**How to Find Slope of the Line ?**

Here we are going to see, finding area of triangle and quadrilateral with vertices.

(1) If θ is the angle of inclination of a non-vertical straight line, then tan θ is called the slope or gradient of the line and is denoted by m.

m = tan θ

(2) To find the slope of a straight line when two points are given

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

(3) To find the slope of a straight line when equation of the line is given

m = -coefficient of x/coefficient of y

(4) If two lines are parallel, then its slopes will be equal.

m_{1} = m_{2}

(5) If two lines are perpendicular, then the product of their slopes

m_{1} x m_{2 }= -1

**Question 1 :**

What is the slope of a line whose inclination with positive direction of x -axis is

(i) 90° (ii) 0

**Solution :**

**(i) θ = ** 90°

m = tan **θ**

**m = tan **90°

m = undefined

**(i) θ = ** 0°

m = tan **θ**

**m = tan **0°

m = 0

**Question 2 :**

What is the inclination of a line whose slope is (i) 0 (ii) 1

**Solution :**

(i) 0

m = 0

tan **θ = 0**

**Hence the required angle of inclination is 0.**

**(ii) 1**

m = 1

tan **θ = 1**

**Hence the required angle of inclination is 45**°.

**Question 3 :**

Find the slope of a line joining the points

(i) (5, 5) with the origin

**Solution :**

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

x_{1} = 5, x_{2} = 0, y_{1} = 5 and y_{2} = 0

m = (0 - 5)/(0 - 5)

m = -5/(-5)

m = 1

(ii) (sin **θ**, -cos **θ**) and (-sin **θ **, cos **θ**)

**Solution :**

x_{1} = sin **θ**, x_{2} = -sin **θ**, y_{1} = -cos **θ** and y_{2} = cos **θ**

m = (cos **θ - (**-cos **θ))/**(-sin **θ - sin** **θ)**

** m = (cos θ + cos θ)/(-sin θ - sin θ)**

** m = (2cos θ)/(-2sin θ)**

** m = -****cos θ/sin θ**

** m = cot ****θ**

**Question 4 :**

What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6, 4) .

**Solution :**

P is the mid-point of the segment joining (4,2) and (-6, 4)

First, let us find the point P.

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

= (4+(-6))/2, (2 + 4)/2

= -2/2, 6/2

= P(-1, 3)

Now, we have to find the slope of the line which is perpendicular to the line joining the points A(5, 1) and P(-1, 3).

Slope of AP x slope of the required line = -1

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

Slope of AP = (3 - 1)/(-1 - 5)

= 2/(-6)

= -1/3

Slope of required line = -1/(-1/3)

= 3

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Slope of the Line".

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