Mathematics STRAIGHT LINES -Introduction , Slope or gradient ,Conditions for parallelism and perpendicularity and Angle between two lines
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### Topics Covered

star Introduction
star Slope or gradient of a Line
star Slope of a line when coordinates of any two points on the line are given
star Conditions for parallelism and perpendicularity of lines in terms of their slopes
star Angle between two lines
star Collinearity of three points

### Introduction

\color{green} ✍️ Let us have a brief recall of coordinate geometry done in earlier classes.

\color{green} ✍️ To recapitulate, the location of the points (6, – 4) and (3, 0) in the XY-plane is shown in Fig 1.
We may note that the point (6, – 4) is at 6 units distance from the y-axis measured along the positive x-axis and at 4 units distance from the x-axis measured along the negative y-axis.

\color{green} ✍️ Similarly, the point (3, 0) is at 3 units distance from the y-axis measured along the positive x-axis and has zero distance from the x- axis.

We also studied there following important formulae:

\color { maroon} ® \color{maroon} ul (" REMEMBER")  \ \ \ \ 1. color{green}("Distance between the points") P \ (x_1, y_1) and Q \ (x_2, y_2) is

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{blue}(PQ=sqrt((x_2-x_1)^2+(y_2-y_1)))

\color { maroon} ® \color{maroon} ul (" REMEMBER")  \ \ \ \ \ 2. The coordinates of a point dividing the line segment joining the points (x_1, y_1) and (x_2, y_2) internally, in the ratio m: n are color{blue} ( (((mx_2+nx_1)/(m+n) , (my_2+ny_1)/(m+n)) )

\color { maroon} ® \color{maroon} ul (" REMEMBER")  \ \ \ \ \ \ 3.  In particular, if m = n, the coordinates of the mid-point of the line segment joining the points (x_1, y_1) and (x_2, y_2) are color{blue} (((x_2+x_1)/2 , (y_2+y_1)/2))

\color { maroon} ® \color{maroon} ul (" REMEMBER") \ \ \ \ \ \ \ 4.  color{green}("Area of the triangle") whose vertices are (x_1, y_1), (x_2, y_2) and (x_3, y_3) is

color{blue} (=1/2| x_1( y_2 − y_3) + x_2( y_3 − y_1) + x_3( y_1 − y_2)|)

"Remark " If the area of the triangle ABC is zero, then three points A, B and C lie on a line, i.e., they are collinear.

### Slope or gradient of a Line

\color{green} ✍️ A line in a coordinate plane forms two angles with the x-axis, which are supplementary.

\color{green} ✍️ The angle (say) θ made by the line l with positive direction of x-axis and measured color(blue)("anti clockwise is called the inclination of the line.") Obviously color(red)(0° ≤ θ ≤ 180°) (Fig 2).

\color{green} ✍️ We observe that lines parallel to x-axis, or coinciding with x-axis, have inclination of 0°. The inclination of a vertical line (parallel to or coinciding with y-axis) is 90°.

\color{purple}ul(✓✓) \color{purple} (" DEFINITION ALERT")

\color{green} ✍️ If θ is the inclination of a line l, then color{green}(tan θ) is called the color(blue)"slope or gradient" of the line l.

\color{green} ✍️ The slope of a line is denoted by color(blue)(m.)

\color { maroon} ® \color{maroon} ul (" REMEMBER") Thus, color(red)(m = tan θ, θ ≠ 90°)

 \color{red} \ox \color{red} \mathbf(COMMON \ CONFUSION)  The slope of a line whose inclination is 90° is not defined.

\color{green} ✍️ It may be observed that the slope of x-axis is zero and slope of y-axis is not defined.

### Slope of a line when coordinates of any two points on the line are given

\color{green} ✍️ Let P(x_1, y_1) and Q(x_2, y_2) be two points on non-vertical line l whose inclination is color(blue)(θ.)

Obviously, x_1 ≠ x_2, otherwise the line will become perpendicular to x-axis and its slope will not be defined. The inclination of the line l may be acute or obtuse.

