Coordinate Geometry

Basics of Coordinate Geometry

Coordinate Geometry represents points, lines, and shapes in a plane using ordered pairs (x, y). The Cartesian plane has two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical).

Important Formulas

• Distance between two points (x₁, y₁) and (x₂, y₂):
  d = √[(x₂ - x₁)² + (y₂ - y₁)²]
• Midpoint of segment joining points (x₁, y₁) and (x₂, y₂):
  M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
• Slope of line through points (x₁, y₁) and (x₂, y₂):
  m = (y₂ - y₁) / (x₂ - x₁)
• Equation of a line (slope-intercept form):
  y = mx + c, where m is slope and c is y-intercept.
• Area of triangle formed by points (x₁, y₁), (x₂, y₂), (x₃, y₃):
  Area = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Previous Year Questions

• SSC CGL 2020: Find the distance between points (3, 4) and (7, 1).
• Distance = √[(7 - 3)² + (1 - 4)²] = √(16 + 9) = √25 = 5
• RRB JE 2021: Find the midpoint of points (-2, 5) and (4, -1).
• Midpoint = ((-2 + 4)/2, (5 + (-1))/2) = (1, 2)
• SSC CHSL 2019: Find the slope of the line passing through points (1, 2) and (4, 6).
• Slope = (6 - 2) / (4 - 1) = 4 / 3
• HSSC JE 2020: Find the equation of the line with slope 3 passing through point (2, 5).
• Using y - y₁ = m(x - x₁): y - 5 = 3(x - 2) → y = 3x - 6 + 5 → y = 3x - 1
• RRB ALP 2019: Calculate the area of the triangle with vertices at (0, 0), (4, 0), and (4, 3).
• Area = (1/2) × base × height = (1/2) × 4 × 3 = 6
• SSC GD 2018: Find the distance between points (-1, -1) and (2, 3).
• Distance = √[(2 - (-1))² + (3 - (-1))²] = √(3² + 4²) = 5
• SSC MTS 2020: Find the coordinates of the midpoint of points (6, 8) and (2, 4).
• Midpoint = ((6 + 2)/2, (8 + 4)/2) = (4, 6)
• RRB NTPC 2021: What is the slope of a line perpendicular to the line y = (2/3)x + 5?
• Perpendicular slope = -3/2 (negative reciprocal)
• SSC CGL 2017: Find the equation of a line passing through (1, -2) and (3, 4).
• Slope m = (4 - (-2)) / (3 - 1) = 6 / 2 = 3. Equation: y - (-2) = 3(x - 1) → y + 2 = 3x - 3 → y = 3x - 5
• HSSC JE 2018: Calculate the distance between points (7, 0) and (0, 24).
• Distance = √[(0 - 7)² + (24 - 0)²] = √(49 + 576) = √625 = 25
• SSC CHSL 2017: Find the area of a triangle with vertices at (2, 3), (4, 7), and (6, 3).
• Area = (1/2) |2(7-3) + 4(3-3) + 6(3-7)| = (1/2) |2×4 + 4×0 + 6×(-4)| = (1/2) |8 + 0 - 24| = (1/2) | -16 | = 8
• RRB JE 2018: Find the coordinates of the point dividing the line segment joining (2, -3) and (4, 1) in the ratio 3:1.
• Coordinates = ((3×4 + 1×2)/(3+1), (3×1 + 1×(-3))/(3+1)) = (14/4, 0/4) = (3.5, 0)
• SSC GD 2017: Find the slope of the line perpendicular to 4x - 3y + 7 = 0.
• Rewrite: 4x - 3y + 7 = 0 → y = (4/3)x + 7/3; slope m = 4/3; perpendicular slope = -3/4
• SSC MTS 2018: Find the area of a triangle formed by points (0, 0), (0, 5), and (5, 0).
• Area = (1/2) × base × height = (1/2) × 5 × 5 = 12.5
• SSC CGL 2016: Find the distance between points (-4, -3) and (4, 3).
• Distance = √[(4 - (-4))² + (3 - (-3))²] = √(8² + 6²) = √(64 + 36) = √100 = 10

Tips & Tricks

• Remember the distance formula is derived from the Pythagoras theorem.
• Slope of vertical lines is undefined; slope of horizontal lines is 0.
• Use the midpoint formula to find the center point between two points easily.
• Area formula is helpful to check if three points are collinear (area = 0 means points are collinear).
• The slope of perpendicular lines are negative reciprocals.