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Quantitative Aptitude

Coordinate Geometry

Ratio Proportion & Variation19 / 51Algebraic Identities & Expressions
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Concepts (2)

Master distance, section, and midpoint formulae to quickly solve coordinate geometry problems involving points, line segments, and centroids for SSC CGL.

Core Formula

Mastering the Distance and Section Formulae is fundamental for Coordinate Geometry in SSC CGL. These formulae enable rapid calculation of distances, division of line segments, and identification of key points like midpoints and centroids.

  1. Distance Formula: Used to find the distance between two points P₁(x₁, y₁) and P₂(x₂, y₂). d = √[(x₂ - x₁)² + (y₂ - y₁)²]

  2. Section Formula (Internal Division): Used to find the coordinates of a point P(x, y) that divides the line segment joining P₁(x₁, y₁) and P₂(x₂, y₂) internally in the ratio m:n. P(x, y) = [(m*x₂ + n*x₁)/(m+n), (m*y₂ + n*y₁)/(m+n)]

  3. Midpoint Formula: A special case of the section formula where the ratio m:n is 1:1. It finds the midpoint M(x, y) of a segment joining P₁(x₁, y₁) and P₂(x₂, y₂). M(x, y) = [(x₁ + x₂)/2, (y₁ + y₂)/2]

  4. Centroid Formula: Used to find the centroid G(x, y) of a triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃). The centroid is the point of intersection of the medians. G(x, y) = [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3]

Worked Example 1 (Distance)

Question: Find the distance between the points A(3, -2) and B(-1, 1).

Solution: Using the distance formula d = √[(x₂ - x₁)² + (y₂ - y₁)²] Here, x₁ = 3, y₁ = -2 and x₂ = -1, y₂ = 1. d = √[(-1 - 3)² + (1 - (-2))²] d = √[(-4)² + (1 + 2)²] d = √[16 + (3)²] d = √[16 + 9] d = √25 d = 5 units

Worked Example 2 (Section Formula)

Question: Find the coordinates of the point that divides the line segment joining P(2, 3) and Q(8, 5) internally in the ratio 1:2.

Solution: Using the section formula P(x, y) = [(m*x₂ + n*x₁)/(m+n), (m*y₂ + n*y₁)/(m+n)] Here, (x₁, y₁) = (2, 3), (x₂, y₂) = (8, 5), and m = 1, n = 2. x = (1*8 + 2*2)/(1+2) = (8 + 4)/3 = 12/3 = 4 y = (1*5 + 2*3)/(1+2) = (5 + 6)/3 = 11/3 The coordinates of the point are (4, 11/3).

Shortcuts & Tricks

  • Visualise for Distance: For simple coordinates, quickly sketch or imagine a right-angled triangle. The horizontal difference is |x₂-x₁| and vertical difference is |y₂-y₁|. Then apply Pythagorean triplet knowledge (e.g., 3,4,5).
  • Midpoint is Average: Remember the midpoint is simply the average of the x-coordinates and the average of the y-coordinates. This is often solvable mentally.
  • Collinearity Check: Three points A, B, C are collinear if the area of the triangle formed by them is zero. A faster method is to check if the slope of AB is equal to the slope of BC. Slope = (y₂ - y₁)/(x₂ - x₁).
  • Points on Axes: If a point lies on the x-axis, its y-coordinate is 0 (e.g., (x, 0)). If it lies on the y-axis, its x-coordinate is 0 (e.g., (0, y)). This simplifies calculations.
  • Centroid Shortcut: The centroid is the average of all x-coordinates and all y-coordinates. No complex ratios involved.

Common Mistakes

  1. Sign Errors: Carelessly handling negative signs, especially in (x₂ - x₁) or (y₂ - y₁) can lead to incorrect results. Double-check subtractions.
  2. Incorrect Ratio Application: In the section formula, students often swap m and n or apply m to x₁ instead of x₂. Remember m multiplies x₂ and n multiplies x₁.
  3. Forgetting Square Root: In the distance formula, remember the final step is to take the square root of the sum of squares. Many stop at d².
  4. Mixing Coordinates: Accidentally using x₁ with y₂ or vice-versa. Always pair x with x and y with y for each point.

Derivation (brief)

  • Distance Formula: This formula is a direct application of the Pythagorean theorem. If you have two points P₁(x₁, y₁) and P₂(x₂, y₂) on a coordinate plane, you can form a right-angled triangle by drawing a horizontal line from P₁ and a vertical line from P₂ (or vice-versa) to meet at a point Q(x₂, y₁). The horizontal leg PQ has length |x₂ - x₁| and the vertical leg P₂Q has length |y₂ - y₁|. The distance P₁P₂ is the hypotenuse, so d² = (x₂ - x₁)² + (y₂ - y₁)², leading to d = √[(x₂ - x₁)² + (y₂ - y₁)²].
  • Section Formula: This formula is derived using the concept of similar triangles. Consider points P₁(x₁, y₁), P₂(x₂, y₂) and P(x, y) dividing P₁P₂ in ratio m:n. Draw perpendiculars from P₁, P, P₂ to the x-axis. This creates similar triangles, allowing us to set up ratios of corresponding sides: (x - x₁)/(x₂ - x) = m/n and (y - y₁)/(y₂ - y) = m/n. Solving these equations for x and y yields the section formula.

