A straight line is a line that is not curved or bent. In this lesson, all the basic and advanced concepts related to straight lines are covered here. This lesson can also be downloaded as a PDF which helps students to refer to the concepts in offline mode.
Table of Contents for Straight Lines
- Definition
- Equation
- Intercept Form
- Point Form
- Slope Point Form
- Relation Between Two Lines
- Angle Between Two Straight Lines
- Length of Perpendicular from a Point on a Line
- Angle Bisector
- Family of Lines
- Concurrency of Three Lines
- Pair of Straight Lines
- Formulas
- Questions
What is a Straight Line?
A line is simply an object in geometry that is characterized under zero width object that extends on both sides. 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.
Download this lesson as PDF:-Straight Lines PDF
Equation of Straight Line
The relation between variables x, y satisfy all points on the curve.
The general equation of straight line is as given below:
ax + by + c = 0 { equation of straight line.
Where x, y are variables and a,b, c are constants.
Slope:-
angle with + ve x-axis
‘tan θ’ is called slope of straight line.
Note 1 – If line is Horizontal, then slope = 0
Note 2 – If line is ⊥ to x-axis, i.e. vertical then slope is undefined.
Slope =
=
Note 3 – If the line is passing through any of two points, then the slope is
Intercept Form
Equation of line with x – intercept as ‘a’ and y – intercept as ‘b’ can be given as
x – co-ordinate of point of intersection of line with x-axis is called x-intercept.
y – intercept will be y-co-ordinate of point of intersection of line with y-axis.
For example,
Along x-axis: x – Intercept = 5 and y – Intercept = 0
Along y-axis: y – Intercept = 5 and x – Intercept = 0
Also,
Length of x – intercept = |x1|
Length of y – intercept = |y1 |
Note: Line passes through the origin, intercept = 0
x – Intercept = 0
y – Intercept = 0
Again,
ON = P
AON = α
Let length of
Learn More: Different Forms Of The Equation Of Line
Point form
Equation of line with slope ‘m’ and which passes through (x1, y1) can be given as
y – y1 = m(x – x1 )
Slope Point form (Equation of a Line with 2 Points)
Equation of a line passing through two points (x1, y1) & (x2, y2) is given as
For example,
Example: Find the equation of the line that passes through the points (-2, 4) and (1, 2).
Solution:
We know general equation of a line passing through two points is:
y = mx + b, where m =
Here m = (2-4)/(1-(-2)) = -2/3
We can find the equation (by solving first for “b”) if we have a point and the slope. So we need to choose one of the points and use it to solve for b. Using the point (–2, 4), we get:
4 = (– 2/3)(–2) + b
4 = 4/3 + b
4 – 4/3 = b
b = 8/3
so, y = ( – 2/3 ) x + 8/3.
Which is the required equation of the line.
Relation between two Lines
Let L1 and L2 be the two lines as
L1 : a1x + b1y + c1 = 0
L2 : a2x + b2y + c2 = 0
- For Parallel lines
Two lines are said to be parallel if the below condition is satisfied,
- For Intersection of lines
Two lines intersect a point if
- For Coincident Lines
Two lines intersect a point if
Angle between Straight lines
Angle =
Special Cases:
Length of Perpendicular from a Point on a Line
The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is
B (x, y) is foot of perpendicular is given by
A’(h, k) is mirror image, given by
Angular Bisector of Straight lines
An angle bisector has equal perpendicular distance from the two given lines. To find the equation of the bisector of the angle between lines, we have
The equation of line L can be given
Family of Lines:
The general equation of the family of lines through the point of intersection of two given lines L1 & L2 is given by L1 +λ L2 = 0
Where λ is a parameter.
Concurrency of Three Lines
Let the lines be
So, the condition for concurrency of lines is
Pair of Straight Lines
Join equation of line L1 & L2 represents P. S. L (a1x + b1y+c1) (a2x+b2y+c2) = 0
i.e. f(x,y) . g(x,y) = 0
Let defines a standard form of equation:-
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represent conics curve equation
Condition for curve of being P.O.S.L Δ = abc + 2fgh – af2 – bg2 – ch2 = 0
If Δ ≠ 0, (i) parabola h2 = ab
(ii) hyperbola h2 < ab
(iii) circle h2 = 0, a = b
(iv) ellipse h2 > ab
Now, lets see how did we get Δ = 0
General equation ax2 + 2gx + 2hxy + by2 + 2fy + c = 0
ax2 + (2g+2hy)x + (by2 + 2fy + c) = 0
we can consider above equation as quadratic equation in x keeping y as constant.
Now, Q(y) has to be perfect square then only we can get two different line equation Q(y) in perfect square for that Δ value of Q(y) should be zero.
