Maths Notes on Co-ordinate Geometry

By Sachin Awasthi|Updated : February 26th, 2021

Co-ordinate Geometry Notes for SSC & Railway Exams 2020. These Notes will help you score good marks in the upcoming SSC & Railway exams 2020.

Co-ordinate Geometry Formulas

1. Distance Formula: Distance between the two points (x1, y1) and B ( x2, y2) is given by:

 

Co-ordinate Geometry Notes for SSC & Railway Exams 2020. These Notes will help you score good marks in the upcoming SSC & Railway exams 2020.

Co-ordinate Geometry Formulas

1. Distance Formula: Distance between the two points (x1, y1) and B ( x2, y2) is given by:

byjusexamprep2. Distance of a point A (x,y) from the origin (0,0) is:

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3. Internal Division: The co-ordinates of the point which divides the straight line joining two given points P (x1, y1) and Q (x2, y2) in the ratio of  l : m are

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4. Mid-point: The co-ordinates of the mid-point of the line joining two given points P (x1, y1) and Q (x2, y2) :

byjusexamprep5.External Division: The co-ordinates of the point which divides the straight line joining the two given points P (x1, y1) and Q (x2, y2) externally in the ratio l: m are

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6. Co-ordinates of Centroid of a Triangle: The co-ordinates of the centroid of a triangle with vertices P (x1, y1), Q (x2, y2) and R (x3, y3) are:

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7. Co-ordinates of the Incentre of a Triangle: The co-ordinates of incentre of a triangle with vertices P (x1, y1) and Q (x2, y2) and R (x3, y3) and sides a, b and c are

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8. Area of Triangle: The area of triangle whose vertices are A (x1, y1), B (x2, y2) and C (x3, y3

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9. This can be represented as solving the matrix:

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Note: The three points P (x1, y1), Q (x2, y2) and R (x3, y3) are collinear if the area of the triangle is zero i.e. 1/2[x1(y2 - y3) - x2(y1 - y3) + x3(y1 - y2) = 0.

10. The equation of x-axis is y = 0

11. The equation of y-axis is x = 0

12. The equation of a straight line parallel to the y-axis at a distance ‘a’ units from it, is x = a

13. The equation of a straight line parallel to the x-axis at distance ‘b’ units from it, is y = b

14. The equation of a straight line passing through the origin (0,0) is y = mx, where m = tan θ and θ is the angle measured in the anti-clockwise direction from the positive direction of the x-axis to the upper part of the line.

15. The general form of the equation is y = mx + c where m = tan θ, θ is the angle formed by x-axis and c is the intercept on the y-axis.

      (a) If c = 0, the straight line passes through the origin and makes an angle θ with the x-axis i.e. y = mx.

      (b) If m = 0, the line is parallel to x-axis and the equation of such a line takes the form y = c

      (c) If θ = 90°, the line is parallel to the y-axis.

16. Intercept Form: x/a +y/b = 1

17. Point-Slope Form: y - y1 = m(x -x1).

18. Two-Point Form : byjusexamprep

19. Length of perpendicular from a point (x1, y1) to the line ax + by + c = 0 is:

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20. The general equation of line is ax + by + c = 0.

     (1) The equation of a line parallel to ax + by + c = 0 is ax + by + k = 0

     (2) The equation of a line perpendicular to ax + by + c = 0 is bx - ay + k = 0

21.  Two lines are parallel if their slopes are equal. Two lines are perpendicular if the product of their slopes is -1, hence two lines with slopes m1 and m2 are perpendicular if m1m2 = -1.

21. Angle θ between two lines y = m1x + c1 and y = m2x + c2 is given by tan θ = 

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22. Two lines y = a1x + b1y + c1 and y = a2x + b2y + c2 are:

      (1) Parallel, if a1/a2 = b1/b2 ≠ c1/c2

      (2) Perpendicular, if a1a2 + b1b2 = 0

      (3) Co-incident, if a1/a2 = b1/b2 = c1/c2

      (4)Intersecting, if they are neither coincident nor parallel. 

23. The equation of a circle with centre (0,0) and radius r is given by x² + y² = r²

24. The equation of a circle with centre (h, k) and radius r is given by (x - h)² + ( y - k)² = r²

25. The equation of tangent to a circle x² + y² = r², at a point (x1,y1) is xx1 + yy1 = r²

 Here are the Important Notes & Short Tricks on Coordinate Geometry. The post is very helpful for the upcoming SSC & Railway Exams.

Coordinate Geometry Short Tricks

  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.

  1. 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 x-axis.

  1. a) (3,5)

b) (0,6)

c) (8,0)

d) (-2, -4)

Solution:  Option (B)

  1. 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)

  1. Slope of line = y2 - y1/x2 - x1 = - 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

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