COORDINATE GEOMETRY - Q&A
EXERCISE 7.1
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)
Solution:
The distance between two points P(x1, y1) and Q(x2, y2) is given by the formula:
Distance = √[ (x2 - x1)2 + (y2 - y1)2 ]
(i) Points: (2, 3) and (4, 1)
Here, x1 = 2, y1 = 3, x2 = 4, y2 = 1
Distance = √[ (4 - 2)2 + (1 - 3)2 ]
= √[ (2)2 + (-2)2 ]
= √[ 4 + 4 ]
= √8
= 2√2 units.
(ii) Points: (- 5, 7) and (- 1, 3)
Here, x1 = -5, y1 = 7, x2 = -1, y2 = 3
Distance = √[ (-1 - (-5))2 + (3 - 7)2 ]
= √[ (-1 + 5)2 + (-4)2 ]
= √[ (4)2 + 16 ]
= √[ 16 + 16 ]
= √32
= 4√2 units.
(iii) Points: (a, b) and (- a, - b)
Here, x1 = a, y1 = b, x2 = -a, y2 = -b
Distance = √[ (-a - a)2 + (-b - b)2 ]
= √[ (-2a)2 + (-2b)2 ]
= √[ 4a2 + 4b2 ]
= √[ 4(a2 + b2) ]
= 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?
Solution:
Let the points be P(0, 0) and Q(36, 15).
Here, x1 = 0, y1 = 0, x2 = 36, y2 = 15
Distance PQ = √[ (36 - 0)2 + (15 - 0)2 ]
= √[ (36)2 + (15)2 ]
= √[ 1296 + 225 ]
= √1521
= 39 units.
Yes, we can find the distance between the two towns A and B.
Assuming town A is at the origin (0, 0) and town B is located at (36, 15) as per the typical example in Section 7.2 of the textbook (where town B is 36 km east and 15 km north of town A):
The distance between them is the same as calculated above.
Distance = 39 km.
3. Determine if the points (1, 5), (2, 3) and (- 2, - 11) are collinear.
Solution:
Let the points be A(1, 5), B(2, 3), and C(-2, -11).
Points are collinear if the sum of the distances of any two segments equals the third segment (e.g., AB + BC = AC).
Distance AB = √[ (2 - 1)2 + (3 - 5)2 ]
= √[ (1)2 + (-2)2 ] = √[ 1 + 4 ] = √5
Distance BC = √[ (-2 - 2)2 + (-11 - 3)2 ]
= √[ (-4)2 + (-14)2 ] = √[ 16 + 196 ] = √212 = 2√53
Distance AC = √[ (-2 - 1)2 + (-11 - 5)2 ]
= √[ (-3)2 + (-16)2 ] = √[ 9 + 256 ] = √265
Here, AB + BC = √5 + √212 ≈ 2.23 + 14.56 = 16.79
AC = √265 ≈ 16.27
Since AB + BC ≠ AC (and no other combination works), the points are not collinear.
Answer: No, the points are not collinear.
4. Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.
Solution:
Let the points be A(5, -2), B(6, 4), and C(7, -2).
An isosceles triangle has at least two sides of equal length.
Length of AB = √[ (6 - 5)2 + (4 - (-2))2 ]
= √[ (1)2 + (6)2 ] = √[ 1 + 36 ] = √37
Length of BC = √[ (7 - 6)2 + (-2 - 4)2 ]
= √[ (1)2 + (-6)2 ] = √[ 1 + 36 ] = √37
Length of AC = √[ (7 - 5)2 + (-2 - (-2))2 ]
= √[ (2)2 + (0)2 ] = √4 = 2
Since AB = BC = √37, two sides are equal.
Answer: Yes, these are the vertices of 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.
Solution:
From the figure (implied coordinates based on standard grid):
Coordinates of A = (3, 4)
Coordinates of B = (6, 7)
Coordinates of C = (9, 4)
Coordinates of D = (6, 1)
To be a square, all four sides must be equal AND the diagonals must be equal.
