Study Materials Available

Access summaries, videos, slides, infographics, mind maps and more

View Materials

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
$0$