Straight-line depreciation is the simplest method for calculating depreciation over time. Under this method, the same amount of depreciation is deducted from the value of an asset for every year of its useful life.

1. STRAIGHT
LINE
DEMONSTRATI
ON
By
Dharmendra Meena

2.  Straight line:- 1. A straight line is an endless one-dimensional figure that has no width. It is a
combination of endless points joined on both sides of a point. A straight line
does not have any curve in it. It can be horizontal, vertical, or slanted. If we
draw an angle between any two points on the straight line, we will always get
a 180-degree.

3. WHAT IS A STRAIGHT LINE?
A straight line is an infinite length line that does not have any curves on it. A straight
line can be formed between two points also but both the ends extend to infinity. A
straight line is a figure formed when two points A (x1, y1) and B (x2, y2) are connected
with the shortest distance between them, and the line ends are extended to infinity.
A B

4. TYPES OF STRAIGHT
LINES
 Straight lines can be of various types. Generally, the straight lines are classified based
on their alignment. Their alignment refers to the angle they form with the x-axis or the
y-axis. According to the alignment of straight lines, they are of the following types:
 Horizontal lines
 Vertical lines
 Oblique or Slanted lines

5. HORIZONTAL LINES
The lines which are drawn horizontally and are parallel
to the x-axis or perpendicular to the y-axis, are called
horizontal lines. They form a 0o or 180o angle with the x-
axis and a 90o or 270o angle with the y-axis.

6. VERTICAL LINES
The lines which are drawn vertically and are parallel to the y-axis, or
perpendicular to the x-axis, are called vertical lines. They form a 90o or 270o
angle with the x-axis and a 0o or 180o angle with the y-axis.

7. OBLIQUE OR SLANTED
LINES
The lines are drawn in a slanting position or form some angle other than 0o, 90o, 180o,
270o, 360o with the horizontal or vertical lines are called oblique or slanting lines.

8. EQUATION OF A STRAIGHT
LINE
An equation of a straight line is a linear equation. A straight line on a cartesian plane
can have different representations based on the known variables, angles, and constants.
The slope of a straight line determines the direction of a straight line and tells how
steep the line is. It is calculated as the difference in y coordinates/difference in x
coordinates, which is also called rise over run. An equation of a straight line is of
various forms.
They are as follows:
General Equation of a Straight Line
The general equation of a straight line can be given as ax + by + c = 0, where
a, b, c are constants, and
x, y are variables.
The slope is –a/b

9. SLOPE AND Y-INTERCEPT
FORM
A straight line having slope m = tanθ where θ is the angle formed by the line with the
positive x-axis, and y-intercept as b is given by: y = mx + b, where m is the slope.

10. SLOPE POINT FORM
A straight line having slope m = tanθ where θ is the angle formed by the line with
the positive x-axis, and passing through a point (x1, y1) is given by: Slope Point
Form as y – y1 = m(x – x1).

11. TWO POINT FORM
A straight line passing through points (x1 , y1) and (x2 , y2) is given by in the two point
form as: y – y1 = [(y2 – y1) / (x2 – x1)] (x – x1).

12. INTERCEPT FORM
A straight line having x-intercept as a and y-intercept as b as shown in the figure
below where point A is on the x-axis (vertical here) and point B is on the y-axis
(horizontal here), is given in the intercept form by x/a + y/b = 1

13. EQUATION OF LINES PARALLEL TO X-AXIS OR
Y-AXIS
 The equation of a line parallel to the x-axis is given by: y = ± a, where a is the distance
of the line from the x-axis. The value of a is + ve if it lies above the x-axis, and –ve if it
lies below the x-axis.
 The equation of a line parallel to the y-axis. Is given by: x = ± b, where b is the
distance of the line from the y-axis. The value of b is +ve if it lies on the right side of
the y-axis, and –ve if it lies on the left side of the y-axis.

14. SLOPE OF A LINE:-
If θ is the angle made by a line with positive direction of x-axis in anticlockwise
direction, then the value of tan θ is called the slope of the line and is denoted by m.
The slope of a line passing through points P(x1 , y1 ) and Q (x2 , y2 ) is given by
m=tanθ=(y2-y1)/(x2-x1)

15. TYPES OF SLOPE
The angle formed by a line with a positive x-axis is the slope of a line. Different lines forms
different angles with the x-axis. A line can have slopes varying from positive, negative, 0, or
even infinite slope.
Zero Slope:- If a line forms a 0o angle with the x-axis, the slope of the line is 0. The slope of a
line is represented by, m = tanθ
Here, θ = 0o . Hence m = tan0 = 0. Therefore, a line with the 0 slope is parallel to the x-axis.

