    # Angle Between Two Straight Lines THEOREM 1:

To find the angle between two given straight lines.

Solution:

Let the two straight lines be AT1 and AT2 which meet at x-axis at T1 and T2 points respectively. Let the equations of the two lines AT1 and AT2 be

y = m1x + c1 and y2 = m2x + c2            (1)

Hence tan AT1X = m1 and tan AT2X = m2

Now ∠T1AT2 = ∠AT1X - ∠AT2X

Hence tan T1AT2 = tan [AT1X - AT2X] Hence the required angle = ∠T1AT2 (2)

If equation (2) is a positive quantity then it is the tangent of the acute angle between the two lines, if (2) is a negative quantity then it is the tangent of the obtuse angle.

Example:

Find the angle between two straight lines y = 6x - 11 and Solution:

The equation of the first line is y = 6x - 11 which implies m1 = 6.

The equation of the second line is hence  THEOREM 2:

To find the condition that two straight lines can be parallel.

Two straight lines are parallel when the angle between them is zero and hence the tangent of this angle is zero. Hence in theorem 1 the equation (2) becomes

m1 = m2 which is the required result.

Example:

Find the equation of a straight line which passes through the point (4, -5) and which is parallel to the straight line 3x + 4y + 5 = 0.

Solution:

Any straight line which is parallel to 3x + 4y + 5 = 0 will be 3x + 4y + c2 = 0 because m1 = m2.

Now given that this line 3x + 4y + c2 = 0 passes through the point (4, -5) hence we get,

3. 4 + 4. (-5) + c2 = 0 or c2 = 8

Putting the value of c2 = 8 we get 3x + 4y + 8 = 0. (Answer)

THEOREM 3:

To find the condition that two straight lines can be perpendicular.

Let the two straight lines be y = m1x + c1 and y2 = m 2x + c2

If the angle between them is θ then by theorem 1, (1)

If the two lines are perpendicular then θ = 90 ο and so tan θ = ∞.

Hence in equation (1) which is only possible if the denominator is zero

Therefore the condition of perpendicularity is

1 + m1m2 = 0 or m1m2 = -1.

Hence the straight line y2= m2x + c2 is perpendicular to y = m1x + c1 if Example:

Find the equation of a straight line which passes through the point (4, -5) and is perpendicular to the straight line 3x + 4y + 5 = 0.

Solution:

Let the equation of a straight line be y = m1x + c

Now given that this line passes through the point (4, -5) hence it becomes, -5 = 4m1 + c         (1)

Given that the above line is perpendicular to the straight line 3x + 4y + 5 = 0         (2)

From (2) we get where Then by the condition of perpendicularity m1m2 = -1.

Hence Or putting which in eq. (1) we get, Hence the required equation will be Or 4x - 3y = 31         (Answer)

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