# Advanced Maths: Short Tricks on Coordinate Geometry pls tell

Kushal Chaudhari
102 Points
5 years ago
Important Short Tricks on Coordinate Geometry
Equation of line parallel to y-axis
X = a

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 = b

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 (x1,y1) & has a slope m

y - y1 = m (x-x1)

Slope of line =
Kushal Chaudhari
102 Points
5 years ago

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)

1. Slope of line = y2 - y1/x2 - x= - coefficient of x/coefficient of y
1. 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°.

1. Equation of two lines parallel to each other

ax + by + c1 = 0

ax + by + c2 = 0

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

1. 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.

1. 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

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

M = (x1 + x2)/2, (y1 + y2)/2

1. 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