16. POSITIVE SLOPE:-
If a line forms an angle that lies between 0o and 90o with the x-axis, the slope of the
line is positive.

17. NEGATIVE SLOPE:-
If a line forms an angle that lies between 90o and 180o with the x-axis, the slope of the
line is negative.

18. INFINITE SLOPE
If a line forms a 90o angle with the x-axis, or the line is parallel to the y-axis, the slope of
the line is not defined or infinite.
As we know, the slope of a line m = tan θ
Here, θ = 90o . Slope m = tan 90o is not defined. Therefore, the line with an infinite slope is
parallel to the y-axis.

19. ANGLE BETWEEN TWO
LINES
The angle θ between the two lines having slopes m1 and m2 is given by
tanθ=±(m1-m2)/1+m1m2
If we take the acute angle between two lines, then
tan θ = |(m1-m2)/1+m1m2|
If the lines are parallel, then m1 = m2 .
If the lines are perpendicular, then m1 m2 = – 1.

20. COLLINEARITY OF THREE
POINTS
If three points P (h, k), Q (x1 , y1 ) and R (x2 ,y2 ) are such that slope of PQ = slope
of QR, i.e.,
(Y2-y1)/(x2-x1)=(k-y1)/(h-x1)
or(h – x1 ) (y2 – y1 ) = (k – y1 ) (x2 – x1 ) then they are said to be collinear.

21. DISTANCE OF A POINT
FROM A LINE
The perpendicular distance (or simply distance) d of a point P (x1 , y1 ) from the line
Ax + By + C = 0 is given by
d=|Ax1+By1+C|/(A^2 +B^2)^1/2
Distance between two parallel lines:- The distance d between two parallel lines y =
mx + c1 and y = mx + c2 is given by
d=|c1-c2|/(1+m^2)^1/2

22. INTERSECTION OF TWO GIVEN
LINES
Two lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are
(i)intersecting if
a1/a2≠ b1/b2
(ii)Parallel and distinct if
(a1/a2 )=(b1/b2)≠(c1/c2)
(iii)Coincident if
(a1/a2)=(b1/b2)=(c1/c2)

23. REMARKS
(i)The points (x1 , y1 ) and (x2 , y2 ) are on the same side of the line or on the
opposite side of the line ax + by + c = 0, if ax1 + by1 + c and ax2 + by2 + c are of the
same sign or of opposite signs respectively.
(ii)The condition that the lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c = 0 are
perpendicular is a1 a2 + b1 b2 = 0.
(iii)The equation of any line through the point of intersection of two lines a1 x + b1 y
+ c1 = 0 and a2 x + b2 y + c2 = 0 is a1 x + b1 y + c1 + k (a2x + b2y + c2 ) = 0. The
value of k is determined from extra condition given in the problem.

24. SOME SOLVED EXAMPLES:-
Example 1 Find the equation of a line which passes through the point (2, 3) and makes
an angle of 30° with the positive direction of x-axis.
Solution Here the slope of the line is m = tanθ = tan 30° = 1/√3 and the given point is
(2, 3). Therefore, using point slope formula of the equation of a line, we have
y – 3 = (1/√ 3) (x – 2)
Or x – √3y + (3√3 – 2) = 0.

25. Example 2 Find the equation of the straight line passing through (1, 2) and
perpendicular to the line x + y + 7 = 0.
Solution
Let m be the slope of the line whose equation is to be found out which is perpendicular
to the line x + y + 7 = 0. The slope of the given line y = (– 1) x – 7 is – 1. Therefore, using
the condition of perpendicularity of lines, we have m × (– 1) = – 1 or m = 1
Hence, the required equation of the line is y – 2= (1) (x – 1)
or y – 2= x – 1
x - y + 1 = 0.

26. Example 3 Find the equation to the straight line passing through the point of
intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line
3x – 5y + 11 = 0.
Solution
First we find the point of intersection of lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 which
is (– 1, – 1).
Also the slope of the line 3x – 5y + 11 = 0 is 3/5 Therefore, the slope of the line
perpendicular to this line is -5/3 .
Hence, the equation of the required line is given by
y + 1 =(-5/3) (x + 1)
Or 5x + 3y + 8 =0

27. Thankyou