List of Points Formulae | Formula Sheet for Points

Points Formulas

Points Formulas play a crucial role in understanding the concept. Memorize the Points Formula by heart and solve basic and advanced problems too easily. We have compiled a good collection of all the Important Formulae for Points to help you understand the concept better. Apply the Formulas in your calculation part and get the results in a matter of seconds.

List of Points Formulae | Formula Sheet for Points

Provided below is the list of some of the common formulas involved in the Concept Points. Access them during your homework or assignments and start doing the problems right way rather than going with traditional methods. Make your calculations simple and fast by employing the Points Formulae in your work.

1. Centres of triangle

If A(x1, y1); B(x2, y2) and C(x3, y3) are vertices of a triangle ABC, then

  • Co-ordinates of centroid G \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
  • Co-ordinates of incentre \(\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\)
  • Circumcentre: The coordinates of circumcentre O(x, y) are obtained by solving the equations given by OA2 = OB2 = OC2
  • Orthocentre: It can be obtained by solving the equation of any two altitudes.

Spacial Cases: Like location in case of equilateral triangle.

2. Area of triangle

(i) Δ = \(\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\x_{2} & y_{2} & 1 \\x_{3} & y_{3} & 1\end{array}\right|\) = \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)
(ii) Δ = \(\frac{1}{2}\)[r1r2sin(θ2 – θ1) + r2r33 – θ2) + r3r1sin(θ1 – θ3)]

3. Area of triangle

= \(\frac{1}{2}\)|(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y4 – x4y3) + (x4y1 – x1y4)|

4. (a) Shift of origin without rotation of axis:
Let P(x, y) are the coordinates of a point in x-y plane and origin is shifted to (h, k) without rotating axis. Let new co-ordinates of same points with respect to new axis are (x’, y’) then relation can be established between new and old co-ordinates x = h + x’, y = k + y’

(b) Rotational Transformations without shifting the origin:
If coordinates of any point P(x, y) with reference to new axis will be (x’, y’) then
Points formulas img 1

5. Reflection (image) of a Point (x, y) with respect to

  • x-axis ⇒ (x, -y)
  • y-axis ⇒ (-x, y)
  • origin ⇒ (-x , -y)
  • line y = x ⇒ (y, x)

6. Some important points

(i) A triangle having vertices (at12, 2at1), (at22, 2at2) and (at32, 2at3), then area is Δ = a2[(t1 – t2) (t2 – t3) (t3 – t1)]
(ii) Area of triangle formed by Co-ordinate axis and the line ax + by + c = 0 is \(\frac{c^{2}}{2 a b}\)
(iii) In a triangle ABC, of D, E, F are midpoint of sides AB, BC and CA then EF = \(\frac{1}{2}\) BC and Δ DEF = \(\frac{1}{4}\) (ΔABC)
Points formulas img 2
(iv) Area of Rhombus formed by ax ± by ± c = 0 is \(\frac{2 c^{2}}{|a b|}\)
(v) To remove the term of xy in the equation ax2 + 2hxy + by2 = 0, the angle θ through which the axis must be turned (rotated) is given by θ = \(\frac{1}{2}\) tan-1\(\left(\frac{2 h}{a-b}\right)\)

7. Area of polygon

The area of the polygon whose vertices are (x1, y1), (x2, y2), (x3, y3) ,…….. (xn, yn) is
A = \(\frac{1}{2}\) |(x2y2 – x2y1) + (x2y3 – x3y2) + …. + (xny1 – x1yn)|


Leave a Reply