Coordinate geometry is a branch of mathematics that uses algebra to study geometry. It involves representing geometric shapes using numbers (coordinates) on a plane.
Key Concepts:
- Cartesian Plane: This is a plane formed by two perpendicular lines, the x-axis and the y-axis, intersecting at the origin (0, 0).
- Coordinates of a Point: Any point on the plane can be represented by an ordered pair (x, y), where x is the distance from the y-axis (abscissa) and y is the distance from the x-axis (ordinate).
- Distance Formula: Used to find the distance between two points on a plane given their coordinates.
- Area of a Triangle: Calculates the area of a triangle given the coordinates of its vertices.
Applications:
- Physics: Representing motion, forces, and fields.
- Engineering: Designing structures and systems.
- Computer Graphics: Creating images and animations.
- Navigation: Determining locations and distances.
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)
Ans :
Distance Formula:
= √[(x₂ – x₁)² + (y₂ – y₁)²]
Let’s calculate the distance for each pair of points:
(i) Points (2, 3) and (4, 1)
- x₁ = 2, y₁ = 3,
- x₂ = 4, y₂ = 1
- D = √[(4 – 2)² + (1 – 3)²] = √[2² + (-2)²] = √8 = 2√2 units
(ii) Points (-5, 7) and (-1, 3)
- x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
- D = √[(-1 + 5)² + (3 – 7)²] = √[4² + (-4)²] = √32 = 4√2 units
(iii) Points (a, b) and (-a, -b)
- x₁ = a, y₁ = b, x₂ = -a, y₂ = -b
- D = √[(-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).
Ans :
Given points: (0, 0) and (36, 15)
Formula:
D = √((x₂ – x₁)² + (y₂ – y₁)²)
Substituting the given points: D = √((36 – 0)² + (15 – 0)²)
= √(36² + 15²)
= √(1296 + 225)
= √1521
= 39 Units
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Ans :
Given points: (1, 5), (2, 3), and (-2, -11)
Let’s denote the points as A(1, 5), B(2, 3), and C(-2, -11).
Calculate the distances:
- AB = √[(2-1)² + (3-5)²] = √5
- BC = √[(-2-2)² + (-11-3)²] = √208
- AC = √[(-2-1)² + (-11-5)²] = √261
Check for collinearity: Since AB + BC ≠ AC (√5 + √208 ≠ √261), the points A, B, and C are not collinear.
4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Ans :
Step 1: We can use the distance formula: d = √((x₂ – x₁)² + (y₂ – y₁)²)
- Distance between (5, -2) and (6, 4): d₁ = √((6-5)² + (4-(-2))²) = √(1 + 36) = √37
- Distance between (6, 4) and (7, -2): d₂ = √((7-6)² + (-2-4)²) = √(1 + 36) = √37
- Distance between (5, -2) and (7, -2): d₃ = √((7-5)² + (-2-(-2))²) = √4 = 2
Step 2: Compare the distances:
We can see that d₁ = d₂ = √37.
Conclusion:
Since two sides of the triangle have equal length (d₁ = d₂), the triangle formed by the given points is an isosceles triangle
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given figure. 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.
Ans :
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)
Ans :
7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Ans :
8. Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Ans :
D = √((x₂ – x₁)² + (y₂ – y₁)²)
In this case:
- D = 10
- (x₁, y₁) = (2, -3)
- (x₂, y₂) = (10, y)
Substituting the values in the distance formula:
10 = √((10 – 2)² + (y – (-3))²)
10 = √(64 + (y + 3)²)
Squaring both sides: 100 = 64 + (y + 3)² (y + 3)² = 36
Taking the square root of both sides: y + 3 = ±6
Case 1: y + 3 = 6 y = 3
Case 2: y + 3 = -6 y = -9
Therefore, the possible 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.
Ans :
Step 1: Finding the value of x
Since Q is equidistant from P and R, PQ = QR.
