Since all four sides are equal (3√2) and both diagonals are equal (6), ABCD is a square. Therefore, Champa is correct.
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)
Let A = (-1, -2), B = (1, 0), C = (-1, 2), D = (-3, 0)
AB = √[(1 - (-1))² + (0 - (-2))²] = √[4 + 4] = √8 = 2√2
BC = √[(-1 - 1)² + (2 - 0)²] = √[4 + 4] = √8 = 2√2
CD = √[(-3 - (-1))² + (0 - 2)²] = √[4 + 4] = √8 = 2√2
DA = √[(-1 - (-3))² + (-2 - 0)²] = √[4 + 4] = √8 = 2√2
AC = √[(-1 - (-1))² + (2 - (-2))²] = √[0 + 16] = 4
BD = √[(-3 - 1)² + (0 - 0)²] = √[16 + 0] = 4
Since all sides are equal and diagonals are equal, it is a square.
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
Let A = (-3, 5), B = (3, 1), C = (0, 3), D = (-1, -4)
AB = √[(3 - (-3))² + (1 - 5)²] = √[36 + 16] = √52
BC = √[(0 - 3)² + (3 - 1)²] = √[9 + 4] = √13
CD = √[(-1 - 0)² + (-4 - 3)²] = √[1 + 49] = √50
DA = √[(-3 - (-1))² + (5 - (-4))²] = √[4 + 81] = √85
Since all sides are different, it is a general quadrilateral.
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Let A = (4, 5), B = (7, 6), C = (4, 3), D = (1, 2)
AB = √[(7 - 4)² + (6 - 5)²] = √[9 + 1] = √10
BC = √[(4 - 7)² + (3 - 6)²] = √[9 + 9] = √18 = 3√2
CD = √[(1 - 4)² + (2 - 3)²] = √[9 + 1] = √10
DA = √[(4 - 1)² + (5 - 2)²] = √[9 + 9] = √18 = 3√2
AC = √[(4 - 4)² + (3 - 5)²] = √[0 + 4] = 2
BD = √[(1 - 7)² + (2 - 6)²] = √[36 + 16] = √52 = 2√13
Since AB = CD and BC = DA, but AC ≠ BD, it is a parallelogram.
7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Let the point on x-axis be P(x, 0). Since it's equidistant from A(2, -5) and B(-2, 9):
PA = PB
√[(x - 2)² + (0 - (-5))²] = √[(x - (-2))² + (0 - 9)²]
(x - 2)² + 25 = (x + 2)² + 81
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
-4x + 29 = 4x + 85
-8x = 56
x = -7
So the required point is (-7, 0).
8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
PQ = 10
√[(10 - 2)² + (y - (-3))²] = 10
√[64 + (y + 3)²] = 10
Squaring both sides:
64 + (y + 3)² = 100
(y + 3)² = 36
y + 3 = ±6
y = 3 or y = -9
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.
QP = QR
√[(5 - 0)² + (-3 - 1)²] = √[(x - 0)² + (6 - 1)²]
√[25 + 16] = √[x² + 25]
√41 = √(x² + 25)
Squaring both sides:
41 = x² + 25
x² = 16
x = ±4
For x = 4:
QR = √[(4 - 0)² + (6 - 1)²] = √[16 + 25] = √41
PR = √[(4 - 5)² + (6 - (-3))²] = √[1 + 81] = √82
For x = -4:
QR = √[(-4 - 0)² + (6 - 1)²] = √[16 + 25] = √41
PR = √[(-4 - 5)² + (6 - (-3))²] = √[81 + 81] = √162 = 9√2
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Let P(x, y) be equidistant from A(3, 6) and B(-3, 4):
PA = PB
√[(x - 3)² + (y - 6)²] = √[(x - (-3))² + (y - 4)²]
(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²
x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16
-6x - 12y + 45 = 6x - 8y + 25
-12x - 4y + 20 = 0
3x + y - 5 = 0
So the required relation is 3x + y - 5 = 0.
Exercise 7.2 Solutions
1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Using the section formula:
x = (m₁x₂ + m₂x₁)/(m₁ + m₂) = (2×4 + 3×(-1))/(2 + 3) = (8 - 3)/5 = 5/5 = 1
y = (m₁y₂ + m₂y₁)/(m₁ + m₂) = (2×(-3) + 3×7)/(2 + 3) = (-6 + 21)/5 = 15/5 = 3
So the required point is (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Let P and Q be the points of trisection, with P dividing AB in ratio 1:2 and Q dividing AB in ratio 2:1.
