Coordinate Geometry — Andhra Pradesh (SSC) Class 9 Mathematics Solutions (Free)
Free step-by-step Andhra Pradesh (SSC) Class 9 Mathematics solutions for "Coordinate Geometry" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Andhra Pradesh (SSC) Class 9 Mathematics solutions for "Coordinate Geometry" — important questions with detailed answers, download P…
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Q1: Plot the following points on a coordinate plane and determine which quadrant each belongs to: A(3, 4), B(-2, 5), C(-3, -2), D(4, -1).
Step 1: Understand quadrants
Quadrant I: x > 0, y > 0
Quadrant II: x < 0, y > 0
Quadrant III: x < 0, y < 0
Quadrant IV: x > 0, y < 0
Step 2: Classify each point
Point A(3, 4): x = 3 > 0, y = 4 > 0 → Quadrant I
Point B(-2, 5): x = -2 < 0, y = 5 > 0 → Quadrant II
Point C(-3, -2): x = -3 < 0, y = -2 < 0 → Quadrant III
Point D(4, -1): x = 4 > 0, y = -1 < 0 → Quadrant IV
Step 3: Plot on coordinate plane
- Mark origin O at (0, 0)
- A is 3 units right,…
Q2: Find the distance between points P(1, 3) and Q(4, 7).
Points: P(1, 3) and Q(4, 7)
Use distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Step 1: Identify coordinates
x₁ = 1, y₁ = 3
x₂ = 4, y₂ = 7
Step 2: Calculate differences
x₂ - x₁ = 4 - 1 = 3
y₂ - y₁ = 7 - 3 = 4
Step 3: Square the differences
(x₂ - x₁)² = 3² = 9
(y₂ - y₁)² = 4² = 16
Step 4: Sum and take square root
d = √(9 + 16) = √25 = 5
Answer: Distance PQ = 5 units
Q3: Find the coordinates of the midpoint of the line segment joining A(-2, 4) and B(6, -2).
Points: A(-2, 4) and B(6, -2)
Use midpoint formula: M = [(x₁ + x₂)/2, (y₁ + y₂)/2]
Step 1: Identify coordinates
x₁ = -2, y₁ = 4
x₂ = 6, y₂ = -2
Step 2: Calculate x-coordinate of midpoint
x_M = (x₁ + x₂)/2 = (-2 + 6)/2 = 4/2 = 2
Step 3: Calculate y-coordinate of midpoint
y_M = (y₁ + y₂)/2 = (4 + (-2))/2 = 2/2 = 1
Step 4: Write midpoint coordinates
M = (2, 1)
Answer: The midpoint is (2, 1)
Q4: Show that the points A(0, 0), B(3, 4), and C(6, 8) are collinear.
Points: A(0, 0), B(3, 4), C(6, 8)
Points are collinear if they lie on the same line.
Method 1: Using slope
Slope between A and B: m_AB = (4 - 0)/(3 - 0) = 4/3
Slope between B and C: m_BC = (8 - 4)/(6 - 3) = 4/3
Slope between A and C: m_AC = (8 - 0)/(6 - 0) = 8/6 = 4/3
Since all three slopes are equal (4/3), the points are collinear.
Method 2: Using area of triangle
If area of triangle ABC = 0, then points are collinear.
Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
= (1/2)|0(4 - 8) + …
Q5: Find the area of a triangle with vertices at A(1, 2), B(4, 2), and C(4, 5).
Vertices: A(1, 2), B(4, 2), C(4, 5)
Method 1: Using coordinate formula
Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Step 1: Identify coordinates
x₁ = 1, y₁ = 2
x₂ = 4, y₂ = 2
x₃ = 4, y₃ = 5
Step 2: Substitute in formula
Area = (1/2)|1(2 - 5) + 4(5 - 2) + 4(2 - 2)|
= (1/2)|1(-3) + 4(3) + 4(0)|
= (1/2)|-3 + 12 + 0|
= (1/2)|9|
= 9/2 = 4.5
Method 2: Geometric observation
Base AB = distance from A(1,2) to B(4,2) = 4 - 1 = 3
Height = perpendicular distance from C(4,5) to line AB = 5 - 2 =…
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