Advanced Examples

Question 1: The vertices of a parallelogram are A(1, 2), B(4, y), C(x, 6) and D(3, 5) taken in order. Find the values of x and y.

Solution: In a parallelogram, the diagonals bisect each other. This means the midpoint of AC is the same as the midpoint of BD. Midpoint of AC: [(1+x)/2, (2+6)/2] = [(1+x)/2, 4] Midpoint of BD: [(4+3)/2, (y+5)/2] = [7/2, (y+5)/2] Equating the x-coordinates: (1+x)/2 = 7/2 => 1+x = 7 => x = 6 Equating the y-coordinates: 4 = (y+5)/2 => 8 = y+5 => y = 3 So, x = 6 and y = 3.

Question 2: Find the ratio in which the point P(4, m) divides the line segment joining A(2, 3) and B(6, -3). Hence find the value of m.

Solution: Let P divide AB in the ratio k:1. Using the section formula for x-coordinate: 4 = (k*6 + 1*2)/(k+1) 4(k+1) = 6k + 2 4k + 4 = 6k + 2 2 = 2k => k = 1 So, the ratio is 1:1. This means P is the midpoint. Now, find m using the y-coordinate with k=1 (or midpoint formula): m = (1*(-3) + 1*3)/(1+1) m = (-3 + 3)/2 = 0/2 = 0 Thus, the ratio is 1:1 and m = 0.

Variation Types

  • Equidistant Points: Problems often ask to find a point on an axis (e.g., x-axis) that is equidistant from two given points. Set up PA² = PB² to avoid square roots and solve for the unknown coordinate.
  • Collinearity with Unknowns: Given three collinear points, one with an unknown coordinate, use the slope equality (y₂ - y₁)/(x₂ - x₁) = (y₃ - y₂)/(x₃ - x₂) to find the unknown.
  • Geometric Figures: Applying midpoint and distance formulae to properties of squares, rectangles, parallelograms, rhombuses (e.g., diagonals bisect each other, opposite sides equal, all sides equal) to find missing vertices or verify properties.
  • External Division: Though less common in SSC CGL, know that for external division in ratio m:n, the formula becomes P(x, y) = [(m*x₂ - n*x₁)/(m-n), (m*y₂ - n*y₁)/(m-n)].

Time-Saving Methods

  • Mental Math for Midpoints: For simple coordinates, average x and y mentally. E.g., midpoint of (1,2) and (3,4) is (2,3).
  • Pythagorean Triplets: Recognize common triplets (3,4,5; 5,12,13; 7,24,25; 8,15,17) for distance calculations. If Δx and Δy are 3 and 4, distance is 5.
  • Slope for Collinearity: Always prefer checking slopes for collinearity over calculating the area of a triangle, which is more prone to calculation errors and takes longer.
  • Parallelogram Property: For finding missing vertices of a parallelogram, remember that the midpoint of its diagonals coincide. This is much faster than using distance formula for opposite sides.
  • Ratio 'k:1' for Unknown Ratios: When asked to find the ratio, assume it's k:1. This simplifies the section formula as (k+1) is the denominator, making algebraic manipulation easier than m:n. If k comes out negative, it implies external division.
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Master slope (m = (y2-y1)/(x2-x1)) and line equations (y=mx+c, point-slope) for speed. Understand parallel (m1=m2) and perpendicular (m1*m2=-1) conditions to ace coordinate geometry questions.

Core Formulas

  1. Slope (m): The steepness of a line. Given two points (x1, y1) and (x2, y2): m = (y2 - y1) / (x2 - x1)

  2. Equation of a Line:

    • Slope-Intercept Form: y = mx + c (where m is slope, c is y-intercept).
    • Point-Slope Form: Given a point (x1, y1) and slope m: y - y1 = m(x - x1).
    • Two-Point Form: Given two points (x1, y1) and (x2, y2): (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).
    • General Form: Ax + By + C = 0. Here, slope m = -A/B.

  3. Parallel Lines: If two lines are parallel, their slopes are equal: m1 = m2.

  4. Perpendicular Lines: If two lines are perpendicular, the product of their slopes is -1: m1 * m2 = -1 (or m2 = -1/m1).

Worked Example 1

Question: Find the slope and equation of the line passing through points (2, 3) and (5, 9).

