The angle 0 between the lines having slopes `m_1` and `m_2` is given by

`tan theta = | (m_2 - m_1)/( 1 + m_1m_2)|`

1. Parallelism of Lines :

If two lines of slopes `m_1` and `m_2` are parallel , then the angle `theta` between them is 0°.

`:. tan theta = tan 0° = 0 => (m_2 - m_1 )/ ( 1 + m_1 m_2) = 0`

`=> m_1 = m_2`

Thus, when two lines are parallel, their slopes are equal.

2. Perpendicularity of Two Lines :

If two lines of slopes `m_1` and `m_2` are perpendicular, then the angle e between them is 90°.

`cot theta = 0`

`=> (1 + m_1m_2 )/( m_1 - m_2 ) = 0 => m_1 * m_2 = -1`

Thus, when two lines are perpendicular, the product of their slopes is -1.

If m is the slope of a line, then the slope of a line perpendicular to it is `( -1/m)`

The equation of a line parallel to a given line ax+ by+ c = 0, is ax +by+ k = 0, where k is a constant.

The equation of a line perpendicular to a given line ax +by+ c = 0, is bx - ay + k = 0, where k is a constant.

The equation of a line in the general form can be written as ax +by+c = 0

1. Slope Intercept Form :

The equation of a line with slope m and making an intercept c on Y-axis is y = mx+c

2. General Form to Slope Intercept Form :

The equation of a line which passes through the point `(x _1, y_1)` and has the slope `'m'` is

`y-y_1=m(x-x_1)`

Let `Q(x_1, y_1)` be the point through which the line passes and let `P(x, y)` be any point on the line. Then, the
slope of the line is

`(y-y_1)/(x-x_1)`

But `m ` is the slope of the line. Therefore

`m=(y-y_1)/(x-x_1) => y -y_1=m(x-x_1)`

Thus, `y - y_1 = m(x - x_1)` is the required equation of the line.

3. Point Slope Form :

The equation of a line which passes through the point `(x_1, y_1 )` and has the slope m, is `(y- y_1) = m(x- x_1)`

4. Two Points Form :

The equation of a line passing through two points `(x_1, y_1)` and `(x_2, y_2 )` is `( y - y_1) = ((y_2 - y_1 )/( x_2 - x_1 ) ) ( x - x_1)`

or ` | (x ,y , 1), ( x_1 , y_1 , 1), (x_2, y_2 , 1)| = 0`

5. Intercept Form of a Line :

The equation of a line which cuts-off intercepts a and b respectively from X and Y -axes, is `x/a + y/b =1`

6. General Equation of a Line to Intercept Form :

The general equation of a line Ax+ By+ C = 0 is `x/(-(C/A) ) + y/(-(C/B)) =1`

7. Normal or Perpendicular Form :

The equation of the straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle a with X -axis is

`xcosalpha + ysinalpha = p`, where `0 < = alpha <= pi`

8. General Equation of a Line to the Normal Form :

The general equation of a line is Ax+ By+ C = 0.

Now, to reduce the general equation of a line to normal form, we first shift the constant term on the RHS and make it positive, if it is not so, then divide both sides by

`sqrt (text(coefficient of x)^2 + text( coefficient of y) ^2)`

`=> (A/sqrt(A^2 +B^2) ) x + (B/sqrt(A^2 + B^2 ) )y = ( (-C)/sqrt (A^2 + B^2 ) )`

9. Distance Point Form :

The equation of the straight line passing through `(x_1 ,y_1 )` and making an angle `theta` with the positive direction of X -axis is

`(x -x_1 ) /(cos theta ) = ( y - y_1 ) /(sin theta ) = r`

where, r is the distance of the point (x, y) on the line from the point `(x_1, y_1 )`.

Let the equations of two lines be

`a_1x + b_1y + c_1 = 0` ... (i)

and `a_2x + b_2y + c_2 = 0` ... (ii)

Suppose these two lines intersect at a point `P(x_1 ,y_1 )`. Then, `(x_1, y_1 )` satisfies each of the given equations

`:. a_1x_1 + b_1y_1 + c_1 = 0`

and `a_2x_1 + b_2y_1 + c_2 = 0`

On solving these two equations by cross-multiplication method, we get

`x_1/(b_1c_2 - b_2c_1) = y_1/(c_1a_2 - c_2 a_1 ) = 1/( a_1b_2 -a_2b_1)`

`=> x_1 = ( b_1c_2 - b_2c_1)/(a_1 b_2 -a_2 b_1 )`

and `y_1 = ( c_1a_2 - c_2 a_1)/(a_1b_2 - a_2b_1)`

which is the required intersection point.

The length of the perpendicular from a point `(x_1 ,y_1)` to a line

`ax + by +c = 0 ` is `| (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`

(i) The length of the perpendicular from the origin to the line ax + by +c = 0 is ` (|c| )/sqrt(a^2 +b^2)`

(ii) Distance between parallel lines `ax+ by+ c_1 = 0` and `ax + by +c_2 = 0` is `(|c_2 - c_1|)/sqrt(a^2 +b^2 )`


Area of the Paralleogram :

Area of the l lgm whose 4 sides are as shown in the fig. using

`A = p_1 p_2 cosec theta` is given by

`|((c_1-c_2)(d_1-d_2))/(m_1-m_2)| =((p_1p_2)/(sin theta))`

The equation of the family of lines passing through the intersection of the lines

`a_1x + b_1y + c_1 = 0`

and `a_2x + b_2 y + c_2 = 0`, is

`(a_1x + b_1y + c_1 )+ lamda (a_2x + b_2y + c_2 ) = 0`

where `lamda ` is a parameter

Conditions for two lines `a_1x + b_1y + c_1 = 0` and `a_2x + b_2y + c_2 = 0` to be coincident, parallel, perpendicular or intersecting, are given below:

(i) Coincident, if `a_1/a_2 = b_1/b_2 = c_1/c_2`

(ii) Parallel , if `a_1/a_2 = b_1/b_2 != c_1/c_2`

(iii) Perpendicular , if `a_1a_2 + b_1 b_2 = 0`

(iv) Intersecting , if `a_1 b_2 - a_2b_1 != 0`

i.e `a_1/a_2 != b_1/b_2`

The equations of the bisectors of the angles between the lines `a_1x + b_1y + c_1 = 0` and `a_2x + b_2y + c_2 = 0` are given by `(a_1x + b_1y + c_1 )/sqrt(a_1^2 + b_1^2 ) = pm (a_2x +b_2 y +c_2)/sqrt (a_2^2 +b_2^2)`

Conditions	Internal (acute) angle bisector	External (obtuse) angle bisector
`a_1 a_2 + b_1 b_2 > 0`	-	+
`a_1 a_2 + b_1 b_2 < 0`	+	-

The three lines are concurrent, if they meet at a point.

Method 1 : Find the point of intersection of any two lines and show that it satisfies the third also.

Method 2 : The three lines `a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0` and `a_3x + b_3y + c_3 = 0`, then they are concurrent if

`|(a_1 , b_1 , c_1 ), ( a_2, b_2 , c_2 ), (a_3, b_3 , c_3 ) | = 0`

Method : 3 The condition for the lines A= 0, B = 0 and C = 0 to be concurrent is that three constants l, m, n (not all zeroes at the same time) can be obtained such that

`l A + m B + n C = 0`