Sides:
AB = √[ (6 - 3)2 + (7 - 4)2 ] = √[ 32 + 32 ] = √[ 9 + 9 ] = √18 = 3√2
BC = √[ (9 - 6)2 + (4 - 7)2 ] = √[ 32 + (-3)2 ] = √[ 9 + 9 ] = √18 = 3√2
CD = √[ (6 - 9)2 + (1 - 4)2 ] = √[ (-3)2 + (-3)2 ] = √[ 9 + 9 ] = √18 = 3√2
DA = √[ (3 - 6)2 + (4 - 1)2 ] = √[ (-3)2 + 32 ] = √[ 9 + 9 ] = √18 = 3√2
All sides are equal.
Diagonals:
AC = √[ (9 - 3)2 + (4 - 4)2 ] = √[ 62 + 02 ] = √36 = 6
BD = √[ (6 - 6)2 + (1 - 7)2 ] = √[ 02 + (-6)2 ] = √36 = 6
Diagonals are also equal.
Since all sides are equal and diagonals are equal, ABCD is a square.
Answer: 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)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, - 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)
AB = √[ (1 - (-1))2 + (0 - (-2))2 ] = √[ 22 + 22 ] = √8 = 2√2
BC = √[ (-1 - 1)2 + (2 - 0)2 ] = √[ (-2)2 + 22 ] = √8 = 2√2
CD = √[ (-3 - (-1))2 + (0 - 2)2 ] = √[ (-2)2 + (-2)2 ] = √8 = 2√2
DA = √[ (-1 - (-3))2 + (-2 - 0)2 ] = √[ 22 + (-2)2 ] = √8 = 2√2
Diagonal AC = √[ (-1 - (-1))2 + (2 - (-2))2 ] = √[ 02 + 42 ] = √16 = 4
Diagonal BD = √[ (-3 - 1)2 + (0 - 0)2 ] = √[ (-4)2 + 0 ] = √16 = 4
Sides are equal and diagonals are equal.
Answer: It is a Square.
(ii) Let A(-3, 5), B(3, 1), C(0, 3), D(-1, -4)
AB = √[ (3 - (-3))2 + (1 - 5)2 ] = √[ 62 + (-4)2 ] = √[ 36 + 16 ] = √52 = 2√13
BC = √[ (0 - 3)2 + (3 - 1)2 ] = √[ (-3)2 + 22 ] = √[ 9 + 4 ] = √13
AC = √[ (0 - (-3))2 + (3 - 5)2 ] = √[ 32 + (-2)2 ] = √[ 9 + 4 ] = √13
Notice that AC + BC = √13 + √13 = 2√13, which equals AB.
This means points A, B, and C are collinear. A quadrilateral cannot be formed if three vertices are collinear.
Answer: No quadrilateral is formed.
(iii) Let A(4, 5), B(7, 6), C(4, 3), D(1, 2)
AB = √[ (7 - 4)2 + (6 - 5)2 ] = √[ 32 + 12 ] = √10
BC = √[ (4 - 7)2 + (3 - 6)2 ] = √[ (-3)2 + (-3)2 ] = √[ 9 + 9 ] = √18 = 3√2
CD = √[ (1 - 4)2 + (2 - 3)2 ] = √[ (-3)2 + (-1)2 ] = √[ 9 + 1 ] = √10
DA = √[ (4 - 1)2 + (5 - 2)2 ] = √[ 32 + 32 ] = √18 = 3√2
Opposite sides are equal (AB = CD and BC = DA). It is a parallelogram.
Now check diagonals:
AC = √[ (4 - 4)2 + (3 - 5)2 ] = √[ 0 + (-2)2 ] = 2
BD = √[ (1 - 7)2 + (2 - 6)2 ] = √[ (-6)2 + (-4)2 ] = √[ 36 + 16 ] = √52
Diagonals are not equal.
Answer: It is a Parallelogram.
7. Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).
Solution:
Let the point on the x-axis be P(x, 0).
Let A = (2, -5) and B = (-2, 9).
According to the problem, PA = PB.
Squaring both sides, PA2 = PB2.
(x - 2)2 + (0 - (-5))2 = (x - (-2))2 + (0 - 9)2
(x - 2)2 + 52 = (x + 2)2 + (-9)2
x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
- 4x + 29 = 4x + 85
- 4x - 4x = 85 - 29
- 8x = 56
x = 56 / -8
x = - 7
Answer: The 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.
