## 1. The slope of line and angles between two lines

Chapter 10

Straight Lines

The slope of line and angles between two lines:

A line is sometimes called a straight line or, more archaically, a right line (Casey 1893), to emphasize that it has no "wiggles" anywhere along its length. While lines are intrinsically one-dimensional objects, they may be embedded in higher dimensional spaces. Harary (1994) called an edge of a graph a "line."

A straight line is just a line with no curves. So, a line that extends to both sides till infinity and has no curves is called a straight line.

The two properties of straight lines in Euclidean geometry are that they have only one dimension, length, and they extend in two directions forever.

Note: 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 of a Line:

A line in a coordinate plane forms two angles with the x-axis, which are supplementary.

The angle (say) q made by the line l with positive direction of x-axis and measured anti clockwise is called the inclination of the line. Obviously £ q £ 180°

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°.

Definition: If q is the inclination of a line l, then tan q is called the slope or gradient of

the line l. The slope of a line whose inclination is 90° is not defined.

The slope of a line is denoted by m.

Thus, m = tan q, q ¹ 90°

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:

Let P(x1, y1) and Q(x2, y2) be two points on non-vertical line l whose inclination is q.

Case I When angle q is acute:

ÐMPQ = q. ... (1)

Therefore, slope of line l = m = tan q.

Case II When angle q is obtuse:

we have ÐMPQ = 180° – q.

Therefore, q = 180° – ÐMPQ.

Now, slope of the line l

m = tan q

= tan ( 180° – ÐMPQ) = – tan ÐMPQ

Example : Find the slope of the lines:

(a) Passing through the points (3, – 2) and (–1, 4),

(c) Passing through the points (3, – 2) and (3, 4),

(c) Making inclination of 30° 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 (3, 4) is

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

(c) Here inclination of the line a = 60°. Therefore, slope of the line is

m = tan 30° = 1 / √3

## 2. Conditions for parallelism and perpendicularity of lines and Collinearity of three points

Conditions for parallelism and perpendicularity of lines and Collinearity of three points:

In a coordinate plane, suppose that non-vertical lines l1 and l2 have slopes m1

and m2, respectively. Let their inclinations be a and b, respectively.

If the line l1 is parallel to l2 (Fig 10.4), then their

inclinations are equal, i.e.,a = b, and hence, tan a = tan b

Therefore m1 = m2, i.e., their slopes are equal.

Conversely, if the slope of two lines l1 and l2 is same,

i.e., m1 = m2.

Then   tan a = tan b.

By the property of tangent function (between 0° and 180°), a = b.

Therefore, the lines are parallel.

Hence, two non vertical lines l1 and l2 are parallel if and only if their slopes are equal.

If the lines l1 and l2 are perpendicular , then b = a + 90°.

Therefore, tan b = tan (a + 90°)

= – cot a = -1/ tan a

i.e., m1= -1/m2   or m1 m2 = -1

Conversely, if m1 m2 = – 1, i.e., tan a tan b = – 1.

Then tan a = – cot b = tan (b + 90°) or tan (b – 90°)

Therefore, a and b differ by 90°.

Thus, lines l1 and l2 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,

Example:  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 = (8 – 6) /(4  – (-2)) = 2 / 6 = 1/ 3

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

m = (24 – 12) /(x – 8) = 12 / (x-8)

Since two lines are perpendicular,

m1 m2 = –1, which gives

• [12 / (x-8)] . [1 / 3] = - 1
• 4 / ( x – 8 ) = - 1
•  x – 8 = - 4
• x = 4

Angle between two lines

Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. If a1 and a2 are the inclinations of lines L1 and L2, respectively. Then m1= tan a1 and m2=tana2

Let q and f be the adjacent angles between the lines L1 and L2 . Then q = α2 - a1  and a1, a2 ¹ 90o

Therefore tan q = tan (a2 – a1) =[ tana2 – tana1] / [1+ tana2 –tana1] = [m2-m1] / [1+ m1m2]

(as 1 + m1m2 ¹ 0)

and  f = 180° – q so that

tan f = tan (180° – q ) = – tan q = - [m2-m1] / [1+ m1m2] (as 1 + m1m2 ¹ 0)

Now, there arise two cases:

which means that q will be obtuse and f will be acute.