Let us take these two cases. Draw perpendicular QR to x-axis and PM perpendicular to RQ as shown .

color{blue}("Case 1 :") color{green}("When angle θ is acute")

In Fig 3 (i), ∠MPQ = θ. ....... (1)

Therefore, slope of line color(purple)(l = m = tan θ.)..........(2)

But in ΔMPQ, we have tanθ =(MQ)/(MP)=(y_2-y_1)/(x_2-x_1)

From equations (1) and (2), we have

\color { maroon} ® \color{maroon} ul (" REMEMBER")  \ \color(red)(m=(y_2-y_1)/(x_2-x_1))

color{blue}("Case 2 :") color{green}("When angle θ is obtuse:")

In Fig 3 (ii), we have ∠MPQ = 180° – θ.

Therefore, θ = 180° – ∠MPQ.

Now, slope of the line color(purple)(l =m = tan θ = tan ( 180° – ∠MPQ))

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = – tan ∠MPQ=-(MQ)/(MP)=- (y_2-y_1)/(x_1-x_2)=(y_2-y_1)/(x_2-x_1)

Consequently, we see that in both the cases the slope m of the line through the points (x_1, y_1) and (x_2, y_2) is given by

\color { maroon} ® \color{maroon} ul (" REMEMBER")  \ \ color(red)(m=(y_2-y_1)/(x_2-x_1))

### Conditions for parallelism and perpendicularity of lines in terms of their slopes

\color{green} ✍️ In a coordinate plane, suppose that non-vertical lines l_1 and l_2 have slopes m_1 and m_2, respectively. Let their inclinations be α and β, respectively.

\color{green} ✍️ color{blue}("If the line") color{blue}(l_1) color{blue}("is parallel to") color{blue}(l_2) (Fig 10 .4), then their inclinations are equal, i.e., color(red)(α = β), and hence, color(red)(tan α = tan β)

Therefore m_1 = m_2, i.e., their slopes are equal.

Conversely, if the slope of two lines l_1 and l_2 is same, i.e.,

m_1 = m_2.

Then tan α = tan β.

By the property of tangent function (between 0° and 180°), α = β. Therefore, the lines are parallel.

Hence, two non vertical lines l_1 and l_2  are parallel if and only if their slopes are equal.

\color{green} ✍️ color{blue}("If the lines" \ \ l_1" and" \ \ l_2" are perpendicular") (Fig10.5), then color(red)(β = α + 90°.)

Therefore, tan β = tan (α + 90°)

= – cot α = - 1/ tanα

m_2 =-1/m_1 or color(red)(m_1*m_2=-1)

Conversely, if m_1 m_2 = – 1, i.e., tan α tan β = – 1.

Then tan α = – cot β = tan (β + 90°) or tan (β – 90°)

Therefore, α and β differ by 90°.

Thus, lines l_1 and l_2 are perpendicular to each other.

Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other,

m_2 =-1/m_1 or color(red)(m_1*m_2=-1)
Q 3146145073

Find the slope of the lines:
(a) Passing through the points (3, – 2) and (–1, 4),
(b) Passing through the points (3, – 2) and (7, – 2),
(c) Passing through the points (3, – 2) and (3, 4),
(d) Making inclination of 60° with the positive direction of x-axis.

Solution:

(a) The slope of the line through (3, – 2) and (– 1, 4) is

 m = (4- (-2) )/(-1-3) = 6/(-4) = -3/2

(b) The slope of the line through the points (3, – 2) and (7, – 2) is

m = (-2 - (-2) )/(7-3) = 0/4= 0

(c) The slope of the line through the points (3, – 2) and (3, 4) is

m = (4- (-2) )/( 3-3) = 6/0 , which is not defined.

(d) Here inclination of the line α = 60°. Therefore, slope of the line is

m = tan 60^o = sqrt 3

### Angle between two lines

\color{green} ✍️ Here we will discuss the angle between two lines in terms of their slopes.

Let L_1 and L_2 be two non-vertical lines with slopes m_1 and m_2, respectively. If α_1 and α_2 are the inclinations of lines L_1 and L_2, respectively. Then

color(blue)(m_1 = tan α_1) and color(blue)(m_2 = tan α_2) .

We know that when two lines intersect each other, they make two pairs of vertically opposite angles such that sum of any two adjacent angles is 180°.