From there D = 0
abc + 2fgh – bg2 – af2 – ch2 = 0
Or
Note:
1. Point of intersection
To find point of intersection of two lines (P.O.S.L), solve the P.O.S.L, factorize it in (L1).(L2) = 0 or f(x, y) . g(u,y) = 0
2. Angle between the lines:
Special cases:
h2 = ab
h2 < ab
h2 > ab
a + b = 0
3. P.O.S.L passing through origin, then
y2 – m2yx – m1xy – m1m2x2 = 0
y2 – (m1 + m2) xy – m1m2x2 = 0
m1 m2 =
Straight Lines Formulas
All Formulas Related to Straight Lines | |
---|---|
Equation of a Straight Line | ax + by + c = 0 |
General form or Standard Form | y = mx + c |
Equation of a Line with 2 Points (Slope Point Form) | (y – y1) = m(x – x1) |
Angle Between Straight lines | \(\begin{array}{l}\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right) \right|\end{array} \) |
Problems on Straight Lines
Question 1:
Find the equation to the straight line which passes through the point (-5, 4) and is such that the portion of it between the axes is divided by the given point in the ratio 1 : 2.
Solution:
Let the required straight line be
Using the given conditions,
But P is (-5, 4)
Hence -5 = 2a/3, 4 = b/3 a = -15/2, b = 12.
Hence the required equation is
Question 2:
Find the equation of the straight line which passes through the point (1, 2) and makes an angle θ with the positive direction of the x-axis where
Solution:
Here
We know that the equation of the straight line passing through the point (x1, y1) having slope m is
y – y1 = m(x – x1)
Therefore the equation of the required line is
Question 3:
Find the equation of the line joining the points (-1, 3) and (4, -2).
Solution:
Equation of the line passing through the points (x1, y1) and (x2, y2) is
Hence equation of the required line will be
Question 4:
Which line is having greatest inclination with positive direction of x-axis?
(i) line joining points (1, 3) and (4, 7)
(ii) line 3x – 4y + 3 = 0
Solution:
(i) Slope of line joining points A(1, 3) and B(4, 7) is
(ii) Slope of line is
Now tan α > tan β. So line (i) has more inclination.
Question 5:
Angle of line positive direction of x-axis is θ. Line is rotated about some point on it in anticlockwise direction by angle 45° and its slope becomes 3. Find the angle θ.
Solution:
Originally slope of line is tan θ = m
Now slope of line after rotation is 3.
Angle between old position and new position of lines is 45°.
∴ we have
1 + 3m = 3 – m
4m = 2
m = 1/2 = tan θ
θ = tan-1(1/2)
Question 6:
If line
Solution:
Slope of line
Slope of line
Since lines are parallel then we have
or
or
or a = 1 or a = -3.
Question 7:
Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear.
Solution:
If points A(x, -1) B(2, 1), and C(4, 5) are collinear, then
Question 8:
The slope of a line is double of the slope of another line. It tangent of the angle between them is
Solution:
Let m1 and m be the slopes of the two given lines such that m1 = 2m
We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then
It is given that the tangent of the angle between the two lines is
∴
Other slope will be -2, -1, 2, 1.
Question 9:
Find equation of the line parallel to the line
Solution:
Line parallel to the line
It passes through the point (-2, 3), so 3(-2) – 4(3) + t = 0 or t = 18.
So equation of line is
Question 10:
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line
Solution:
Let (a, b) be the coordinates of the foot of the perpendicular from the point (-1, 3) to the line
Slope of the line joining (-1, 3) and (a, b)
Slope of the line
Since these two lines are perpendicular, m1m2 = -1
∴
Point (a, b) lies on line
∴
On solving equations (1) and (2), we obtain
Thus, the required coordinates of the foot of the perpendicular are
Question 11:
Three lines
Solution:
Distance between the two parallel lines is
Since the distance between sides A and B = distance between sides B and C,
Hence the fourth side of the two squares is
(i)
Question 12:
For the straight lines
(i) bisector of the obtuse angle between them,
(ii) bisector of the acute angle between them,
(iii) bisector of the angle which contains (1, 2).
Solution:
Equations of bisectors of the angles between the given lines are
If θ is the angle between the line
Hence
(i) The bisector of the obtuse angle is
(ii) The bisector of the acute angle is
(iii) For the point (1, 2)
Hence equation of the bisector of the angle containing the point (1, 2) is
Question 13:
Find the value of λ if
Solution:
The given equation
If abc + 2fgh – af2 – bg2 – ch2 = 0 i.e., if
Question 14:
If one of the lines of the pair
Solution:
Bisector of the angle between the positive directions of the axes is y = x.
Since it is one of the lines of the given pair of lines
We have
Question 15:
If the angle between the two lines represented by
Solution:
The angle between the lines
Question 16:
The pair of lines
Solution:
The given equation of pair of straight lines can be rewritten as
After rotation, the separate equations are
y = tan 90° x and y = tan 60° x
or x = 0 and
The combined equation in the new position is
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Frequently Asked Questions
Give the general equation of a straight line.
The general equation of a straight line is ax + by + c = 0. (x, and y are variables and a, b, c are constants.)
What do you mean by the slope of a straight line?
The angle formed by a line with a positive x-axis is the slope of a line. It is denoted by tan θ.
Give the equation of the straight line passing through (x1, y1) with slope m.
Equation of the straight line passing through (x1, y1) with slope m is given by y – y1 = m(x – x1 ).