Short Tricks Coordinate Geometry

Short Tricks on Coordinate Geometry – Easy Tips for SSC & Competitive Exams

Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Coordinate Geometry. The post is very helpful for the upcoming SSC CGL Exam. Looking for short tricks on Coordinate Geometry to ace SSC, Banking, Railways, and other competitive exams? You’re in the right place! GovernmentAdda provides easy-to-understand formulas and shortcuts to help you solve Coordinate Geometry problems quickly and accurately.

What is Coordinate Geometry?

Coordinate Geometry, also known as Analytical Geometry, deals with the study of geometric figures using algebraic equations. It plays a crucial role in various SSC, Banking, and Railway exams.

Key Concepts in Coordinate Geometry

• Cartesian Plane (X-axis & Y-axis)
• Distance Formula
• Section Formula (Midpoint & Ratio Division)
• Slope of a Line
• Equation of a Line
• Area of Triangle & Quadrilateral

Short Tricks on Coordinate Geometry

1. Distance Between Two Points

Formula:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}

Shortcut Trick:

• If points are on the same axis, just subtract the values.
  • Example: Distance between (3,0) and (7,0) = |7-3| = 4.
• If the points are diagonally placed, use squares and apply the direct formula.

2. Midpoint of a Line Segment

Formula:

M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Shortcut Trick:

• Directly average the x-values and y-values.
  • Example: Midpoint of (2,4) and (6,8) = (4,6).

3. Section Formula (Dividing Line in Ratio m:n)

Formula:

P=(mx2+nx1m+n,my2+ny1m+n)P = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}\right)

Shortcut Trick:

• If the ratio is 1:1, the point is the midpoint.
• If ratio m:n is given, apply the formula quickly using cross multiplication.

4. Slope of a Line

Formula:

m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}

Shortcut Trick:

• If x-values are same, slope is undefined.
• If y-values are same, slope is zero.
• If points are (0,0) and (x,y), slope = y/x (just divide the coordinates).

5. Equation of a Line

Formula:

• Slope-Intercept Form: y=mx+cy = mx + c
• Two-Point Form: y−y1=y2−y1x2−x1(x−x1)y – y_1 = \frac{y_2 – y_1}{x_2 – x_1} (x – x_1)

Shortcut Trick:

• If a line passes through origin (0,0), equation is y = mx.
• If perpendicular to x-axis: x = constant.
• If perpendicular to y-axis: y = constant.

6. Area of a Triangle (Given 3 Points)

Formula:

A=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣A = \frac{1}{2} | x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) |

Shortcut Trick:

• If three points are collinear, the area = 0.
• Avoid full expansion; apply determinant-like calculations.

Coordinate Geometry Tricks for Quick Calculation

✔ Memorize important formulas using mnemonics.
✔ Use elimination method in MCQs to find the correct answer faster.
✔ Check symmetry in points to simplify calculations.
✔ For collinear points, put them in area formula – if 0, they are collinear.

Conclusion

Using these short tricks on Coordinate Geometry, you can solve problems faster and improve accuracy in competitive exams. Keep practicing and apply these techniques to score better in SSC, Banking, and other government exams.

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Important Short Tricks on Coordinate Geometry

1. Equation of line parallel to y-axis

X = b

For Example:     A Student plotted four points on a graph. Find out which point represents the line parallel to y-axis.

1. a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

Solution:  Option (C)

2. Equation of line parallel to x-axis

Y = a

For Example:     A Student plotted four points on a graph. Find out which point represents the line parallel to x-axis.

1. a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

Solution:  Option (B)

3. Equations of line

a) Normal equation of line

ax + by + c = 0

b) Slope – Intercept Form

y = mx + c            Where, m = slope of the line & c = intercept on y-axis

For Example: What is the slope of the line formed by the equation 5y – 3x – 10 = 0?

Solution: 5y – 3x – 10 = 0, 5y = 3x + 10

Y = 3/5 x + 2

Therefore, slope of the line is = 3/5

c) Intercept Form

x/A + y/B = 1, Where, A & B are x-intercept & y-intercept respectively

For Example: Find the area of the triangle formed the line 4x + 3 y – 12 = 0, x-axis and y-axis?

Solution: Area of triangle is = ½ * x-intercept * y-intercept.

Equation of line is 4x + 3 y – 12 = 0

4x + 3y = 12,

4x/12 + 3y/12 = 1

x/3 + y/4 = 1

Therefore area of triangle = ½ * 3 * 4 = 6

d) Trigonometric form of equation of line, ax + by + c = 0

x cos θ + y sin θ = p,

Where, cos θ = -a/ √(a2 + b2) ,  sin θ = -b/ √(a2 + b2) & p = c/√(a2 + b2)

e) Equation of line passing through point (x₁,y₁) & has a slope m

y – y₁ = m (x-x₁)

4. Slope of line = y₂ – y₁/x₂ – x₁ = – coefficient of x/coefficient of y

5. Angle between two lines

Tan θ = ± (m₂ – m₁)/(1+ m₁m₂)   where, m₁ , m₂ = slope of the lines

Note:    If lines are parallel, then tan θ = 0

If lines are perpendicular, then cot θ = 0

For Example: If 7x – 4y = 0 and 3x – 11y + 5 = 0 are equation of two lines. Find the acute angle between the lines?

Solution: First we need to find the slope of both the lines.

7x – 4y = 0

⇒ y = 74x

Therefore, the slope of the line 7x – 4y = 0 is 74

Similarly, 3x – 11y + 5 = 0

⇒ y = 311x + 511

Therefore, the slope of the line 3x – 11y + 5 = 0 is = 311

Now, let the angle between the given lines 7x – 4y = 0 and 3x – 11y + 5 = 0 is θ

Now,

Tan θ = ± (m₂ – m₁)/(1+ m₁m₂) = ±[(7/4)−(3/11)]/[1+(7/4)*(3/11)] = ± 1

Since θ is acute, hence we take, tan θ = 1 = tan 45°

Therefore, θ = 45°

Therefore, the required acute angle between the given lines is 45°.

6. Equation of two lines parallel to each other

ax + by + c₁ = 0

ax + by + c₂ = 0

Note:    Here, coefficient of x & y are same.

7. Equation of two lines perpendicular to each other

ax + by + c₁ = 0

bx – ay + c₂ = 0

Note:    Here, coefficient of x & y are opposite & in one equation there is negative sign.

8. Distance between two points (x₁, y₁), (x₂, y₂)

D = √ (x₂ – x₁)² + (y₂ – y₁)²

For Example: Find the distance between (-1, 1) and (3, 4).

Solution: D = √ (x₂ – x₁)² + (y₂ – y₁)²

= √ (3 – (-1))² + (4 – 1)² = √(16 + 9) = √25 = 5

9. The midpoint of the line formed by (x₁, y₁), (x₂, y₂)

M = (x₁ + x₂)/2, (y₁ + y₂)/2

10. Area of triangle whose coordinates are (x₁, y₁), (x₂, y₂), (x₃, y₃)

½ I x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) I

For Example: Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5).

Solution: We have (x₁, y₁) = (1, 1), (x₂, y₂) = (2, 3) and (x₃, y₃) = (4, 5)

Area of Triangle = ½ I x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) I

=1/2 I (1(3−5) +2(5−1) + 4(1−3)) I

=1/2 I(−2+8−8) =1/2 (−2) I = I−1I = 1