Using the distance formula:
- PQ = √((5-0)² + (-3-1)²) = √(25 + 16) = √41
- QR = √((x-0)² + (6-1)²) = √(x² + 25)
Equating PQ and QR:
√41 = √(x² + 25)
Squaring both sides:
41 = x² + 25
x² = 16
x = ±4
Step 2: Finding QR and PR
We already found QR = √41.
For x = 4, R(4, 6)
- PR = √((4-5)² + (6-(-3))²) = √(1 + 81) = √82
For x = -4, R(-4, 6)
- PR = √((-4-5)² + (6-(-3))²) = √(81 + 81) = 9√2
Therefore, the values of x are 4 and -4, QR = √41, and PR can be √82 or 9√2 depending on the value of x.
10. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Ans :
Let P(x, y) be any point on the locus. Given: PA = PB
Using the distance formula:
- PA = √[(x – 3)² + (y – 6)²]
- PB = √[(x + 3)² + (y – 4)²]
Since PA = PB, we can equate their squares:
(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
Expanding and simplifying: x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16 -12x – 4y + 20 = 0
Dividing by -4: 3x + y – 5 = 0
Therefore, the relation between x and y 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.
Ans :
Given:
- (x₁, y₁) = (-1, 7)
- (x₂, y₂) = (4, -3)
- m : n = 2 : 3
Substituting the values in the section formula:
x = (24 + 3(-1)) / (2 + 3) = (8 – 3) / 5 = 1
y = (2*(-3) + 3*7) / (2 + 3) = (-6 + 21) / 5 = 3
Therefore, the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3 are (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Ans :
Given:
- (x₁, y₁) = (4, -1)
- (x₂, y₂) = (-2, -3)
For the first point of trisection (dividing the line in the ratio 1:2):
- m = 1, n = 2
Substituting the values in the section formula:
- x = (1*(-2) + 2*4) / (1 + 2) = 2
- y = (1*(-3) + 2*(-1)) / (1 + 2) = -5/3
For the second point of trisection (dividing the line in the ratio 2:1):
- m = 2, n = 1
Substituting the values in the section formula:
- x = (2*(-2) + 1*4) / (2 + 1) = 0
- y = (2*(-3) + 1*(-1)) / (2 + 1) = -7/3
Therefore, the coordinates of 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 given
figure below. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th 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?
Ans :
Finding the Positions of the Flags
- Niharika:
Runs 1/4th the distance AD on the 2nd line.
- Distance along AD = (1/4) * 100 = 25 m
- Position of green flag = (2, 25) (2nd line, 25 m distance)
- Preet:
Runs 1/5th the distance AD on the 8th line.
- Distance along AD = (1/5) * 100 = 20 m
- Position of red flag = (8, 20) (8th line, 20 m distance)
Finding the Distance Between Flags
Using the distance formula: d = √((x₂ – x₁)² + (y₂ – y₁)²)
- Distance between green and red flags = √((8-2)² + (20-25)²) = √(36 + 25) = √61 m
Finding the Position for the Blue Flag
The blue flag should be at the midpoint of the line joining the green and red flags.
- Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
- Midpoint = ((2+8)/2, (25+20)/2) = (5, 22.5)
Therefore, Rashmi should post the blue flag at 22.5 m on the 5th line.
4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Ans :
Let the ratio be k:1.
Using the section formula, we can find the coordinates of the point dividing the line segment in the ratio k:1.
x-coordinate of the point = (kx₂ + x₁) / (k + 1)
y-coordinate of the point = (ky₂ + y₁) / (k + 1)
Substituting the given values:
-1 = (k6 + (-3)) / (k + 1)
6 = (k(-8) + 10) / (k + 1)
Solving the first equation for k: -k – 1 = 6k – 3 7k = 2 k = 2/7
Therefore, the ratio in which the line segment is divided is 2:7.
5. Find the ratio in which 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.
Ans :
Let the ratio be k:1.
Since the point of division lies on the x-axis, its y-coordinate is 0.
