Concept & Reference:
Coordinate geometry is an algebraic tool for studying geometry of figures. It helps us to study geometry using algebra, and understand algebra with the help of geometry. The distance of a point from the y-axis is called its x-coordinate (abscissa), and the distance from the x-axis is called its y-coordinate (ordinate). (Page 99)
DDistance Formula
Formula & Explanation:
The distance between two points P(x₁, y₁) and Q(x₂, y₂) is given by:
PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
Distance of a point P(x, y) from the origin O(0, 0) is OP = √(x² + y²). (Page 102)
1Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (-a, -b)
Solution & Reference:
Using the distance formula: PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
2Find 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?
Solution & Reference:
Distance = √[(36-0)² + (15-0)²] = √[1296 + 225] = √1521 = 39 km
Yes, this is the distance between towns A and B as discussed in Section 7.2 where town B is 36 km east and 15 km north of town A. (Page 105, Exercise 7.1 Q2)
3Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution & Reference:
Let A(1,5), B(2,3), C(-2,-11)
AB = √[(2-1)² + (3-5)²] = √[1 + 4] = √5 ≈ 2.24
BC = √[(-2-2)² + (-11-3)²] = √[16 + 196] = √212 ≈ 14.56
AC = √[(-2-1)² + (-11-5)²] = √[9 + 256] = √265 ≈ 16.28
Since AB + BC ≠ AC (2.24 + 14.56 = 16.8 ≠ 16.28), the points are not collinear. (Page 105, Exercise 7.1 Q3)
4Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution & Reference:
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 is isosceles. (Page 105, Exercise 7.1 Q4)
5In 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 & Reference:
From Fig. 7.8, coordinates are 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
Diagonals: AC = √[(9-3)² + (4-4)²] = √36 = 6
BD = √[(6-6)² + (1-7)²] = √36 = 6
Since all sides are equal and diagonals are equal, ABCD is a square. Champa is correct. (Page 105, Exercise 7.1 Q5)
6Name 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 & Reference:
(i) Let A(-1,-2), B(1,0), C(-1,2), D(-3,0)
AB = √[(1+1)² + (0+2)²] = √[4+4] = √8
BC = √[(-1-1)² + (2-0)²] = √[4+4] = √8
CD = √[(-3+1)² + (0-2)²] = √[4+4] = √8
DA = √[(-1+3)² + (-2-0)²] = √[4+4] = √8
Diagonals: AC = √[(-1+1)² + (2+2)²] = √16 = 4
BD = √[(-3-1)² + (0-0)²] = √16 = 4
All sides equal and diagonals equal → Square
(ii) Points don't form a specific quadrilateral as sides are unequal.
(iii) 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
CD = √[(1-4)² + (2-3)²] = √[9+1] = √10
DA = √[(4-1)² + (5-2)²] = √[9+9] = √18
AB = CD and BC = DA → Parallelogram
(Page 105-106, Exercise 7.1 Q6)
7Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution & Reference:
Let P(x, 0) be the point on x-axis.
AP = BP ⇒ AP² = BP²
(x-2)² + (0+5)² = (x+2)² + (0-9)²
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
-4x + 29 = 4x + 85
-8x = 56 ⇒ x = -7
Required point is (-7, 0). (Page 106, Exercise 7.1 Q7)
8Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
10Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution & Reference:
Let P(x,y) be equidistant from A(3,6) and B(-3,4)
AP = BP ⇒ AP² = BP²
(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
This is the required relation. (Page 106, Exercise 7.1 Q10)
SSection Formula
Formula & Explanation:
The coordinates of point P(x, y) which divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁:m₂ are:
x = (m₁x₂ + m₂x₁)/(m₁ + m₂), y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
Mid-point formula (ratio 1:1): ((x₁+x₂)/2, (y₁+y₂)/2)
(Page 107)
1Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
Solution & Reference:
Using section formula:
x = (2×4 + 3×(-1))/(2+3) = (8-3)/5 = 1
y = (2×(-3) + 3×7)/(2+3) = (-6+21)/5 = 3
Required point is (1, 3). (Page 111, Exercise 7.2 Q1)
2Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution & Reference:
Let P and Q be points of trisection.
