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Distance and Angle between the Two Lines

A line is formed by joining two points. Let there are points P and Q having coordinates (x1, y1) and (x2, y2) then the equation of line passing through these two points is

y-y1 = m (x – x1)

where m is the slope of the line.

m = (y2-y1)/(x2-x1); if x1 is not equal to x2

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if x1 = x2 then m is undefined and the line will be vertical and if y1=y1 then the slope m will be 0 and line will be horizontal line.

Equation having form ax + by = c have the slope = -a/b.

Rearrange the equation

By = -ax + c

y = -a/b x + c

Angle between the Two Lines:

  1. If a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are equations of two lines, the angle ? between them is given by

Cos ? = (a1a2+b1b2)/{(a12+b12)(a22+b22)}1/2

2. If m1 and m2 are the slopes of two lines, the angle ? between them is given by

Tan ? = | m1 – m2|/|1 + m1m1|

Note:

1. Condition for parallel lines:

a1/a2 = b1/b2 or m1 = m2

2. Condition for perpendicular lines:

a1a2 + b1b2 = 0 or m1 m2 = -1

Solved Examples:

here is few free geometry problems

Example 1: Show that the lines 2x – 3y – 4 = 0 and -8x + 12y – 6 = 0 are parallel.

Solution: a1 = 2, a2 = -8, b1 = -3, b2 = 12

a1/a2 = 2/-8 = -1/4

b1/b2 = -3/12 = -1/4

a1/a2 = b1/b2

Example 2: find the angle between the lines 3x + 5y – 9=0 and 10x – 6y + 7=0.

Solution: a1 = 3, b1 = 5, c1= -9

a2 = 10, b2 = -6, c2 = 7

Cos ? = (a1a2+b1b2)/{(a12+b12)(a22+b22)}1/2

= (3*10 + 5*-6)/{ (32 + 52)(102 + (-6)2}1/2

= (30-30)/{36*136}1/2

= 0

? = 90?

Distance between Two Lines:

  1. If ax + by + c = 0 is the equation of a line, the perpendicular distance from a point (x1, y1) to this line is given by: |ax1 + by1 + c|/ (a2 + b2)1/2
  2. The distance between two parallel straight lines ax + by + c1 = 0 and ax + by + c2 =0 is given by: |c1 – c2|/(a2 + b2)1/2

Solved Examples:

Example 3: Find the perpendicular distance of the point (2, 2) from the line 4x + 3y – 4 = 0.

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

d = |ax1 + by1 + c|/ (a2 + b2)1/2

So d = |4*2 + 3*2 – 4|/(42 + 32)1/2

= |8 + 6 – 4|/(16 + 9)1/2

= |10|/(25)1/2

= 10/5

= 2

Example 4: Find the value of k, if (x + y – 1) – k(3x – 7y + 12) = 0 is parallel to y-axis.

Solution: Rearrange the equation:

(1 – 3k)x + (1 + 7k)y – (1 + 12k) = 0 ----------------------- (1)

Slope = - (1 – 3k)/(1 + 7k) = (3k – 1)/(1 + 7k)

Since eqn (1) is parallel to y-axis

Slope = infinity

1 + 7k = 0

k = -1/7