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
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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
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