Important Straight Line Math Formulas For JEE

Here is a list of all formulas from the topic Straight Line. It is very important to revise these before your JEE main and JEE Advanced exam. Students are recommended to go through these formulas on a daily basis so that they will remember these for the exams.

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. Area of a Triangle:

Area of triangle ABC

\(\begin{array}{l}=\frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{vmatrix}\end{array} \)

7. 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

8. Condition of collinearity of three points:

\(\begin{array}{l}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{vmatrix} = 0\end{array} \)

9. 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) Determinant form:

\(\begin{array}{l}\begin{vmatrix} x & y & 1\\ x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1 \end{vmatrix} = 0\end{array} \)

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

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

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

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

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

x intercept = -c/a

y intercept = -c/b

10. Angle between two straight lines in terms of their slopes:

\(\begin{array}{l}\tan \theta = \left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |\end{array} \)

11. 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.

If ax + by + c₁ = 0 and ax + by + c₂ = 0 are two parallel lines then the distance between these two parallel lines

\(\begin{array}{l}=\left | \frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}} \right |\end{array} \)

12. Perpendicular lines:

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

13. 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 as ax₁+ by₁+ c and ax₂+ by₂+ c are of the same or opposite sign, respectively.

14. 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

\(\begin{array}{l}\left | \frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right |\end{array} \)

15. 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)

16. Bisectors of the angles between two lines:

\(\begin{array}{l}\frac{ax+by+c}{\sqrt{a^{2}+b^{^{2}}}} = \pm \frac{a’x+b’y+c’}{\sqrt{a’^{2}+b’^{^{2}}}}\end{array} \)

17. Methods to discriminate between the acute angle bisector and the obtuse angle bisector:

If aa’ + bb’<0, cc’ > 0, then the equation of the bisector of acute angle is:

\(\begin{array}{l}\frac{ax+by+c}{\sqrt{a^{2}+b^{^{2}}}} = + \frac{a’x+b’y+c’}{\sqrt{a’^{2}+b’^{^{2}}}}\end{array} \)

If aa’ + bb’> 0, cc’ > 0, then the equation of the bisector of obtuse angle is:

\(\begin{array}{l}\frac{ax+by+c}{\sqrt{a^{2}+b^{^{2}}}} = \frac{a’x+b’y+c’}{\sqrt{a’^{2}+b’^{^{2}}}}\end{array} \)

18. Discriminate between the bisector of the angle containing a point:

To discriminate between the bisector of the angle containing the origin and that of the angle not containing the origin. Rewrite the equations, ax + by + c = 0 and a’x + b’y + c’ = 0 such that the constant terms c and c’ are positive.

Then;

\(\begin{array}{l}\frac{ax+by+c}{\sqrt{a^{2}+b^{^{2}}}} = – \frac{a’x+b’y+c’}{\sqrt{a’^{2}+b’^{^{2}}}}\end{array} \)

gives the equation of the bisector of the angle not containing the origin.

In general, equation of the bisector which contains the point is

\(\begin{array}{l}\frac{ax+by+c}{\sqrt{a^{2}+b^{^{2}}}} = + \frac{a’x+b’y+c’}{\sqrt{a’^{2}+b’^{^{2}}}}\end{array} \)

\(\begin{array}{l}\frac{ax+by+c}{\sqrt{a^{2}+b^{^{2}}}} = – \frac{a’x+b’y+c’}{\sqrt{a’^{2}+b’^{^{2}}}}\end{array} \)

according to a + b + c and a’ + b’ + c’ having the same sign or otherwise.

19. Condition of congruency: of three straight lines a_ix + b_iy + c_i = 0, i = 1, 2, 3 is

\(\begin{array}{l}\begin{vmatrix} a_{1} & b_{1} & c_{1}^{}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3}\\ \end{vmatrix} = 0\end{array} \)

20. Family of straight lines:

The equation of a family of straight lines passing through the point of intersection of the lines,

L₁ = a₁x + b₁y + c₁ = 0 and L₂ = a₂x + b₂y + c₂ = 0 is given by L₁ + kL₂ = 0

21. A pair of straight lines through origin: ax²+ 2hxy + by² = 0

If θ is the acute angle between the pair of straight lines, then

\(\begin{array}{l}\tan \theta = \left | \frac{2\sqrt{h^{2}-ab}}{a+b} \right |\end{array} \)

22. General equation of second degree representing a pair of straight lines:

ax²+ 2hxy + by²+ 2gx + 2fy + c = 0 represents a pair of straight lines if abc +2fgh – af²– bg²– ch² = 0

i.e. if

\(\begin{array}{l}\begin{vmatrix} a & h &g \\ h& b & f\\ g & f & c \end{vmatrix} = 0\end{array} \)

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