For P:
x = (1×(-2) + 2×4)/(1 + 2) = (-2 + 8)/3 = 6/3 = 2
y = (1×(-3) + 2×(-1))/(1 + 2) = (-3 - 2)/3 = -5/3
For Q:
x = (2×(-2) + 1×4)/(2 + 1) = (-4 + 4)/3 = 0/3 = 0
y = (2×(-3) + 1×(-1))/(2 + 1) = (-6 - 1)/3 = -7/3
So the points of trisection are (2, -5/3) and (0, -7/3).
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 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 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?
Assuming AD = 100m (since there are 100 flower pots placed 1m apart):
Niharika's position: 2nd line, 1/4 of AD = 25m from A. Coordinates: (2, 25)
Preet's position: 8th line, 1/5 of AD = 20m from A. Coordinates: (8, 20)
Distance between flags:
d = √[(8 - 2)² + (20 - 25)²] = √[36 + 25] = √61 ≈ 7.81m
Midpoint between flags:
x = (2 + 8)/2 = 5, y = (25 + 20)/2 = 22.5
So Rashmi should post her blue flag on the 5th line at 22.5m from A.
4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Let the ratio be k:1. Using section formula:
-1 = (6k + (-3)×1)/(k + 1)
-1(k + 1) = 6k - 3
-k - 1 = 6k - 3
-7k = -2
k = 2/7
So the ratio is 2:7.
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.
Let the ratio be k:1. Since the point lies on x-axis, y-coordinate is 0.
0 = (5k + (-5)×1)/(k + 1)
0 = (5k - 5)/(k + 1)
5k - 5 = 0
k = 1
So the ratio is 1:1.
Coordinates of division point:
x = (1×(-4) + 1×1)/(1 + 1) = (-4 + 1)/2 = -3/2
y = 0 (since it's on x-axis)
So the point is (-3/2, 0).
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
In a parallelogram, diagonals bisect each other.
Midpoint of AC = Midpoint of BD
((1 + x)/2, (2 + 6)/2) = ((4 + 3)/2, (y + 5)/2)
((1 + x)/2, 4) = (7/2, (y + 5)/2)
Equating coordinates:
(1 + x)/2 = 7/2 ⇒ 1 + x = 7 ⇒ x = 6
4 = (y + 5)/2 ⇒ 8 = y + 5 ⇒ y = 3
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).
Centre is the midpoint of diameter AB.
(2, -3) = ((x + 1)/2, (y + 4)/2)
(x + 1)/2 = 2 ⇒ x + 1 = 4 ⇒ x = 3
(y + 4)/2 = -3 ⇒ y + 4 = -6 ⇒ y = -10
So A is (3, -10).
8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
AP = 3/7 AB ⇒ AP:PB = 3:4
x = (3×2 + 4×(-2))/(3 + 4) = (6 - 8)/7 = -2/7
y = (3×(-4) + 4×(-2))/(3 + 4) = (-12 - 8)/7 = -20/7
So P is (-2/7, -20/7).
9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Let the points be P, Q, R dividing AB into four equal parts.
P divides AB in ratio 1:3:
P = ((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)
Q is the midpoint of AB (ratio 1:1):
Q = ((-2+2)/2, (2+8)/2) = (0, 5)
R divides AB in ratio 3:1:
R = ((3×2 + 1×(-2))/(3+1), (3×8 + 1×2)/(3+1)) = ((6-2)/4, (24+2)/4) = (4/4, 26/4) = (1, 13/2)
So the points are (-1, 7/2), (0, 5), and (1, 13/2).
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 = 1/2 (product of its diagonals)]
Let A(3, 0), B(4, 5), C(-1, 4), D(-2, -1)
Diagonals: AC and BD
AC = √[(-1 - 3)² + (4 - 0)²] = √[16 + 16] = √32 = 4√2
BD = √[(-2 - 4)² + (-1 - 5)²] = √[36 + 36] = √72 = 6√2
Area = 1/2 × AC × BD = 1/2 × 4√2 × 6√2 = 1/2 × 4×6×2 = 24 square units
Key Formulas
Distance Formula
Distance between P(x₁, y₁) and Q(x₂, y₂):
PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
Distance from origin O(0, 0) to P(x, y):
OP = √(x² + y²)
Section Formula
Coordinates of point dividing line segment joining A(x₁, y₁) and B(x₂, y₂)
in ratio m₁ : m₂:
x = (m₁x₂ + m₂x₁)/(m₁ + m₂)
y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
Midpoint formula (when m₁ : m₂ = 1 : 1):
x = (x₁ + x₂)/2
y = (y₁ + y₂)/2
Area of Triangle
Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
Area = 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Collinearity of Points
Three points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are collinear if:
Area of triangle ABC = 0
OR
Slope of AB = Slope of BC = Slope of AC
Centroid of Triangle
Centroid of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)