A line is formed by joining two points. Let there are points P and Q having coordinates (x_{1}, y_{1}) and (x_{2}, y_{2}) then the equation of line passing through these two points is

y-y_{1} = m (x – x_{1})

where m is the slope of the line.

m = (y_{2}-y_{1})/(x_{2}-x_{1}); if x_{1} is not equal to x_{2}

if x_{1} = x_{2} then m is undefined and the line will be vertical and if y_{1}=y_{1} 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:

- If a
_{1}x + b_{1}y + c_{1}= 0 and a_{2}x + b_{2}y + c_{2}= 0 are equations of two lines, the angle ? between them is given by

Cos ? = (a_{1}a_{2}+b_{1}b_{2})/{(a_{1}^{2}+b_{1}^{2})(a_{2}^{2}+b_{2}^{2})}^{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 ? = (a_{1}a_{2}+b_{1}b_{2})/{(a_{1}^{2}+b_{1}^{2})(a_{2}^{2}+b_{2}^{2})}^{1/2}

= (3*10 + 5*-6)/{ (3^{2} + 5^{2})(10^{2} + (-6)^{2}}^{1/2}

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

= 0

? = 90?

## Distance between Two Lines:

- If ax + by + c = 0 is the equation of a line, the perpendicular distance from a point (x
_{1}, y_{1}) to this line is given by: |ax_{1}+ by_{1}+ c|/ (a^{2}+ b^{2})^{1/2} - The distance between two parallel straight lines ax + by + c
_{1}= 0 and ax + by + c_{2}=0 is given by: |c_{1}– c_{2}|/(a^{2}+ b^{2})^{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(x_{1}, y_{1}) to the line ax + by + c = 0 is given by:

d = |ax_{1} + by_{1} + c|/ (a^{2} + b^{2})^{1/2}

So d = |4*2 + 3*2 – 4|/(4^{2} + 3^{2})^{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