P divides AB in ratio 1:2:
x = (1×(-2) + 2×4)/(1+2) = (-2+8)/3 = 2
y = (1×(-3) + 2×(-1))/3 = (-3-2)/3 = -5/3
Q divides AB in ratio 2:1:
x = (2×(-2) + 1×4)/3 = (-4+4)/3 = 0
y = (2×(-3) + 1×(-1))/3 = (-6-1)/3 = -7/3
Points are (2, -5/3) and (0, -7/3). (Page 111, Exercise 7.2 Q2)
3To 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 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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?
Solution & Reference:
AD = 100m (100 flower pots at 1m distance)
Niharika's position: 2nd line, 1/4 × 100 = 25m → (2, 25)
Preet's position: 8th line, 1/5 × 100 = 20m → (8, 20)
Distance = √[(8-2)² + (20-25)²] = √[36+25] = √61 m
Midpoint (Rashmi's position): ((2+8)/2, (25+20)/2) = (5, 22.5)
She should post the blue flag at (5, 22.5). (Page 111, Exercise 7.2 Q3)
4Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution & Reference:
Let ratio be k:1
x = (k×6 + 1×(-3))/(k+1) = -1
6k - 3 = -k -1 ⇒ 7k = 2 ⇒ k = 2/7
Ratio = 2:7
Verify with y-coordinate: (k×(-8) + 1×10)/(k+1) = (-16/7 + 10)/(2/7 + 1) = (54/7)/(9/7) = 6 ✓
(Page 111, Exercise 7.2 Q4)
5Find 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.
Solution & Reference:
Let ratio be k:1. Point on x-axis has y=0.
y = (k×5 + 1×(-5))/(k+1) = 0
5k - 5 = 0 ⇒ k = 1
Ratio = 1:1 (midpoint)
x = (1×(-4) + 1×1)/2 = -3/2
Point of division: (-3/2, 0). (Page 111, Exercise 7.2 Q5)
6If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution & Reference:
In 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 = 7/2 ⇒ 1+x = 7 ⇒ x = 6
8/2 = (y+5)/2 ⇒ 8 = y+5 ⇒ y = 3
(Page 111, Exercise 7.2 Q6)
7Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Solution & Reference:
Centre is midpoint of 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
Coordinates of A are (3, -10). (Page 111, Exercise 7.2 Q7)
8If 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.
Solution & Reference:
AP:PB = 3:4 (since AP = 3/7 AB ⇒ PB = 4/7 AB)
x = (3×2 + 4×(-2))/(3+4) = (6-8)/7 = -2/7
y = (3×(-4) + 4×(-2))/7 = (-12-8)/7 = -20/7
Coordinates of P are (-2/7, -20/7). (Page 111, Exercise 7.2 Q8)
9Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution & Reference:
Let P, Q, R divide AB into four equal parts.
P divides AB in ratio 1:3:
x = (1×2 + 3×(-2))/4 = (2-6)/4 = -1
y = (1×8 + 3×2)/4 = (8+6)/4 = 3.5
Q is midpoint: ((-2+2)/2, (2+8)/2) = (0, 5)
R divides AB in ratio 3:1:
x = (3×2 + 1×(-2))/4 = (6-2)/4 = 1
y = (3×8 + 1×2)/4 = (24+2)/4 = 6.5
Points: (-1, 3.5), (0, 5), (1, 6.5). (Page 111, Exercise 7.2 Q9)
10Find 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)]
Solution & Reference:
Diagonals: AC and BD
A(3,0), C(-1,4): AC = √[(-1-3)² + (4-0)²] = √[16+16] = √32 = 4√2
B(4,5), D(-2,-1): 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 × 48 = 24 sq units
(Page 111, Exercise 7.2 Q10)