Thus, the acute angle (say q) between lines L1 and L2 with slopes m1 and m2,

respectively, is given by

……………………(1)

The obtuse angle (say f) can be found by using  f =1800q.

Example: If P (-2, 1), Q (2, 3) and R (-2, -4) are three points, find the angle between the straight lines PQ and QR.

The slope of PQ is given by

m = ( y2 – y1 ) / (x2 – x1)

m =( 3 – 1 ) / (2 – (-2 ))

m= 2/4

Therefore, m1=1/2

The slope of QR is given by

m= (−4−3) / (−2−2)

m= 7/4

Therefore, m2 = 7/4

Substituting the values of m2 and m1 in the formula for the angle between two lines when we know the slopes of two sides, we have,

tan θ=± (m2 – m1 ) / (1+m1m2)

tan θ=± ((7/4) – (1/2) ) / (1+ (1/2)(7/4))

tan θ=± (2/3)

Therefore,  θ = tan -1 (â…”)

Co linearity of three points

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. Hence, if A, B and C are three points in the XY-plane, then they will lie on a line, i.e., three points are collinear  if and only if

slope of AB = slope of BC.

Example: Three points P (h, k), Q (x1, y1) and R (x2, y2) lie on a line. Show that

(h x1) (y2 – y1) = (k y1) (x2 – x1).

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

Slope of PQ = Slope of QR,

or   (h x1) (y2 – y1) = (k y1) (x2 – x1).

## 3. Various Forms of the Equation of a Line

Various Forms of the Equation of a Line:

Different forms of equations of a straight line

• Point-slope form equation of line. Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and. ...
• Two-point form equation of line. ...
• Slope-intercept form equation of line. ...
• Intercept form. ...
• Normal form.

Horizontal and vertical lines:

If a horizontal line L is at a distance a from thex-axis then ordinate of every point lying on the line is either a or – a .

Therefore, equation of the line L is either y = a or y = a. Choice of sign will depend upon the position of the line according as the line is above or below the y-axis.

Similarly,

the equation of a vertical line at a distance b from the y-axis is either x = b or x = – b

Example 6 Find the equations of the lines parallel to axes and passing through (– 2, 3).

Solution Position of the lines is shown in the Figure. The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through

(– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3)

is x = – 2.

Point-slope form

Suppose that P0 (x0, y0) is a fixed point on a non-vertical line L, whose slope is m. Let P (x, y) be an arbitrary point on L  .

Then, by the definition, the slope of L is given by

is the equation of line.

Example : Find the equation of the line through (– 2, 3) with slope – 4.

Solution: Here m = – 4 and given point (x0 , y0) is (– 2, 3).

By slope-intercept form above formula , equation of the given line is

y – 3 = – 4 (x + 2) or

4x + y + 5 = 0, which is the required equation.

Two-point form

Let the line L passes through two given points P1 (x1, y1) and P2 (x2, y2).

Let P (x, y) be a general point on L . The three points P1, P2 and P are collinear, therefore, we have slope of P1P = slope of P1P2

is the equation of the line passing through the points (x1, y1) and (x2, y2).

Example: Write the equation of the line through the points (1, –1) and (3, 5).

Solution : Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form  above formula

for the equation of the line, we have

• –3x + y + 4 = 0, which is the required equation.

Slope-intercept form Sometimes a line is known to us with its slope and an

intercept on one of the axes. We will now find equations of such lines.

Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin

. The distance c is called the y-intercept of the line L. Obviously, coordinates of the point where the line meet the y-axis are (0, c). Thus, L has slope m and passes through a fixed point (0, c).

Therefore, by point-slope form, the equation of L is

y - c = m( x - 0 ) or y = mx + c

Thus, the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if

y = mx +c ......................(3)

Note that the value of c will be positive or negative according as the intercept is made

on the positive or negative side of the y-axis, respectively.

Case II Suppose line L with slope m makes x-intercept d. Then equation of L is

y = m(x – d) …………………... (4)

Students may derive this equation themselves by the same method as in Case I.

