Solved Exercises on Coordinate Geometry Problems in Grade 7

Master coordinate geometry problems: distance formula, midpoint, slope, and geometric applications through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Distance Formula
Exercise 1
Find the distance between points A(2, 3) and B(6, 7).
Definition:

Distance Formula: The distance between two points (x₁, y₁) and (x₂, y₂) is: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

Coordinate Plane: A two-dimensional plane formed by the intersection of x and y axes.

Distance Calculation Method:
  1. Identify the coordinates of both points
  2. Assign them to (x₁, y₁) and (x₂, y₂)
  3. Substitute values into the distance formula
  4. Calculate the differences and apply the formula
  5. Simplify to get the final answer
Point A
(2, 3)
Point B
(6, 7)
Distance
4√2 units
Step 1: Identify coordinates

Point A: (x₁, y₁) = (2, 3)

Point B: (x₂, y₂) = (6, 7)

Step 2: Apply distance formula

d = √[(x₂-x₁)² + (y₂-y₁)²]

d = √[(6-2)² + (7-3)²]

Step 3: Calculate differences

(6-2) = 4 and (7-3) = 4

Step 4: Square the differences

4² = 16 and 4² = 16

Step 5: Add and take square root

d = √(16 + 16) = √32 = √(16×2) = 4√2 ≈ 5.66 units

Distance = 4√2 units ≈ 5.66 units
Final Answer:

The distance between points A(2,3) and B(6,7) is 4√2 units.

Applied Rules:

• Distance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²]

• Pythagorean Theorem: Distance formula is derived from it

• Order of Operations: Parentheses, exponents, square root

2 Midpoint Formula
Exercise 2
Find the midpoint of the line segment connecting points C(-3, 4) and D(5, -2).
Definition:

Midpoint Formula: The midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is: \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Midpoint: The point that divides a line segment into two equal parts.

Point C
(-3, 4)
Point D
(5, -2)
Midpoint
(1, 1)
Step 1: Identify coordinates

Point C: (x₁, y₁) = (-3, 4)

Point D: (x₂, y₂) = (5, -2)

Step 2: Apply midpoint formula

Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)

Step 3: Calculate x-coordinate of midpoint

x = (-3 + 5)/2 = 2/2 = 1

Step 4: Calculate y-coordinate of midpoint

y = (4 + (-2))/2 = 2/2 = 1

Step 5: Combine coordinates

Midpoint = (1, 1)

Midpoint = (1, 1)
Final Answer:

The midpoint of the line segment connecting C(-3,4) and D(5,-2) is (1,1).

Applied Rules:

• Midpoint Formula: Average of x-coordinates and average of y-coordinates

• Average: Sum of values divided by number of values

• Order of Operations: Calculate numerator first, then divide

3 Slope Calculation
Exercise 3
Find the slope of the line passing through points E(1, 2) and F(4, 8). Describe the line's direction.
Definition:

Slope Formula: The slope of a line through points (x₁, y₁) and (x₂, y₂) is: \(m = \frac{y_2-y_1}{x_2-x_1}\)

Slope: A measure of the steepness and direction of a line.

Point E
(1, 2)
Point F
(4, 8)
Slope
m = 2
Step 1: Identify coordinates

Point E: (x₁, y₁) = (1, 2)

Point F: (x₂, y₂) = (4, 8)

Step 2: Apply slope formula

m = (y₂ - y₁)/(x₂ - x₁)

m = (8 - 2)/(4 - 1)

Step 3: Calculate numerator and denominator

Numerator: 8 - 2 = 6

Denominator: 4 - 1 = 3

Step 4: Divide to find slope

m = 6/3 = 2

Step 5: Interpret the result

Since m > 0, the line rises from left to right (positive slope).

Slope = 2 (positive, rising line)
Final Answer:

The slope of the line through E(1,2) and F(4,8) is 2. The line rises from left to right.

Applied Rules:

• Slope Formula: m = (change in y)/(change in x)

• Positive Slope: Line rises from left to right

• Negative Slope: Line falls from left to right

Rules and methods, laws,...
\(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Distance Formula
Midpoint Formula
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Average of coordinates
Slope Formula
\(m = \frac{y_2-y_1}{x_2-x_1}\)
Rate of change
Equation of Line
\(y = mx + b\)
Slope-intercept form
Key Definitions:

Coordinate Plane: A two-dimensional surface defined by x and y axes intersecting at origin (0,0)

Ordered Pair: (x, y) representing a point's position with x as horizontal distance and y as vertical distance