Solution:
Given Distance PQ = 10.
PQ2 = 100.
Using distance formula:
(10 - 2)2 + (y - (-3))2 = 100
(8)2 + (y + 3)2 = 100
64 + (y + 3)2 = 100
(y + 3)2 = 100 - 64
(y + 3)2 = 36
Taking square root on both sides:
y + 3 = ± 6
Case 1: y + 3 = 6 ⇒ y = 3
Case 2: y + 3 = -6 ⇒ y = -9
Answer: The values of y are 3 and -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.
Solution:
Given QP = QR.
Squaring both sides, QP2 = QR2.
(5 - 0)2 + (-3 - 1)2 = (x - 0)2 + (6 - 1)2
52 + (-4)2 = x2 + 52
25 + 16 = x2 + 25
16 = x2
x = ± 4
So, x can be 4 or -4.
Case 1: When x = 4, R is (4, 6).
Distance QR = √[ (4 - 0)2 + (6 - 1)2 ] = √[ 16 + 25 ] = √41
Distance PR = √[ (4 - 5)2 + (6 - (-3))2 ] = √[ (-1)2 + 92 ] = √[ 1 + 81 ] = √82
Case 2: When x = -4, R is (-4, 6).
Distance QR = √[ (-4 - 0)2 + (6 - 1)2 ] = √[ 16 + 25 ] = √41
Distance PR = √[ (-4 - 5)2 + (6 - (-3))2 ] = √[ (-9)2 + 92 ] = √[ 81 + 81 ] = √162 = 9√2
Answer: x = ± 4. QR = √41. PR = √82 (if x=4) or 9√2 (if x=-4).
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let P(x, y) be equidistant from A(3, 6) and B(-3, 4).
So, PA = PB ⇒ PA2 = PB2.
(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
Canceling x2, y2, and 9 from both sides:
- 6x - 12y + 36 = 6x - 8y + 16
- 6x - 6x - 12y + 8y + 36 - 16 = 0
- 12x - 4y + 20 = 0
Divide by -4:
3x + y - 5 = 0
Answer: The relation is 3x + y - 5 = 0.
EXERCISE 7.2
1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2 : 3.
Solution:
Let the points be A(-1, 7) and B(4, -3). The ratio m1:m2 = 2:3.
Using the Section Formula:
x = (m1x2 + m2x1) / (m1 + m2)
y = (m1y2 + m2y1) / (m1 + m2)
x = (2(4) + 3(-1)) / (2 + 3) = (8 - 3) / 5 = 5 / 5 = 1
y = (2(-3) + 3(7)) / (2 + 3) = (-6 + 21) / 5 = 15 / 5 = 3
Answer: The coordinates of the point are (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, - 1) and (- 2, - 3).
Solution:
Let A(4, -1) and B(-2, -3) be the points.
Trisection means dividing the line into 3 equal parts. Let points P and Q trisect AB.
P divides AB in the ratio 1:2.
Coordinates of 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
So, P is (2, -5/3).
Q divides AB in the ratio 2:1.
Coordinates of 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, Q is (0, -7/3).
Answer: 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 1 m each. 100 flower pots have been placed at a distance of 1 m 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:
Total distance AD = 100 pots * 1 m = 100 m.
Niharika's position (Green Flag):
x-coordinate (line number) = 2
y-coordinate (distance) = 1/4 of AD = (1/4) * 100 = 25 m.
Coordinates G(2, 25).
Preet's position (Red Flag):
x-coordinate (line number) = 8
y-coordinate (distance) = 1/5 of AD = (1/5) * 100 = 20 m.
Coordinates R(8, 20).
Distance between flags (GR):
Distance = √[ (8 - 2)2 + (20 - 25)2 ]
= √[ 62 + (-5)2 ]
= √[ 36 + 25 ]
= √61 meters.
Rashmi's position (Blue Flag):
Rashmi posts the flag halfway (midpoint) between G and R.
x = (2 + 8) / 2 = 10 / 2 = 5
y = (25 + 20) / 2 = 45 / 2 = 22.5
Answer: The distance between the flags is √61 m. Rashmi should post her flag on the 5th line at a distance of 22.5 m.
4. Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6).
Solution:
Let the ratio be k:1.
Let A(-3, 10) and B(6, -8). The dividing point is P(-1, 6).
Using Section Formula for x-coordinate:
-1 = (k(6) + 1(-3)) / (k + 1)
-1(k + 1) = 6k - 3
-k - 1 = 6k - 3
-1 + 3 = 6k + k
2 = 7k
k = 2/7
So the ratio is 2:7.
(We can verify with y-coordinate: y = (2(-8) + 7(10)) / 9 = (-16 + 70) / 9 = 54 / 9 = 6. Correct).
Answer: 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.
Solution:
Let the ratio be k:1.
Since the point lies on the x-axis, its y-coordinate is 0. Let the point be P(x, 0).
Using Section Formula for y-coordinate:
y = (m1y2 + m2y1) / (m1 + m2)
0 = (k(5) + 1(-5)) / (k + 1)
0 = 5k - 5
5k = 5
k = 1
So the ratio is 1:1 (Midpoint).
Now, find the x-coordinate:
x = (1(-4) + 1(1)) / (1 + 1)
x = (-4 + 1) / 2
x = -3 / 2
Answer: The ratio is 1 : 1. The coordinates of the point are (- 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.
Solution:
Let the vertices be A(1, 2), B(4, y), C(x, 6), and D(3, 5).
The diagonals of a parallelogram bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD.
Midpoint of AC = [ (1 + x) / 2 , (2 + 6) / 2 ] = [ (1 + x) / 2 , 4 ]
Midpoint of BD = [ (4 + 3) / 2 , (y + 5) / 2 ] = [ 7 / 2 , (y + 5) / 2 ]
Equating the coordinates:
For x:
(1 + x) / 2 = 7 / 2
1 + x = 7
x = 6
For y:
4 = (y + 5) / 2
8 = y + 5
y = 3
Answer: x = 6 and 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).
Solution:
Let the coordinates of A be (x, y).
The centre of the circle is the midpoint of the diameter AB.
Centre O(2, -3) is the midpoint of A(x, y) and B(1, 4).
Using Midpoint Formula:
2 = (x + 1) / 2
4 = x + 1
x = 3
-3 = (y + 4) / 2
-6 = y + 4
y = -10
Answer: The coordinates of point A are (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.
Solution:
Given AP = (3/7) AB.
This means AB is divided into 7 parts, and AP takes 3 parts.
PB = AB - AP = AB - (3/7)AB = (4/7)AB.
Ratio AP : PB = 3 : 4.
Now find P(x, y) dividing A(-2, -2) and B(2, -4) in ratio 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
Answer: The coordinates of P are (- 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.
Solution:
To divide AB into 4 equal parts, we need 3 points: P, Q, R.
Q is the midpoint of AB.
P is the midpoint of AQ.
R is the midpoint of QB.
1. Find Q (Midpoint of AB):
Qx = (-2 + 2) / 2 = 0
Qy = (2 + 8) / 2 = 5
Q is (0, 5).
2. Find P (Midpoint of A(-2, 2) and Q(0, 5)):
Px = (-2 + 0) / 2 = -1
Py = (2 + 5) / 2 = 7/2
P is (-1, 7/2).
3. Find R (Midpoint of Q(0, 5) and B(2, 8)):
Rx = (0 + 2) / 2 = 1
Ry = (5 + 8) / 2 = 13/2
R is (1, 13/2).
Answer: 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)]
Solution:
Let the vertices be A(3, 0), B(4, 5), C(-1, 4), and D(-2, -1).
Area = (1/2) * Diagonal 1 * Diagonal 2.
Diagonal 1 (AC):
AC = √[ (-1 - 3)2 + (4 - 0)2 ] = √[ (-4)2 + 42 ] = √[ 16 + 16 ] = √32 = 4√2
Diagonal 2 (BD):
BD = √[ (-2 - 4)2 + (-1 - 5)2 ] = √[ (-6)2 + (-6)2 ] = √[ 36 + 36 ] = √72 = 6√2
Area = (1/2) * (4√2) * (6√2)
Area = (1/2) * 24 * 2
Area = 24 square units.
Answer: The area of the rhombus is 24 square units.