Inclination of a line: 1. Slope of a Straight line 2. Slopes of parallel lines 3. Slopes of perpendicular lines

**Inclination of a line**

The inclination of a line or the angle of inclination of a line is the angle
which a straight line makes with the positive direction of *X* axis
measured in the counter-clockwise direction to the part of the line above the *X*
axis. The inclination of the line is usually denoted by *θ*.

**Note**

·
The inclination of *X* axis and every line parallel to *X*
axis is 0°.

·
The inclination of *Y* axis and every line parallel to *Y*
axis is 90°.

While laying roads one must know how steep the road will be.
Similarly, when constructing a staircase, we should consider its steepness. For
the same reason, anyone travelling along a hill or a bridge, feels hard
compared to travelling along a plain road.

All these examples illustrate one important aspect called
“Steepness”. The measure of steepness is called slope or gradient.

The concept of slope is important in economics because it is used to measure the rate at which the demand for a product changes in a given period of time on the basis of its price. Slope comprises of two factors namely steepness and direction.

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*.

Therefore the slope of the straight line is *m* = tan *θ*
, 0 ≤ *θ* ≤ 180° , *θ* ≠ 90°

Slope *m *= tan *θ*

The slope of the line through (*x* _{1} , *y*_{1}
) and (*x* _{2} , *y*_{2} ) with *x *_{1}*
≠ x*_{2 }is .

Note

The slope of a vertical line is undefined

**Values of slopes**

Two non-vertical lines are parallel if and only if their slopes are equal.

Let *l*_{1} and *l*_{2} be two
non-vertical lines with slopes *m*_{1}* *and* m*_{2}*
*respectively.

Let the inclination of the lines with positive direction of *X*
axis be *θ*_{1} and *θ*_{2} respectively.

Assume, *l*_{1} and *l*_{2}are parallel

*θ*_{1} =* θ*_{2} (Since, *θ*_{1} , *θ*_{2}
are corresponding angles)

tan *θ*_{1}= tan* θ*_{2}

*m*_{1}=* m*_{2}

Hence, the slopes are equal.

Therefore, non-vertical parallel lines have equal slopes.

**Conversely**

Let the slopes be equal, then *m*_{1} =* m*_{2}

tan *θ*_{ 1}= tan* θ*_{ 2}

*θ*_{1}=* θ*_{2 }(since 0 ≤ *θ*_{1} , *θ*_{2}
≤ 180°)

That is the corresponding angles are equal.

Therfore, *l*_{1} and *l*_{2} are
parallel

Thus, non-vertical lines having equal slopes are parallel.

Hence, non vertical lines are parallel if and only if their slopes are equal.

Two non-vertical lines with slopes *m*_{1} and *m*_{2}
are perpendicular if and only if *m*_{1}*m*_{2}* *= −1

Let *l*_{1} and *l*_{2} be two
non-vertical lines with slopes *m*_{1} and *m*_{2} ,
respectively. Let their inclinations be *θ*_{1} and *θ*_{2}
respectively.

Then *m*_{1} = tan *θ*_{1}
and *m*_{2} = tan *θ*_{2}

First we assume that, *l*_{1} and *l*_{2}
are perpendicular to each other.

Then ∠*ABC* = 90° − *θ*_{1}
(sum of angles of ΔABC is 180° )

Now measuring slope of *l*_{2} through angles *θ*_{2}
and 90° − *θ*_{1} , which are opposite to each other, we get

tan *θ*_{2 }= −tan(90° − *θ*_{1} )

tan *θ*_{1} . tan *θ*_{2 }= −1

*m*_{1}* *.*m*_{2 }= −1

Thus, when the line *l*_{1} is perpendicular to line *l*_{2}
then *m*_{1}*m*_{2} = −1 .

**Conversely,**

Let *l*_{1} and *l*_{2} be two
non-vertical lines with slopes *m*_{1} and *m*_{2}
respectively, such that *m*_{1}*m*_{2}* *= −1*
*.

Since *m*_{1 }= tan *θ*_{ 1} , *m*_{2
}= tan *θ*_{2}

We have tan *θ*_{1} tan *θ*_{2 }= −1

tan *θ*_{1} = −1/ tan *θ*_{2}

tan *θ*_{1} = −cot *θ*_{2}

tan *θ*_{1} = −tan(90° - *θ*_{2})

tan *θ*_{1} = tan(-(90° - *θ*_{2})) =
tan(*θ*_{2 }- 90°)

*θ*_{1} = *θ*_{2 }- 90° (since 0 ≤ *θ*_{1, }*θ*_{2
}≤ 180)

*θ*_{2} = 90° + *θ*_{1}

But in ΔABC, *θ*_{2} = ∠C + *θ*_{1}

Therefore, ∠C = 90°

**Note **

Let** ***l*_{1}** **and** ***l*_{2}**
**be two lines with well-defined slopes** ***m*_{1}** **and**
***m*_{2}** **respectively, then

*(i) l*_{1}* *is parallel to* l*_{2}* *if and only if* m*_{1}*
*=* m*_{2}* *.