Slope: The steepness of a line, calculated as rise over run

Complete Methodology:
  1. Analyze the Problem: Determine what is being asked (distance, midpoint, slope, etc.)
  2. Identify Given Information: Locate coordinates and relevant data
  3. Select Appropriate Formula: Match the problem to the correct formula
  4. Substitute Values: Carefully plug coordinates into the formula
  5. Calculate Step-by-Step: Follow order of operations
  6. Verify Results: Check if the answer makes sense geometrically
Tip 1: Always subtract in the same order (second point minus first point) for consistent results.
Tip 2: For horizontal lines, slope = 0; for vertical lines, slope is undefined.
Tip 3: Remember that distance is always positive, while slope can be positive or negative.
Tip 4: Visualize the points on the coordinate plane to verify your calculations.
Common Errors: Mixing up x and y coordinates, incorrect order of subtraction, arithmetic mistakes.
Exam Preparation: Memorize formulas, practice with various coordinate combinations, understand geometric interpretations.
Solution: Exercises 4 to 5
4 Ratio Division
Exercise 4
Find the point that divides the line segment joining P(2, 3) and Q(8, 9) in the ratio 2:3.
Definition:

Section Formula: A point dividing the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n is: \(\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)

Ratio Division: Dividing a line segment into parts with a specific proportional relationship.

Points
P(2,3), Q(8,9)
Ratio
2:3
Division Point
(32/5, 33/5)
Step 1: Identify coordinates and ratio

Point P: (x₁, y₁) = (2, 3)

Point Q: (x₂, y₂) = (8, 9)

Ratio m:n = 2:3

Step 2: Apply section formula

Division point = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))

Step 3: Calculate x-coordinate

x = (2×8 + 3×2)/(2+3) = (16 + 6)/5 = 22/5 = 4.4

Step 4: Calculate y-coordinate

y = (2×9 + 3×3)/(2+3) = (18 + 9)/5 = 27/5 = 5.4

Step 5: Express as coordinate pair

Division point = (22/5, 27/5) or (4.4, 5.4)

Division point = (22/5, 27/5) or (4.4, 5.4)
Final Answer:

The point that divides the line segment joining P(2,3) and Q(8,9) in the ratio 2:3 is (22/5, 27/5).

Applied Rules:

• Section Formula: For ratio m:n, point = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))

• Weighted Average: Each coordinate is weighted by the opposite segment

• Proportional Division: The ratio determines the relative position along the segment

5 Area of Triangle
Exercise 5
Find the area of triangle with vertices A(1, 2), B(4, 5), and C(2, 7) using the coordinate geometry method.
Definition:

Shoelace Formula: For a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the area is: \(\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)

Triangle Area: Half the base times the height, or calculated using coordinates.

Vertices
A(1,2), B(4,5), C(2,7)
Shoelace
Formula
Area
6 units²
Step 1: Assign coordinates

(x₁, y₁) = (1, 2), (x₂, y₂) = (4, 5), (x₃, y₃) = (2, 7)

Step 2: Apply shoelace formula

Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

Step 3: Calculate each term

Term 1: x₁(y₂-y₃) = 1(5-7) = 1(-2) = -2

Term 2: x₂(y₃-y₁) = 4(7-2) = 4(5) = 20

Term 3: x₃(y₁-y₂) = 2(2-5) = 2(-3) = -6

Step 4: Sum the terms

Sum = -2 + 20 + (-6) = 12

Step 5: Calculate area

Area = ½|12| = ½ × 12 = 6 square units

Area = 6 square units
Final Answer:

The area of triangle ABC with vertices A(1,2), B(4,5), and C(2,7) is 6 square units.

Applied Rules:

• Shoelace Formula: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

• Absolute Value: Ensures positive area regardless of vertex order

• Vertex Order: Can be clockwise or counterclockwise

Detailed Summary: Coordinate Geometry Fundamentals
\(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Distance Formula
Key definitions:

Coordinate Plane: A two-dimensional surface defined by x and y axes intersecting at origin (0,0)

Ordered Pair: (x, y) representing a point's position with x as horizontal distance and y as vertical distance

Distance: The length of the shortest path between two points

Midpoint: The point equidistant from both endpoints of a line segment

Slope: The rate of change of y with respect to x, indicating the steepness and direction of a line

Complete methodology:
  1. Understand the Problem: Identify what is being asked (distance, midpoint, slope, area, etc.)
  2. Extract Given Information: Note all coordinates and relevant parameters
  3. Select the Appropriate Formula: Match the problem to the correct coordinate geometry formula
  4. Substitute Values Carefully: Ensure coordinates are placed in the correct positions
  5. Follow Order of Operations: Calculate inside parentheses first, then exponents, etc.
  6. Verify Your Answer: Check if the result makes geometric sense
Tip 1: Always subtract in the same order: (second coordinate - first coordinate).
Tip 2: For horizontal lines, slope = 0; for vertical lines, slope is undefined.
Tip 3: The midpoint is always between the two given points.
Tip 4: Distance is always positive; slope can be positive, negative, zero, or undefined.
Tip 5: When calculating area using coordinates, use the shoelace formula for accuracy.
Common errors: Mixing up x and y coordinates, incorrect order of subtraction, arithmetic mistakes, forgetting absolute value.
Exam preparation: Memorize formulas, practice with various coordinate combinations, understand geometric interpretations, verify results visually.
Formulas to know by heart:

