Important Straight Line Formulas For JEE

Here is a rundown of all equations from the point Straight Line. It is vital to change these before your JEE primary and JEE Advanced test. Understudies are prescribed to go through these recipes consistently so they will recall these for the tests.

Formulas for the straight line

1. Distance formula:

d = √[(x₂-x₁)²+(y₂-y₁)²]

2. Section Formula:

x = (mx₂+nx₁)/(m+n)

y = (my₂+ny₁)/(m+n)

3. Centroid:

G = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]

4. Incentre:

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

5. Excentre:

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

6. Slope formula:

(i) Line joining two points (x₁, y₁) and (x₂, y₂), m = (y_{1 –} y₂) / (x_{1 –} x₂)

(ii) Slope of a line ax+by+c = 0 is -coefficient of x/coefficient of y = -a/b

7. Equation of a straight line in various forms:

(i) Point Slope form: y-y₁ = m(x – x₁)

(ii) Slope intercept form: y = mx + c

(iii) Two point form: y-y₁ = {(y₂ – y₁) / (x₂ – x₁)} × (x-x₁)

(iv) Intercept form: (x/a) + (y/b) = 1

(v) Perpendicular / Normal form: x cos α +y sin α = p

(vi) Parametric form: x = x₁+ r cos θ , y = y₁ + r sin θ

(vii) Symmetric form: (x – x₁)/cos θ = (y – y₁) / sin θ = r

(viii) General form: ax + by + c = 0

x intercept = -c/a

y intercept = -c/b

8. Parallel lines:

Two lines ax+by+c = 0 and a’x+b’y+c’ = 0 are parallel if a/a’ = b/b’ ≠ c/c’.

Thus any line parallel to ax+by+c = 0 is of the type ax+by+k = 0, where k is a parameter.

9. Perpendicular lines:

Two lines ax+by+c = 0 and a’x+b’y+c’ = 0 are perpendicular if aa’+bb’ = 0

10. Position of the points (x₁, y₁) and (x₂, y₂) relative to the line ax+by+c = 0:

In general, two points (x₁, y₁) and (x₂, y₂) will lie on the same side or opposite side of ax+by+c = 0 according to ax₁+by₁+c and ax₂+by₂+c are of the same or opposite sign respectively.

11. Length of the perpendicular from a point on a line :

The length of the perpendicular from a point (x₁, y₁) to a line ax + by + c = 0 is

12. Reflection of a point about a line:

(i) Foot of the perpendicular from a point on the line is (x-x₁)/a = (y-y₁)/b = -(ax₁+by₁+c)/(a²+b²)

(ii)Image of (x₁, y₁) in the line ax+by+c = 0 is (x-x₁)/a = (y-y₁)/b = -2 (ax₁+by₁+c)/(a²+b²)

FAQs:

Is straight line significant for JEE?

Yes, it is still in JEE prospectus. It is one of the significant subjects of Coordinate Geometry, so don't miss it.

Are recipes significant for JEE?

The JEE Main significant recipes can help applicants in different ways: It helps in saving time for the test. Makes the estimations simpler. Decreases the gamble of errors.