Using the section formula, we can find the y-coordinate of the point of division:
- y = (ky₂ + y₁) / (k + 1)
Where:
- (x₁, y₁) = (1, -5)
- (x₂, y₂) = (-4, 5)
- y = 0 (as the point lies on the x-axis)
Substituting the values:
- 0 = (k*5 + (-5)) / (k + 1)
- 5k – 5 = 0
- k = 1
Therefore, the ratio in which the line segment is divided is 1:1.
Now, let’s find the x-coordinate of the point of division using the section formula for the x-coordinate:
- x = (kx₂ + x₁) / (k + 1) = (1*(-4) + 1*1) / (1 + 1) = -3/2
So, the coordinates of the point of division are (-3/2, 0).
In conclusion, the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis in the ratio 1:1, and the point of division 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.
Ans :
Midpoint formula:
- Midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2)
Midpoint of AC (O):
- ((1+x)/2, (2+6)/2) = ((1+x)/2, 4)
Midpoint of BD (O):
- ((4+3)/2, (y+5)/2) = (7/2, (y+5)/2)
Since O is the midpoint of both AC and BD, their coordinates are equal:
- (1+x)/2 = 7/2
- 1+x = 7
- x = 6
- 4 = (y+5)/2
- 8 = y+5
- y = 3
Therefore, the coordinates of the fourth vertex C are (6, 6) and the value of y is 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).
Ans :
Let the coordinates of point A be (x, y).
Since (2, -3) is the midpoint of AB, we can use the midpoint formula:
- Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)
Applying the midpoint formula to points A(x, y) and B(1, 4):
- (2, -3) = ((x + 1)/2, (y + 4)/2)
Equating corresponding coordinates:
- 2 = (x + 1)/2
- -3 = (y + 4)/2
Solving for x and y:
- x + 1 = 4
- x = 3
- y + 4 = -6
- y = -10
Therefore, 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.
Ans :
Since P divides AB in the ratio 3:4 (as AP:PB = 3:7), we can use the section formula to find the coordinates of P.
If a point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂) in the ratio m:n, then:
- x = (mx₂ + nx₁) / (m + n)
- y = (my₂ + ny₁) / (m + n)
Substituting the given values:
- x₁ = -2, y₁ = -2
- x₂ = 2, y₂ = -4
- m = 3, n = 4
Calculating the coordinates of P:
- x = (32 + 4(-2)) / (3 + 4) = (6 – 8) / 7 = -2/7
- y = (3*(-4) + 4*(-2)) / (3 + 4) = (-12 – 8) / 7 = -20/7
Therefore, the coordinates of point 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.
Ans :
Since we need to divide the line segment into four equal parts, we are essentially finding the midpoints of different segments.
Step 1: Find the midpoint of AB (let’s call it P).
- Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)
- P = ((-2 + 2)/2, (2 + 8)/2) = (0, 5)
Step 2: Find the midpoint of AP (let’s call it Q).
- Q = ((-2 + 0)/2, (2 + 5)/2) = (-1, 7/2)
Step 3: Find the midpoint of PB (let’s call it R).
- R = ((0 + 2)/2, (5 + 8)/2) = (1, 13/2)
Therefore, the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts 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)]
Ans :
Step 1: Find the coordinates of the diagonals
- Diagonal 1: Join (3, 0) and (-1, 4)
- Diagonal 2: Join (4, 5) and (-2, -1)
Step 2: Calculate the length of the diagonals
- We can use the distance formula: d = √((x₂ – x₁)² + (y₂ – y₁)²)
Diagonal 1:
- Length = √((-1 – 3)² + (4 – 0)²) = √(16 + 16) = √32 = 4√2
Diagonal 2:
- Length = √((-2 – 4)² + (-1 – 5)²) = √(36 + 36) = √72 = 6√2
Step 3:
Area = (1/2) * (4√2) * (6√2) = (1/2) * 48 = 24 square units
