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 (x_{1}, y_{1}) and B ( x_{2}, y_{2}) is given by:
2. Distance of a point A (x,y) from the origin (0,0) is:
3. Internal Division: The co-ordinates of the point which divides the straight line joining two given points P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) in the ratio of l : m are
4. Mid-point: The co-ordinates of the mid-point of the line joining two given points P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) :
5.External Division: The co-ordinates of the point which divides the straight line joining the two given points P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) 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 (x_{1}, y_{1}), Q (x_{2}, y_{2}) and R (x_{3}, y_{3}) are:
7. Co-ordinates of the Incentre of a Triangle: The co-ordinates of incentre of a triangle with vertices P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) and R (x_{3}, y_{3}) and sides a, b and c are
8. Area of Triangle: The area of triangle whose vertices are A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3})
9. This can be represented as solving the matrix:
Note: The three points P (x_{1}, y_{1}), Q (x_{2}, y_{2}) and R (x_{3}, y_{3}) are collinear if the area of the triangle is zero i.e. 1/2[x_{1}(y_{2} - y_{3}) - x_{2}(y_{1} - y_{3}) + x_{3}(y_{1} - y_{2}) = 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 - y_{1} = m(x -x_{1}).
18. Two-Point Form :
19. Length of perpendicular from a point (x_{1}, y_{1}) to the line ax + by + c = 0 is:
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 m_{1} and m_{2} are perpendicular if m_{1}m_{2} = -1.
21. Angle θ between two lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2} is given by tan θ =
22. Two lines y = a_{1}x + b_{1}y + c_{1} and y = a_{2}x + b_{2}y + c_{2} are:
(1) Parallel, if a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
(2) Perpendicular, if a_{1}a_{2} + b_{1}b_{2} = 0
(3) Co-incident, if a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
(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 (x_{1},y_{1}) is xx_{1} + yy_{1} = 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
- 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)
- 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 (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
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