Solution:

  • Step 1: Calculate slope (m). Using m = (y2 - y1) / (x2 - x1) with (x1, y1) = (2, 3) and (x2, y2) = (5, 9): m = (9 - 3) / (5 - 2) = 6 / 3 = 2.
  • Step 2: Use Point-Slope Form. Using point (2, 3) and m = 2: y - 3 = 2(x - 2) y - 3 = 2x - 4 y = 2x - 1. (Alternatively, using slope-intercept form: 3 = 2(2) + c => c = -1, so y = 2x - 1).

Worked Example 2

Question: Find the equation of a line perpendicular to 2x + 3y = 6 and passing through the point (1, -2).

Solution:

  • Step 1: Find the slope of the given line (m1). Rewrite 2x + 3y = 6 in slope-intercept form y = mx + c: 3y = -2x + 6 y = (-2/3)x + 2. So, m1 = -2/3.
  • Step 2: Find the slope of the perpendicular line (m2). For perpendicular lines, m1 * m2 = -1: (-2/3) * m2 = -1 m2 = 3/2.
  • Step 3: Use Point-Slope Form with m2 and the given point (1, -2). y - (-2) = (3/2)(x - 1) y + 2 = (3/2)x - 3/2 Multiply by 2 to clear fractions: 2(y + 2) = 3(x - 1) 2y + 4 = 3x - 3 3x - 2y - 7 = 0.

Shortcuts & Tricks

  • Slope from General Form: For Ax + By + C = 0, slope m = -A/B. For a perpendicular line, the slope is B/A.
  • Equation of Parallel Line: If Ax + By + C = 0 is given, a parallel line will have the form Ax + By + K = 0. Substitute the given point to find K.
  • Equation of Perpendicular Line: If Ax + By + C = 0 is given, a perpendicular line will have the form Bx - Ay + K = 0 (swap coefficients, change sign of one). Substitute the given point to find K.
  • Quick Check: In MCQs, if you find an equation, quickly check if it satisfies the given point and slope conditions.

Common Mistakes

  1. Incorrect Slope Calculation: Swapping x and y coordinates or mismanaging signs in the (y2 - y1) / (x2 - x1) formula. Always be consistent with (x1, y1) and (x2, y2).
  2. Confusing Parallel and Perpendicular Conditions: Using m1 = m2 for perpendicular lines or m1 * m2 = -1 for parallel lines. Remember: Parallel = Same slope; Perpendicular = Negative reciprocal slope.
  3. Algebraic Errors: Mistakes in simplifying equations, especially when dealing with fractions or negative signs. Double-check your calculations.

Derivation (brief)

The concept of slope (gradient) is fundamentally about the 'rise over run' of a line. Imagine a right-angled triangle formed by a line segment and lines parallel to the x and y axes. The 'rise' is the vertical change (Δy = y2 - y1) and the 'run' is the horizontal change (Δx = x2 - x1). The slope, m, is defined as this ratio: m = Δy / Δx. Geometrically, the slope is also tan(θ), where θ is the angle the line makes with the positive x-axis. This constant ratio ensures the line is straight.

Advanced Examples

Question: Find the area of the triangle formed by the lines y = x, x + y = 6, and the x-axis.

Solution:

  • Step 1: Find the vertices of the triangle.
    • Intersection of y = x and x + y = 6: Substitute y=x into the second equation: x + x = 6 => 2x = 6 => x = 3. Since y = x, y = 3. Vertex A = (3, 3).
    • Intersection of y = x and the x-axis (y = 0): 0 = x. Vertex B = (0, 0).
    • Intersection of x + y = 6 and the x-axis (y = 0): x + 0 = 6 => x = 6. Vertex C = (6, 0).
  • Step 2: Calculate the area of the triangle. The base of the triangle lies on the x-axis, from (0, 0) to (6, 0). Length of base = |6 - 0| = 6 units. The height of the triangle is the perpendicular distance from vertex A (3, 3) to the x-axis, which is the y-coordinate of A. Height = 3 units. Area = (1/2) * base * height = (1/2) * 6 * 3 = 9 square units.

Variation Types

  • Intercept Form: x/a + y/b = 1, where 'a' is the x-intercept (point where line crosses x-axis) and 'b' is the y-intercept (point where line crosses y-axis). This form is highly useful when intercepts are directly given or easily found.
  • Distance between Parallel Lines: For two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, the perpendicular distance between them is given by the formula: d = |C1 - C2| / sqrt(A^2 + B^2).

Time-Saving Methods

  • Option Verification: In multiple-choice questions, if you're short on time or stuck, try plugging the coordinates of the given points into the answer options. The correct equation must satisfy all given conditions.
  • Visual Estimation: For simpler problems, a quick sketch can help you visualize the slopes and intercepts, potentially eliminating incorrect options or guiding your calculation.
  • Direct Formula Application: Memorize the shortcut forms for parallel (Ax + By + K = 0) and perpendicular (Bx - Ay + K = 0) lines. This saves time from calculating slopes and then using point-slope form separately.
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