• Distance Formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

• Midpoint Formula: \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

• Slope Formula: \(m = \frac{y_2-y_1}{x_2-x_1}\)

• Section Formula: \(\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)

• Triangle Area: \(\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)

Exercise with Visualization: Line Relationships
Exercise 6: Parallel and Perpendicular Lines
Consider the following line segments on the coordinate plane:
Line 1: A(1,1) to B(4,4), slope = 1
Line 2: C(2,1) to D(5,4), slope = 1
Line 3: E(1,2) to F(4,-1), slope = -1

Analysis: The visualization shows how slopes determine line relationships.

  • Lines 1 and 2: Same slope (m=1), therefore parallel
  • Lines 1 and 3: Product of slopes = 1×(-1) = -1, therefore perpendicular
  • Parallel lines have equal slopes
  • Perpendicular lines have slopes that multiply to -1

Questions & Answers

Question: I often get confused about when to use the distance formula versus the midpoint formula. How do I know which one to use?

Answer: The key is to identify what the question is asking for:

  • Use Distance Formula when: You need to find the length between two points, or when the question asks for "distance," "length," or "how far apart."
  • Use Midpoint Formula when: You need to find the point halfway between two points, or when the question asks for "midpoint," "center," or "halfway point."

Think of it this way: Distance measures how far apart things are, while midpoint finds the exact middle point between them.

Example: If asked "What is the distance from A to B?" → Use distance formula. If asked "What point is exactly in the middle of A and B?" → Use midpoint formula.

Question: Why does the distance formula have a square root and squares? Where does it come from?

Answer: The distance formula comes directly from the Pythagorean Theorem! Here's how:

  • When you have two points on a coordinate plane, you can form a right triangle by drawing horizontal and vertical lines
  • The horizontal leg has length |x₂ - x₁| and the vertical leg has length |y₂ - y₁|
  • The distance between the points is the hypotenuse of this right triangle
  • By the Pythagorean Theorem: c² = a² + b², so d² = (x₂-x₁)² + (y₂-y₁)²
  • Taking the square root gives us: d = √[(x₂-x₁)² + (y₂-y₁)²]

So the squares represent the squared lengths of the legs of the right triangle, and the square root gives us the hypotenuse (the actual distance).

Question: What happens if I switch the order of the points when calculating slope? Does it matter?

Answer: It doesn't matter as long as you're consistent! The slope will be the same regardless of which point you call (x₁, y₁) and which you call (x₂, y₂).

For example, with points A(1,2) and B(4,5):

  • If A is (x₁, y₁) and B is (x₂, y₂): m = (5-2)/(4-1) = 3/3 = 1
  • If B is (x₁, y₁) and A is (x₂, y₂): m = (2-5)/(1-4) = -3/-3 = 1

Notice that both the numerator and denominator change signs, so the fraction stays the same. However, make sure to subtract consistently: (y₂-y₁)/(x₂-x₁).

What matters is that you subtract the coordinates of the same point in both numerator and denominator.

Question: When would I ever need to use the section formula in real life?

Answer: The section formula has many practical applications:

  • Architecture and Construction: Finding points that divide beams or walls in specific ratios
  • Computer Graphics: Calculating positions for animation or rendering
  • Surveying: Locating points along property boundaries
  • Navigation: Finding waypoints along a route
  • Economics: Finding weighted averages in financial models

For example, if you're building a bridge and need to place support posts that divide the span in a 2:3 ratio, you'd use the section formula to calculate the exact location.

Even though you might not explicitly use the formula, the concept of dividing a line segment proportionally is fundamental in many fields.

Question: How can I quickly check if my coordinate geometry calculations are correct?

Answer: Here are several verification strategies:

  1. Visual Estimation: Plot the points roughly and see if your answer makes sense geometrically
  2. Reverse Calculation: Use your result to work backwards and see if you get back to the original values
  3. Alternative Method: Solve the problem using a different approach
  4. Boundary Checks: For midpoint, ensure it's between the two original points

For distance, check that it's positive and reasonable compared to the coordinate differences. For midpoint, verify it's roughly centered between the original points.

You can also use online graphing tools to verify your manual calculations. The more you practice, the better your intuition for reasonable answers will become!