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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**

**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.

- a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

**Solution:** Option (C)

**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.

- a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

**Solution:** Option (B)

**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 (x _{1},y_{1}) & has a slope m**

y – y_{1} = m (x-x_{1})

**Slope of line = y**_{2}– y_{1}/x_{2}– x_{1 }= – coefficient of x/coefficient of y

**Angle between two lines**

Tan θ = ± (m_{2} – m_{1})/(1+ m_{1}m_{2}) where, m_{1} , m_{2} = 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 = ± (m_{2} – m_{1})/(1+ m_{1}m_{2}) = ±[(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°.

**Equation of two lines parallel to each other**

ax + by + c_{1} = 0

ax + by + c_{2} = 0

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

**Equation of two lines perpendicular to each other**

ax + by + c_{1} = 0

bx – ay + c_{2} = 0

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

**Distance between two points (x**_{1}, y_{1}), (x_{2}, y_{2})

D = √ (x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

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

**Solution: **D = √ (x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

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

**The midpoint of the line formed by (x**_{1}, y_{1}), (x_{2}, y_{2})

M = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

**Area of triangle whose coordinates are (x**_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

½ I x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) I

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

**Solution:** We have (x_{1}, y_{1}) = (1, 1), (x_{2}, y_{2}) = (2, 3) and (x_{3}, y_{3}) = (4, 5)

Area of Triangle = ½ I x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) 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