Let θ and φ be the adjacent angles between the lines L_1 and L_2 (Fig10.6). Then

 color(blue)(θ = α_2 – α_1) and α_1, α_2 ≠ 90°

Therefore color(red)(tan θ = tan (α_2 – α_1) = (tan α_2 - tan α_1 )/(1+ tan α_1 tan α_2) = (m_2 - m_1)/(1+ m_1 m_2)) (as 1 + m_1 m_2 ≠ 0)

and color(blue)(φ = 180° – θ) so that

color(red)(tan φ = tan (180^o - θ ) = - tan θ = -(m_2 - m_1 )/( 1+ m_1 m_2))  , as 1 + m_1 m_2 ≠ 0

Now, there arise two cases:

color(red)(text(Case I)) : If  color(green)((m_2 - m_1 )/( 1+ m_1 m_2)) is color(blue)("positive"), then tan θ will be positive and tan φ  will be negative, which means θ will be acute and φ will be obtuse.

color(red)(text(Case II)) : If  color(green)((m_2 - m_1 )/(1+ m_1 m_2)) is color(blue)("negative"), then tan θ will be negative and tan φ will be positive, which means tha t θ will be obtuse and φ will be acute.

Thus, the acute angle (say θ) between lines L_1 and L_2 with slopes m_1 and m_2, respectively, is given by

color(red)(tan theta = | (m_2- m_1 )/(1+ m_1 m_2) |) , as 1 + m_1 m_2 ≠ 0 .... (1)

The obtuse angle (say φ) can be found by using φ =180^o – θ.
Q 3116245170

If the angle between two lines is  pi/4 and slope of one of the lines is 1/2 , find the slope of the other line.

Solution:

We know that the acute angle θ between two lines with slopes m_1 and m_2

is given by  tan theta = | (m_2-m_1)/( 1+ m_1 m_2) | .............(1)

Let m_1 = 1/2 , m_2 = m  and theta = pi/4

Now, putting these values in (1), we get

tan pi/4 = | ( m-1/2 )/(1+1/2 m) |  or  1 = | (m-1/2)/( 1+1/2 m ) | ,

which gives  (m-1/2)/(1+1/2 m ) =1 or  - ( m -1/2)/(1+1/2 m) = -1 .

Therefore  m =3 or  m = -1/3

Hence, slope of the other line is

3  or -1/3 Fig 10.7 explains the
Q 3166245175

Line through the points (–2, 6) and (4, 8) is perpendicular to the line
through the points (8, 12) and (x, 24). Find the value of x.

Solution:

Slope of the line through the points (– 2, 6) and (4, 8) is

 m_1 = (8-6)/(4- (-2) ) = 2/6 = 1/3

Slope of the line through the points (8, 12) and (x, 24) is

m_2= (24-12)/(x-8) = 12/(x-8)

Since two lines are perpendicular,

m_1 m_2 = –1, which gives

1/3 xx 12/(x-8) = -1 or x =4

### Collinearity of three points

\color{green} ✍️ We know that slopes of two parallel lines are equal. If two lines having the same slope pass through a common point, then two lines will coincide.

\color{green} ✍️ Hence, if A, B and C are three points in the XY-plane, then they will lie on a line, i.e.,

color(blue)("three points are collinear") (Fig 10.8)

if and only if color(red)"slope of AB = slope of BC."
Q 3186245177

Three points P (h, k), Q (x_1, y_1) and R (x_2, y_2) lie on a line. Show that
(h – x_1) (y_2 – y_1) = (k – y_1) (x_2 – x_1) .

Solution:

Since points P, Q and R are collinear, we have

Slope of PQ = Slope of QR, i.e.,  (y_1 k)/(x_1 - h) = (y_2 - y_1 )/(x_2 - x_1)

or  (k-y_1)/(h-x_1) = (y_2 - y_1)/( x_2 - x_1) ,

or  (h-x_1) (y_2 - y_1) = (k- y_1) (x_2 -x_1)
Q 3176345276

In Fig 10.9, time and distance graph of a linear motion is given.
Two positions of time and distance are recorded as, when T = 0, D = 2 and when
T = 3, D = 8. Using the concept of slope, find law of motion, i.e., how distance
depends upon time.

Solution:

Let (T, D) be any point on the line, where D denotes the distance at time
T. Therefore, points (0, 2), (3, 8) and (T, D) are collinear so that

 (8-2)/(3-0) = (D-8)/(T-3) or  6 (T-3) = 3 (D-8)

or D = 2 (T+1)

which is the required relation.