By Vijeta Bhatt|Updated : February 1st, 2022

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 BBA and IPM Entrance Exams.

Important Short Tricks on Coordinate Geometry

1. Equation of line parallel to the y-axis

X = a

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

a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

Solution:  Option (C)

1. 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 the x-axis.

a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

Solution:  Option (B)

1. Equations of line

a) Normal equation of the 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

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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

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