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Short Tricks on Coordinate Geometry

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 2015.

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