*(ii) l*_{1} is perpendicular to *l*_{2} if and only if *m*_{1}*m*_{2
}= −1.

**Example 5.8**

(i) What is the slope of a line whose inclination is 30° ?

(ii) What is the inclination of a line whose slope is √3?

*Solution*

(i) Here *θ *=* *30°

Slope *m* = tan *θ*

Therefore, slope *m* = tan 30° = 1 /√3

(ii) Given *m* = √3 let *θ* be the inclination of the
line

tan *θ* = √3

We get, *θ *=* *60°

**Example 5.9**

Find the slope of a line joining the given points

(i) (-6, 1) and (-3, 2)
(ii) (-1/3, ½) , and (2/7, 3/7) (iii) (14, 10) and (14, -6)

*Solution*

(i) (-6, 1) and (-3, 2)

The slope

(iii) (14, 10) and (14, - 6)

The slope is undefined.

The line** ***r*** **passes through the points** **(–2, 2)** **and** **(5, 8)** **and the line** ***s*** **passes through the points (–8, 7) and (–2, 0).
Is the line *r* perpendicular to *s* ?

That is,*m*_{1}*m*_{2 }= −1

Therefore, the line *r* is perpendicular to line *s*.

**Example 5.11**

The line** ***p*** **passes through the points** **(3,** **-** **2)** **,** **(12, 4)** **and the line** ***q*** **passes** **through the points (6, - 2) and (12, 2) . Is *p*
parallel to *q* ?

*Solution*

Thus, slope of line *p *= slope of line *q.*

Therefore, line *p* is parallel to the line *q.*

**Example 5.12**

Show that the points** **(-2, 5)** **,** **(6,** **-1) and (2, 2) are collinear.

*Solution *

The vertices are* **A*(-2, 5)** **,

We get, Slope of *AB *= Slope of *BC*

Therefore, the points *A, B, C* all lie in a same straight
line.

Hence the points *A*, *B* and *C* are collinear.

Let** ***A*(1,** **-** **2)** **,** ***B*(6,** **-** **2)** **,** ***C*(5, 1)** **and** ***D*(2, 1)** **be four points

(i) Find the slope of the line segments (a) *AB
*(b) *CD*

(ii) Find the slope of the line segments (a) *BC (b) AD*

(iii) What can you deduce from your answer.

(iii) The slope of AB and CD are equal so AB, CD are parallel.

Similarly the lines AD and BC are not parallel, since their slopes
are not equal. So, we can deduce that the quadrilateral ABCD is a trapezium.

**Example 5.14**

Consider the graph representing growth of population (in crores).
Find** **the slope of the line *AB*
and hence estimate the population in the year 2030?

*Solution*

The points* **A*(2005, 96)** **and

Let the growth of population in 2030 be *k* crores.

Assuming that the point *C* (2030,*k*) is on *AB*,

we have, slope of *AC* = slope of *AB*

*k* - 96 = 10

*k* = 106

Hence the estimated population in 2030 = 106 Crores.

Without using Pythagoras theorem, show that the points** **(1,** **-** **4)** **,** **(2,** **-** **3)** **and (4, - 7) form a
right angled triangle.

Let the given points be *A*(1, - 4) , *B*(2, - 3) and *C*(4,*
*-* *7).

Slope of *AB* × slope of *AC *= (1)(−1) = −1

AB is perpendicular to AC. ∠*A* = 90°

Therefore, Δ*ABC* is a right angled triangle.

Prove analytically that the line segment joining the mid-points of
two sides**
**of a triangle is
parallel to the third side and is equal to half of its length.

Let* **P*** **(

Let *S* be the mid-point of *PQ* and *T* be the mid-point
of *PR*

Therefore, *ST* is parallel to *QR*. (since, their
slopes are equal)

Also

Thus
ST is parallel to QR and half of it.

*Note*

This example illustrates how a geometrical result can be proved
using coordinate Geometry.

Tags : Slopes | Coordinate Geometry , 10th Mathematics : UNIT 5 : Coordinate Geometry

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10th Mathematics : UNIT 5 : Coordinate Geometry : Inclination of a line | Slopes | Coordinate Geometry

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