Intercept – form

Suppose a line L makes x-intercept a and y-intercept b on theaxes. Obviously L meets x-axis at the point (a, 0) and y-axis at the point (0, b) . By two-point form of the equation of the line, we have,

y – 0 =   b-00-a( x – a)

• ay = – bx + ab
• xa  + yb  = 1

is equation of the line making intercepts a and b on x-and y-axis, respectively

Normal form

Suppose a non-vertical line is known to us with following data:

(i) Length of the perpendicular (normal) from origin to the line.

(ii) Angle which normal makes with the positive direction of x-axis.

Let L be the line, whose perpendicular distance from origin O be OA = p and the

angle between the positive x-axis and OA be ÐXOA = w.

Draw perpendicular AM on the x-axis in each case.

In each case, we have OM = p cos w and MA = p sin w, so that the coordinates of the

point A are (p cos w, p sin w).

Further, line L is perpendicular to OA. Therefore

The slope of the line L

the equation of the line L at point A(pcos w, psin w ) is

x cos w + y sin w = p.

Hence, the equation of the line having normal distance p from the origin and angle w

## 4. General Equation of a Line

General Equation of a Line

General equation of a line is Ax + By + C = 0. The inclination of angle θ to a line with a positive direction of X-axis in the anti-clockwise direction, the tangent of angle θ is said to be slope or gradient of the line and is denoted by m

General equation of first degree in two variables,

Ax + By + C = 0, where A, B and C are real constants such that A and B are not zero

simultaneously. Graph of the equation Ax + By + C = 0 is always a straight line.

Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero

simultaneously is called general linear equation or general equation of a line.

Different forms of Ax + By + C = 0 The general equation of a line can be

reduced into various forms of the equation of a line.

If B = 0, then x = - C/A which is a vertical line whose slope is undefined and x-intercept  is  -C/A.

If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0, which is a line passing through the origin and, therefore, has zero intercepts on the axes.

(Q1) The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.
Solution:

The given equation 2x – 6y + 3 = 0 can be represented in slope-intercept form as:

y = x/3 + 1/2

Comparing it with y = mx + c,
Slope of the line, m = 1/3

Also, the above equation can be re-framed in intercept form as;

x/a + y/b = 1

2x – 6y = -3

x/(-3/2) – y/(-1/2) = 1

Thus, x-intercept is given as a = -3/2 and y-intercept as b = 1/2.

(Q2) The equation of a line is given by, 13x – y + 12 = 0. Find the slope and both the intercepts.

Solution: The given equation 13x – y + 12 = 0 can be represented in slope-intercept form as:

y = 13x + 12
Comparing it with y = mx + c,
Slope of the line, m = 13

Also, the above equation can be re-framed in intercept form as;

x/a + y/b = 1

13x – y = -12

x/(-12/13) + y/12 = 0

Thus, x-intercept is given as a = -12/13 and y-intercept as b = 12.

Example: Show that two lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0, where b1,b2 ¹ 0 are:

## 5. Distance of a Point From a Line and Distance between two parallel lines

Distance of a Point from a Line and Distance between two parallel lines

The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1, y1) is d. Draw a perpendicular PM from the point P to the line L

Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) is given by

Distance between two parallel lines

We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form

y = mx + c1 ... (1)

and

y = mx + c2 ... (2)

Line (1) will intersect x-axis at the point A(-c/m,0)

Distance between two lines is equal to the length of the perpendicular from point A to line (2). Therefore, distance between the lines (1) and (2) is

If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0,

then above formula will take the form

Example:

Example: If the lines 2x + y - 3 = 0, 5x + ky - 3 = 0 and 3x - y - 2 = 0 are concurrent, find the value of k.

Solution Three lines are said to be concurrent, if they pass through a common point,

i.e., point of intersection of any two lines lies on the third line. Here given lines are

2x + y – 3 = 0 ... (1)

5x + ky – 3 = 0 ... (2)

3x y – 2 = 0 ... (3)

Solving (1) and (3) by cross-multiplication method, we get

or  x= 1, y= 1

Therefore, the point of intersection of two lines is (1, 1). Since above three lines are

concurrent, the point (1, 1) will satisfy equation (2) so that

5.1 + k .1 – 3 = 0 or k = – 2.