ANGLE BETWEEN TWO STRAIGHT LINES

Two straight lines in a plane would either be parallel or coincide or intersect. Normally when two straight lines intersect, they form two angles at the point of intersection. One is an acute angle and another is an obtuse angle or equal. Both these angles would be supplements (Sum equals 180°) of each other. By definition, when we say ‘angle between two straight lines’ we mean the acute angle between the two lines.

Example 1 :

Find the angle between the straight lines :

3x - 2y + 9 = 0

2x + y - 9 = 0

Solution :

Write each equation in slope intercept form.

Formula to find angle between the two straight lines : 

θ = tan^-1|(m₁ - m₂)/(1 + m₁m₂)|

Substitute m₁ = 3/2 and m₂ = -2.

= tan^-1|(3/2 - 2)/(1 + (3/2)(-2))|

= tan^-1|(3/2) + 2))/(1 - 3)|

= tan^-1|(7/2)/(- 2)|

= tan^-1|-7/4|

= tan^-1(7/4)

Example 2 :

Show that the following two straight lines are parallel.

2x + y - 9 = 0

2x + y - 10 = 0

Solution :

Write each equation in slope intercept form.

Formula to find angle between the two straight lines : 

θ = tan^-1|(m₁ - m₂)/(1 + m₁m₂)|

Substitute m₁ = -2 and m₂ = -2.

= tan^-1|(-2 + 2)/(1 + (-2)(-2))|

= tan^-1|0/(1 + 4)|

= tan^-1|0|

= 0°

Since the angle between the given two straight lines is 0°, the lines are parallel.

Example 3 :

Show that the following two straight lines are perpendicular. 

2x + 3y - 9 = 0

3x - 2y + 10 = 0

Solution :

Write each equation in slope intercept form.

Formula to find angle between the two straight lines : 

θ = tan^-1|(m₁ - m₂)/(1 + m₁m₂)|

Substitute m₁ = -2/3 and m₂ = 3/2.

= tan^-1|(-2/3 + 3/2)/(1 + (-2/3)(3/2))|

= tan^-1|(5/9)/(1 - 1)|

= tan^-1|(5/9)/0|

= tan^-1(∞)

= 90°

Since the angle between the given two straight lines is 90°, the lines are parallel.

Example 4 :

Find the equations of the lines through the point (4, 3) that make an angle of 45 degree with the line for

6x + y - 5 = 0

Solution :

From the given information, we know that angle created by the line and the line 6x + y - 5 = 0 is 45 degree.

y = -6x + 5

m₁ = -6

Slope of the required line be m₂

Formula to find angle between the two straight lines : 

θ = tan^-1|(m₁ - m₂)/(1 + m₁m₂)|

45 = tan^-1|(-6 - m₂)/(1 + (-6)m₂)|

tan 45 = |(-6 - m₂)/(1 - 6m₂)|

1 = (-6 - m₂)/(1 - 6m₂)

1 - 6m₂ = -6 - m₂

- 6m₂ + m₂ = -6 - 1

-5m₂ = -7

m₂ = 7/5

Equation of the line :

y - y₁ = m(x - x₁)

y - 3 = (7/5)(x - 4)

5(y - 3) = 7(x - 4)

5y - 15 = 7x - 28

7x - 5y - 28 + 15 = 0

7x - 5y - 13 = 0

Example 5 :

If one diagonal of a square is along the line 8x – 15y = 0 and one of its vertex is at (1, 2), then find the equation of sides of the square passing through this vertex.

Solution :

Let ABCD be the given square and the coordinates of the vertex D be (1, 2). We are required to find the equations of its sides DC and AD.

Given that BD is along the line 8x – 15y = 0, so its slope is 8/15.

The angles made by BD with sides AD and DC is 45°. Let the slope of DC be m. Then

tan 45 = m - (8/15) / (1 + (8m/15))

1 = m - (8/15) / (1 + (8m/15))

1 + (8m/15) = m - (8/15)

1 + 8/15 = m - (8m/15)

23/15 = (15m - 8m)/15

23/15 = 7m/15

m = (23/15) x (15/7)

m = 23/7

Equation of the side DC :

y - y₁ = m(x - x₁)

y - 2 = (23/7)(x - 1)

y = (23/7) x - (23/7) + 2

y = (23/7) x + (14 - 23)/7

y = (23/7) x - 9/7

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