Coordinate Geometry — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)

Distance Formula: The distance d between two points P(x₁, y₁) and Q(x₂, y₂) in a coordinate plane is: d = √[(x₂ - x₁)² + (y₂ - y₁)²] This formula is derived from the Pythagorean theorem. Problem: Find distance between A(3, 4) and B(6, 8) Step 1: Identify the coordinates. Point A: x₁ = 3, y₁ = 4 Point B: x₂ = 6, y₂ = 8 Step 2: Substitute into the distance formula. d = √[(6 - 3)² + (8 - 4)²] Step 3: Simplify inside the square root. d = √[(3)² + (4)²] d = √[9 + 16] d = √25 d = 5 units Answer: Distance AB = 5 units

Section Formula (Internal Division): If point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m:n, then: P = ((m·x₂ + n·x₁)/(m + n), (m·y₂ + n·y₁)/(m + n)) Given: A(2, 3), B(8, 9) Ratio = 2:1 (m:n = 2:1) Step 1: Identify the values. x₁ = 2, y₁ = 3 x₂ = 8, y₂ = 9 m = 2, n = 1 Step 2: Apply the section formula for x-coordinate. x = (m·x₂ + n·x₁)/(m + n) x = (2·8 + 1·2)/(2 + 1) x = (16 + 2)/3 x = 18/3 x = 6 Step 3: Apply the section formula for y-coordinate. y = (m·y₂ + n·y₁)/(m + n) y = (2·9 + 1·3)/(2 + 1) y = (18 + 3)/3 y = 21/3 y = 7 Answer: The point is P(6, 7) Verification: Point (6, 7) lies on AB and divides it in ratio 2:1.

Given: Points A(0, 0), B(3, 4), C(6, 8) Prove: These points are collinear (lie on the same line) Method 1: Using Slope Step 1: Find the slope of AB. Slope of AB = (y₂ - y₁)/(x₂ - x₁) = (4 - 0)/(3 - 0) = 4/3 Step 2: Find the slope of BC. Slope of BC = (8 - 4)/(6 - 3) = 4/3 Step 3: Check if slopes are equal. Slope of AB = Slope of BC = 4/3 Also, B lies on both lines, and A and C are on the same side of B. Step 4: Conclusion. Since the slopes are equal and B is a common point, the three points are collinear. Method 2: Using Distance Formula Step 1: Find AB. AB = √[(3 - 0)² + (4 - 0)²] = √[9 + 16] = √25 = 5 Step 2: Find BC. BC = √[(6 - 3)² + (8 - 4)²] = √[9 + 16] = √25 = 5 Step 3: Find AC. AC = √[(6 - 0)² + (8 - 0)²] = √[36 + 64] = √100 = 10 Step 4: Check the condition. AB + BC = 5 + 5 = 10 = AC ✓ Since AB + BC = AC, point B lies on the line segment AC. Therefore, A, B, and C are collinear. ✓

Area of Triangle using Coordinate Formula: For a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃): Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| Given: A(1, 2), B(4, 6), C(5, 1) x₁ = 1, y₁ = 2 x₂ = 4, y₂ = 6 x₃ = 5, y₃ = 1 Step 1: Substitute into the formula. Area = (1/2)|1(6 - 1) + 4(1 - 2) + 5(2 - 6)| Step 2: Calculate each term. = (1/2)|1(5) + 4(-1) + 5(-4)| = (1/2)|5 - 4 - 20| = (1/2)|-19| = (1/2)(19) = 9.5 square units Answer: Area of triangle ABC = 9.5 square units or 19/2 square units Alternative method verification using the formula: Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| We can also write this as: Area = (1/2)|x₁y₂ - x₁y₃ + x₂y₃ - x₂y₁ + x₃y₁ - x₃y₂| = (1/2)|1(6) - 1(1) + 4(1) - 4(2) + 5(2) - 5(6)| = (1/2)|6 - 1 + 4 - 8 + 10 - 30| = (1/2)|-19| = 9.5 ✓

