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Important Short Tricks on Coordinate Geometry
1. Equation of line parallel to y-axis
X = b
For Example: A Student plotted four points on a graph. Find out which point represents the
line parallel to y-axis.
1. a) (3,5) b) (0,6) c) (8,0) d) (-2, -4)
Solution: Option (C)
2. 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.
1. a) (3,5) b) (0,6) c) (8,0) d) (-2, -4)
Solution: Option (B)
3. 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

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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)
4. Slope of line = y2 – y1/x2 – x1 = – coefficient of x/coefficient of y
5. 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?

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Solution: First we need to find the slope of both the lines.
7x – 4y = 0
⇒ y = 7/4x
Therefore, the slope of the line 7x – 4y = 0 is 7/4
Similarly, 3x – 11y + 5 = 0
⇒ y = 3/11x + 5/11
Therefore, the slope of the line 3x – 11y + 5 = 0 is = 3/11
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°.
6. Equation of two lines parallel to each other.
ax + by + c1 = 0
ax + by + c2 = 0
Note: Here, coefficient of x & y are same.
7. Equation of two lines perpendicular to each other.
ax + by + c1 = 0

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bx – ay + c2 = 0
Note: Here, coefficient of x & y are opposite & in one equation there is negative sign.
8. 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
9. The midpoint of the line formed by (x1, y1), (x2, y2)
M = (x1 + x2)/2, (y1 + y2)/2
10. Area of triangle whose coordinates are (x1, y1), (x2, y2), (x3, y3)
½ [ x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) ]
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 [ x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) ]
=1/2 [ (1(3−5) +2(5−1) + 4(1−3)) ]
=1/2 (−2+8−8) =1/2 (−2) ]= -1
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