**Slopes of Parallel Lines : **

If two non vertical straight lines are parallel, then slopes of the lines will be equal.

If two non vertical straight lines with slopes m_{1} and m_{2}, are parallel, then

m_{1} = m_{2}

On the other hand, if the slopes of two straight lines are equal, then the lines will be parallel.

**Slopes of Perpendicular Lines :**

If two non vertical straight lines are perpendicular, then the product of slopes of the lines will be equal to '-1'.

If two non vertical straight lines with slopes m_{1} and m_{2}, are perpendicular, then

m_{1} m_{2} = –1

On the other hand, if the product of slopes of two straight lines is equal to '-1', then the lines will be perpendicular.

**Note :**

The straight lines x-axis and y-axis are perpendicular to each other. But, the condition m_{1}m_{2} = –1 is not true. Because the slope of the x-axis is zero and the slope of the y-axis
is not defined.

**Example 1 :**

The line joining the points A (-2 , 3) and B (a , 5) is parallel to the line joining the points C (0 , 5) and D (-2 , 1). Find the value of a.

**Solution :**

Since the line joining the points AB and CD are parallel, the slope of those two lines will be equal.

**Slope of AB :**

A(-2 , 3) ==> (x_{1}, y_{1}), B(a , 5) ==> (x_{2}, y_{2})

Slope (m_{1}) = (y_{2} - y_{1})/(x_{2} - x_{1})

= (5 - 3) / (a - (-2))

= 2 / (a + 2)

**Slope of CD :**

C(0 , 5) ==> (x_{1}, y_{1}), D(-2 , 1) ==> (x_{2}, y_{2})

Slope (m_{2}) = (y_{2} - y_{1})/(x_{2} - x_{1})

= (1 - 5) / (-2 - 0)

= -4 / (-2)

= 2

Slope of AB = Slope of CD

2 / (a + 2) = 2

Multiply by (a +2) on both sides

2 = 2(a + 2)

2 = 2a + 4

Subtract 4 on both sides

2 - 4 = 2a

2a = -2

Divide by 2 on both sides,

a = -1

**Example 2 :**

The line joining the points A(0, 5) and B (4, 2) is perpendicular to the line joining the points C (-1, -2) and D (5, b). Find the value of b.

**Solution :**

Since the line joining the points AB and CD are perpendicular, the product of slopes will be equal to -1.

**Slope of AB :**

A (0, 5) ==> (x_{1}, y_{1}), B (4, 2) ==> (x_{2}, y_{2})

Slope (m_{1}) = (y_{2} - y_{1})/(x_{2} - x_{1})

= (2 - 5) / (4 - 0)

= -3 / 4

**Slope of CD :**

C (-1, -2) ==> (x₁, y₁), D (5 , b) ==> (x₂, y₂)

Slope (m_{2}) = (y_{2} - y_{1})/(x_{2} - x_{1})

= (b - (-2)) / (5 - (-1))

= (b + 2) / (5 + 1)

= (b + 2) / 6

Slope of AB x Slope of CD = -1

(-3/4) x (b + 2)/6 = -1

(b + 2)/8 = 1

Multiply by 8 on both sides,

b + 2 = 8

Subtract by 2 on both sides

b = 8 - 2

b = 6

**Example 3 :**

The vertices of triangle ABC are A(1, 8), B(-2, 4), C(8, -5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.

**Solution :**

Mid point of the side AB = M

mid point = (x_{1} + x_{2})/2 , (y_{1} + y_{2})/2

= [1 + (-2)]/2 , (8 + 4)/2

= (-1/2, 6)

Mid point of the side AC = N

mid point = (x_{1} + x_{2})/2 , (y_{1} + y_{2})/2

= [1 + 8]/2 , (8 - 5)/2

= (9/2, 3/2)

Slope of BC :

= (y_{2} - y_{1})/(x_{2} - x_{1})

= (-5 -4) / (8 + 2)

= -9 / 10 ------(1)

Slope of MN :

= (y_{2} - y_{1})/(x_{2} - x_{1})

= ((3/2) - 6) / ((9/2) + (1/2))

= (-9/2) / (10/2)

= -9 / 10 ------(2)

So, the sides MN and BC are parallel.

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