Midpoint Formula: If M is the midpoint of AB where A(x₁, y₁) and B(x₂, y₂), then: M = ((x₁ + x₂)/2, (y₁ + y₂)/2) Given: M(2, 3) is the midpoint A(0, 1) Find: B(x₂, y₂) Step 1: Apply the midpoint formula for the x-coordinate. 2 = (0 + x₂)/2 4 = 0 + x₂ x₂ = 4 Step 2: Apply the midpoint formula for the y-coordinate. 3 = (1 + y₂)/2 6 = 1 + y₂ y₂ = 5 Answer: B(4, 5) Verification: Midpoint of A(0, 1) and B(4, 5) = ((0 + 4)/2, (1 + 5)/2) = (2, 3) ✓

Property of Parallelogram: In a parallelogram, the diagonals bisect each other. If ABCD is a parallelogram with diagonals AC and BD, then the midpoint of AC equals the midpoint of BD. Given: A(1, 2), B(4, 3), C(6, 6) Find: D(x, y) Assuming the vertices are in order A, B, C, D: Step 1: Find the midpoint of diagonal AC. Midpoint of AC = ((1 + 6)/2, (2 + 6)/2) = (7/2, 4) = (3.5, 4) Step 2: This midpoint must equal the midpoint of diagonal BD. Midpoint of BD = ((4 + x)/2, (3 + y)/2) Step 3: Set the midpoints equal. (4 + x)/2 = 7/2 (3 + y)/2 = 4 Step 4: Solve for x. 4 + x = 7 x = 3 Step 5: Solve for y. 3 + y = 8 y = 5 Answer: D(3, 5) Verification: Midpoint of AC = (3.5, 4) Midpoint of BD = ((4 + 3)/2, (3 + 5)/2) = (7/2, 8/2) = (3.5, 4) ✓ The diagonals bisect each other, confirming ABCD is a parallelogram.

Given: Point A(5, -3), Origin O(0, 0) Find: Distance OA Step 1: Apply the distance formula. Distance = √[(x₂ - x₁)² + (y₂ - y₁)²] Step 2: Substitute values. OA = √[(5 - 0)² + (-3 - 0)²] OA = √[(5)² + (-3)²] OA = √[25 + 9] OA = √34 units Answer: Distance from A to origin = √34 units ≈ 5.83 units

Median: A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Given: A(-1, 2), B(3, -4), C(5, -2) Find: Length of median from A to midpoint of BC Step 1: Find the midpoint of BC. Let M be the midpoint of BC. M = ((3 + 5)/2, (-4 + (-2))/2) M = (8/2, -6/2) M = (4, -3) Step 2: Find the distance from A to M (this is the median). AM = √[(4 - (-1))² + (-3 - 2)²] AM = √[(4 + 1)² + (-5)²] AM = √[(5)² + (-5)²] AM = √[25 + 25] AM = √50 AM = 5√2 units Answer: Length of median from A = 5√2 units ≈ 7.07 units

Given: Points P(6, -1), Q(5, 0), R(4, 1) Show: The points are collinear Method: Using the Area Formula Three points are collinear if the area of the triangle formed by them is zero. Step 1: Apply the area formula. Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| Step 2: Substitute the coordinates. With P(6, -1), Q(5, 0), R(4, 1): Area = (1/2)|6(0 - 1) + 5(1 - (-1)) + 4((-1) - 0)| Area = (1/2)|6(-1) + 5(2) + 4(-1)| Area = (1/2)|-6 + 10 - 4| Area = (1/2)|0| Area = 0 Step 3: Conclusion. Since the area is zero, the points P, Q, and R are collinear. Alternative verification using slopes: Slope of PQ = (0 - (-1))/(5 - 6) = 1/(-1) = -1 Slope of QR = (1 - 0)/(4 - 5) = 1/(-1) = -1 Slope of PR = (1 - (-1))/(4 - 6) = 2/(-2) = -1 Since all slopes are equal and the points are connected, P, Q, and R are collinear. ✓