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# Inclination of a line

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°.

## 1. Slope of a Straight line

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.

## Definition

If θ is the angle of inclination of a non -vertical straight line, then tanθ 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°

### To find the slope of a straight line when two points are given

Slope m = tan θ The slope of the line through (x 1 , y1 ) and (x 2 , y2 ) with x 1 ≠ x2 is .

Note

The slope of a vertical line is undefined

Values of slopes ## 2. Slopes of parallel lines

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

Let l1 and l2 be two non-vertical lines with slopes m1 and m2 respectively.

Let the inclination of the lines with positive direction of X axis be θ1 and θ2 respectively. Assume, l1 and l2are parallel

θ1 = θ2   (Since, θ1 , θ2  are corresponding angles)

tan θ1= tan θ2

m1= m2

Hence, the slopes are equal.

Therefore, non-vertical parallel lines have equal slopes.

Conversely

Let the slopes be equal, then m1 = m2

tan θ 1= tan θ 2

θ1= θ2 (since 0 ≤ θ1 , θ2  ≤ 180°)

That is the corresponding angles are equal.

Therfore, l1  and l2 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.

## 3. Slopes of perpendicular lines

Two non-vertical lines with slopes m1 and m2 are perpendicular if and only if m1m2 = −1

Let l1 and l2 be two non-vertical lines with slopes m1 and m2 , respectively. Let their inclinations be θ1 and θ2 respectively. Then m1  = tan θ1  and m2  = tan θ2

First we assume that, l1 and l2 are perpendicular to each other.

Then ABC = 90° − θ1  (sum of angles of ΔABC is 180° )

Now measuring slope of l2 through angles θ2 and 90° − θ1 , which are opposite to each other, we get

tan θ2 = −tan(90° − θ1 ) tan θ1 . tan θ2 = −1

m1 .m2 = −1

Thus, when the line l1 is perpendicular to line l2  then m1m2  = −1 .

Conversely,

Let l1 and l2 be two non-vertical lines with slopes m1 and m2 respectively, such that m1m2 = −1 .

Since m1 = tan θ 1 , m2 = tan θ2

We have tan θ1 tan θ2 = −1

tan θ1 = −1/ tan θ2

tan θ1 = −cot θ2

tan θ1 = −tan(90° - θ2)

tan θ1 = tan(-(90° - θ2)) = tan(θ2 - 90°)

θ1 = θ2 - 90°         (since 0 ≤ θ1, θ2 ≤ 180)

θ2 = 90° + θ1

But in ΔABC, θ2 = C + θ1

Therefore, C = 90°

Note

Let l1 and l2 be two lines with well-defined slopes m1  and m2  respectively, then

(i) l1 is parallel to l2 if and only if m1 = m2 .

(ii) l1 is perpendicular to l2 if and only if m1m2 = −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.

### Example 5.10

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 ?

### Solution That is,m1m2 = −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) , B(6, -1) and C(2, 2). 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.

### Example 5.13

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

### Solution (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 B(2015, 100) are on the line AB. 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.

### Example 5.15

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

### Solution

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, ΔABC is a right angled triangle.

### Example 5.16

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.

### Solution

Let P (a,b) Q (c,d) and R(e, f ) be the vertices of a triangle.

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