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 Important Algebraic Formulae 

In this article, we will study various important algebraic formulae which will help you solve the complex mathematical problems in a comparatively easy way.

While preparing for exams like IIT-JEE you need to be very quick with your calculations and specially for a subject like mathematics, you need to have formulas on your tips to solve complex problems in no time. To help you with the algebraic formulas, in the article, we have collaborated on all the important algebraic formulas which will help you to solve complex algebraic problems of algebra very quickly. 

 Let us see what all are important algebraic formulas that one should keep in mind before appearing for the exam:

1: Formulas of Straight Lines 

  • The distance formula is used to find the distance between 2 points in a straight line.

d = √[(x2–x1)2+(y2–y1)2]

  • The section formula is used when a point divides a line segment internally. 

x = (mx2+nx1)/(m+n)

y = (my2+ny1)/(m+n)

  • The formula to find centroid is used to derive the point where all the medians meet. 

G = [(x1+x2+x3)/3, (y1+y2+y3)/3]

  • The formula to find incentre is used to find the intersection point of all the 3 angle bisectors of a triangle

I = {(ax1+bx2+cx3)/(a+b+c), (ay1+by2+cy3) / (a+b+c)}

  • Formula to find excentre: 

I1 = {(–ax1+bx2+cx3)/(–a+b+c), (–ay1+by2+cy3)/(–a+b+c)}

  •  Equation of a straight line in various forms:
    • Point slope form: y – y1 = m(x – x1)
    • Slope intercept form: y = mx + c
    • Two point form: y – y1 = {(y2 – y1) / (x2 – x1)} × (x-x1)
    • Intercept form: (x/a) + (y/b) = 1
    • Perpendicular / normal form: x cos α +y sin α = p
    • Parametric form: x = x1+ r cos θ , y = y1 + r sin θ
    • Symmetric form: (x – x1)/cos θ = (y – y1) / sin θ = r
    • General form: ax + by + c = 0

x intercept = –c/a

y intercept = –c/b


2: Formulas of Parabola 

  • Equation of standard parabola: While studying the standard equation of a parabola, we assume the focus at (a,0), where a>0 and directrix x = –a is y2 = 4ax. We also assume that the vertex of the parabola is (0,0) and the axis is y = 0. 

Length of latus rectum = 4a, ends of the latus rectum are L(a,2a), and L’(a,–2a).

  • Parametric representation of a parabola: 

x = at2 and y = 2at

  • Position of a point relative to a parabola: Below given is the condition where we can locate a point whether it lies outside or inside a parabola. 

The point (x1,y1) lies outside, on, or inside the parabola y = 4ax according to y12–4ax1>, = or < 0


3: Formulas for Solution of Triangle

  • Sine Rule:

a/sin A = b/sin B = c/sin C

  • 2. Cosine Formula:

             In any triangle ABC,

  • cos A = (b2+c2–a2)/2bc
  • cos B = (c2+a2–b2)/2ca
  • cos C = (a2+b2–c2)/2ab


  • Projection Formula:

In any triangle ABC,

  • a = b cos C + c cos B
  • b = c cos A + a cos C
  • c = a cos B + b cos A 


  • The distances of the special points from vertices and sides of triangle:
    • Circumcentre (O): OA = R and Oa= R cos A
    • Incentre (I): IA = r cosec(A/2) and Ia = r
    • Excentre (I1): I1A = r1 cosec(A/2)
    • Orthocentre: HA = 2R cos A and Ha = 2R cos B cos C

 4: Formulas Related to Integration 

  • ∫ sin x dx = –cos x + c
  • ∫ cos x dx = sin x + c
  • ∫ sec 2x dx = tan x + c
  • ∫ cosec 2x dx = –cot x + c
  • ∫ sec x tan x dx = sec x + c
  • ∫ cosec x cot x dx = –cosec x + c
  • ∫ sin (ax+b)dx = –(1/a) cos (ax+b) + c
  • ∫ cos (ax+b) dx = (1/a) sin (ax+b) + c
  • ∫ tan (ax+b)dx = (1/a) ln sec (ax+b) + c
  • ∫ cot (ax+b)dx = (1/a) ln sin (ax+b) + c
  • ∫ sec 2(ax+b)dx = (1/a) tan (ax+b) + c
  • ∫ cosec 2(ax+b) dx = -(1/a) cot (ax+b) + c
  • ∫ sec (ax+b) ⋅ tan (ax+b) dx = (1/a) sec (ax+b) + c
  • ∫ cosec (ax+b) ⋅ cot (ax+b) dx = –(1/a) cosec (ax+b) + c


Mathematics is all about understanding the concept and solving problems with the best possible methods. Thus, these were some of the important algebraic formulas which will help you secure a very good score not only for the IIT-JEE exams but also the elementary examinations. We have tried to incorporate all the formulas regarding straight lines, parabola, solution to triangle, and integration too. These formulas will not only help you solve the problems quickly, but will also help you understand the topic in a much better way. 

Just like these algebraic formulas one should also devote some time in learning the formulas of differential calculus which will also help in solving the problems in a much faster way. 


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