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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.

[Image of rhombus diagonal formula]

Quick Navigation:
Quick Review Flashcards - Click to flip and test your knowledge!
Question
What pair of reference lines is required to locate the position of a point on a plane?
Answer
Coordinate axes.
Question
The distance of a point from the y-axis is known as its x-coordinate or _____.
Answer
Abscissa
Question
The distance of a point from the x-axis is known as its y-coordinate or _____.
Answer
Ordinate
Question
What is the general form of the coordinates of any point lying on the x-axis?
Answer
(x, 0)
Question
What is the general form of the coordinates of any point lying on the y-axis?
Answer
(0, y)
Question
What geometric shape is formed when a linear equation of the form ax + by + c = 0 is represented graphically?
Answer
A straight line.
Question
What geometric shape is produced by the graph of the equation y = ax^2 + bx + c where a \ne 0?
Answer
A parabola.
Question
Coordinate geometry is primarily developed as an algebraic tool for studying _____.
Answer
Geometry of figures.
Question
State the distance formula used to find the distance between points P(x_1, y_1) and Q(x_2, y_2).
Answer
\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Question
If two points A(4, 0) and B(6, 0) lie on the x-axis, how is the distance AB calculated?
Answer
By finding the difference between the x-coordinates (6 - 4).
Question
Formula: The distance of a point P(x, y) from the origin O(0, 0) is _____.
Answer
\sqrt{x^2 + y^2}
Question
Why is only the positive square root taken when calculating distance between two points?
Answer
Because distance is always non-negative.
Question
What theorem is used as the basis for deriving the distance formula in a coordinate plane?
Answer
Pythagoras Theorem.
Question
To prove that three points P, Q, and R form a triangle, the sum of any two distances between them must be _____.
Answer
Greater than the third distance.
Question
According to the converse of Pythagoras theorem, if PQ^2 + PR^2 = QR^2, what is the measure of \angle P?
Answer
90^\circ
Question
To show that four points are vertices of a square, one must prove all sides are equal and both _____ are equal.
Answer
Diagonals
Question
What is the specific geometric condition for a quadrilateral to be a square if all four sides are equal?
Answer
One angle must be 90^\circ.
Question
Term: Collinear points
Answer
Definition: Points that lie on the same straight line.
Question
How can the distance formula determine if three points A, B, and C are collinear?
Answer
Check if the sum of two segments equals the third (AB + BC = AC).
Question
A point (x, y) that is equidistant from two points A and B lies on the _____ of the segment AB.
Answer
Perpendicular bisector
Question
What is the x-coordinate of any point that lies on the y-axis?
Answer
0
Question
What is the y-coordinate of any point that lies on the x-axis?
Answer
0
Question
State the section formula for the x-coordinate of point P dividing AB in ratio m_1:m_2.
Answer
x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}
Question
State the section formula for the y-coordinate of point P dividing AB in ratio m_1:m_2.
Answer
y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}
Question
Which similarity criterion is used to derive the section formula?
Answer
AA (Angle-Angle) similarity criterion.
Question
If a point P divides a line segment in the ratio k:1, what is the x-coordinate of P?
Answer
\frac{kx_2 + x_1}{k + 1}
Question
If a point P divides a line segment in the ratio k:1, what is the y-coordinate of P?
Answer
\frac{ky_2 + y_1}{k + 1}
Question
Formula: The coordinates of the mid-point of a line segment joining (x_1, y_1) and (x_2, y_2) are _____.
Answer
\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
Question
The mid-point of a line segment divides it in the ratio _____.
Answer
1 : 1
Question
What is meant by the 'points of trisection' of a line segment?
Answer
Points that divide the segment into three equal parts.
Question
In what ratio does the first point of trisection (closer to A) divide the line segment AB?
Answer
1 : 2
Question
In what ratio does the second point of trisection (closer to B) divide the line segment AB?
Answer
2 : 1
Question
If a point lies on the y-axis and divides a segment, which coordinate is set to zero to find the ratio?
Answer
The abscissa (x-coordinate).
Question
What property of parallelogram diagonals is used to find a missing vertex coordinate?
Answer
The diagonals bisect each other.
Question
If points A, B, C, and D are vertices of a parallelogram, the mid-point of AC is the same as the mid-point of _____.
Answer
BD
Question
How is the distance between two flags on different vertical lines calculated if their coordinates are known?
Answer
By applying the distance formula.
Question
Formula: Area of a rhombus (given diagonals d_1 and d_2) is _____.
Answer
\frac{1}{2} \times d_1 \times d_2
Question
What does it mean if a point P divides the line AB 'externally'?
Answer
Point P lies on the line AB but outside the segment AB.
Question
To find the coordinates of a point A when the centre and point B of a circle's diameter are known, which formula is used?
Answer
Mid-point formula.
Question
If AP = \frac{3}{7}AB for point P on segment AB, what is the ratio AP:PB?
Answer
3 : 4
Question
To divide a line segment into four equal parts, one would find the mid-point of the segment and the mid-points of the resulting _____.
Answer
Two halves
Question
What is the primary algebraic tool for finding the coordinates of a point that divides a segment into a given ratio?
Answer
The section formula.
Question
In Example 4, the relation x - y = 2 represents the _____ of the segment joining (7, 1) and (3, 5).
Answer
Perpendicular bisector
Question
What are the two components of the distance formula's radicand?
Answer
The square of the difference in x and the square of the difference in y.
Question
If a point is equidistant from A(2, -5) and B(-2, 9) and lies on the x-axis, what is its y-coordinate?
Answer
0
Question
Concept: Abscissa
Answer
Definition: The horizontal coordinate (x) of a point in a two-dimensional system.
Question
Concept: Ordinate
Answer
Definition: The vertical coordinate (y) of a point in a two-dimensional system.
Question
What field of study uses coordinate geometry for tracking tectonic plate movements?
Answer
Seismology
Question
How many units is the distance between C(0, 3) and D(0, 8)?
Answer
5 units.
Question
If P(x_1, y_1) and Q(x_2, y_2) are the same point, what is the value of the distance PQ?
Answer
0
Question
Which coordinate is constant for all points on a line parallel to the x-axis?
Answer
The y-coordinate (ordinate).
Question
Which coordinate is constant for all points on a line parallel to the y-axis?
Answer
The x-coordinate (abscissa).
Question
In the section formula derivation, if P divides AB internally, m_1 and m_2 represent the lengths _____.
Answer
PA and PB
Question
What is the distance between (2, 3) and (4, 1)?
Answer
\sqrt{8} (or 2\sqrt{2}) units.
Question
Is the distance formula \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} equivalent to the standard distance formula?
Answer
Yes, because squaring a negative number yields the same positive result.
Question
In the Classroom exercise (Fig. 7.8), what shape are Champa and Chameli discussing?
Answer
A square.
Question
If the distance between P(2, -3) and Q(10, y) is 10 units, what is the setup equation using the distance formula?
Answer
(10 - 2)^2 + (y - (-3))^2 = 10^2
Question
The coordinate system where geometry is studied using algebra is called _____.
Answer
Coordinate geometry
Question
What is the x-coordinate of the origin?
Answer
0
Question
What is the y-coordinate of the origin?
Answer
0