A straight line is a line that is not curved or bent. All the basic and advanced concepts related to straight lines are covered here on this page. This lesson can also be downloaded as a PDF which helps students to refer to the concepts in offline mode.
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What Is a Straight Line?
A line is a geometry object characterised 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 to infinity and has no curves is called a straight line.
Download this lesson as PDF:-Straight Lines PDF
Equation of a Straight Line
The general equation of a straight line is given below:
ax + by + c = 0Β
Where x and y are variables, a, b, and c are constants.
Slope:-
The equation of a straight line in slope-intercept form is given by:
y = mx + c
Here, m denotes the slope of the line, and c is the y-intercept.
When the angle with + ve x-axis βtan ΞΈβ is called the slope of a straight line.
Note 1 β If the line is horizontal, then slope = 0
Note 2 β If the line is perpendicular to the x-axis, i.e., vertical, then the slope is undefined.
Slope = 1/0 = β = tan β‘Ο/2
Note 3 β If the line is passing through any of two points, then the slope is
Intercept Form
The equation of the line with x-intercept as βaβ and y-intercept as βbβ can be written as;Β
- x β coordinate of the point of intersection of the line with the x-axis is called the x-intercept
- y-intercept will be the y-coordinate of the point of intersection of the line with the y-axis
For example,
Along the x-axis: x β Intercept = 5 and y β Intercept = 0
Along y-axis: y β Intercept = 5 and x β Intercept = 0
Also,
Length of x-intercept = |x_{1}|
Length of y-intercept = |y_{1} |
Note: Line passes through the origin, intercept = 0
x β Intercept = 0
y β Intercept = 0
Again,
ON = P
β AON = Ξ±
Let the length of the perpendicular from origin to straight line be βPβ and let this perpendicular make an angle with + vex- axis βΞ±β, then the equation of a line can be:
Learn More: Different Forms of the Equation of a Line
Point form
Equation of line with slope βmβ and which passes through (x_{1}, y_{1}) can be given as
y – y_{1Β }= m(x – x_{1 })
Slope Point form (Equation of a Line with 2 Points)
Equation of a line passing through two points (x_{1}, y_{1}) & (x_{2}, y_{2}) is given as
Example: Find the equation of the line that passes through the points (-2, 4) and (1, 2).
Solution:
We know that the general equation of a line passing through two points is:
Now,
(y_{2} – y_{1})/(x_{2} – x_{1}) = (2 – 4)/(1 – (-2)) = -2/3
Thus, the equation of the line is:
y – 4 = (-2/3)[x – (-2)]
3(y – 4) = -2(x + 2)
3y – 12 = -2x – 4
2x + 3y – 8 = 0
Which is the required equation of the line.
Relation between Two Lines
Let L_{1} and L_{2} be the two lines as
L_{1}Β : a_{1}x + b_{1}y + c_{1} = 0
L_{2 }Β : a_{2}x + b_{2}y + c_{2} = 0
- For parallel lines
Two lines are said to be parallel if the below condition is satisfied,
- For intersecting lines
Two lines intersect at a point if
- For coincident lines
Two lines coincide if
Angle between Straight Lines
Special Cases:
Length of Perpendicular from a Point on a Line
The length of the perpendicular from P(x_{1}, y_{1}) on ax + by + c = 0 is
B (x, y) is the foot of perpendicular is given by
Aβ(h, k) is mirror image, given by
Angular Bisector of Straight lines
An angle bisector has an equal perpendicular distance from the two given lines.
The equation of line L can be given as
Family of Lines:
The general equation of the family of lines through the point of intersection of two given lines, L_{1} & L_{2,} is given by L_{1} +Ξ» L_{2} = 0
Where Ξ» is a parameter.
Concurrency of Three Lines
Let the lines be
So, the condition for the concurrency of lines is
Pair of Straight Lines
Join equation of lines L1 & L2 represents P. S. L (a_{1}x + b_{1}y+c_{1}) (a_{2}x+b_{2}y+c_{2}) = 0
i.e. f(x,y) . g(x,y) = 0
Let’s defines a standard form of the equation:-
ax^{2 }+ by^{2 }+ 2hxy + 2gx + 2fy + c = 0 represent conics curve equation
Condition for curve of being P.O.S.L Ξ = abc + 2fgh β af^{2} β bg^{2} β ch^{2} = 0
If Ξ β 0, (i) parabola h^{2} = ab
(ii) hyperbola h^{2} < ab
(iii) circle h^{2} = 0, a = b
(iv) ellipse h^{2} > ab
Now, let’s see how did we get Ξ = 0
General equation ax^{2} + 2gx + 2hxy + by^{2} + 2fy + c = 0
ax^{2} + (2g+2hy)x + (by^{2} + 2fy + c) = 0
we can consider the above equation as a quadratic equation in x, keeping y constant.
Now, Q(y) has to be a perfect square, and only then, we can get two different line equations Q(y) in the perfect square for that Ξ value of Q(y) should be zero.
From there D = 0
abc + 2fgh β bg^{2} β af^{2} β ch^{2} = 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 (L_{1}).(L_{2}) = 0 or f(x, y) . g(u,y) = 0
2. Angle between the lines
Special cases:
h^{2} = ab β lines are either parallel or coincident
h^{2} < ab β imaginary line
h^{2} > ab β Two distinct lines
a + b = 0 β perpendicular line
3. P.O.S.L passing through the origin, then
β (y β m_{1}x) (y β m_{2}x) = 0
y^{2} β m_{2}yx β m_{1}xy β m_{1}m_{2}x^{2} = 0
y^{2} β (m_{1} + m_{2}) xy β m_{1}m_{2}x^{2} = 0
β ax^{2} + 2hxy + by^{2} = 0
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 – y_{1}) = m(x – x_{1})
Here, m = (y_{2} – y_{1})/(x_{2} – x_{1}) |
Angle between Straight Lines | Β \(\begin{array}{l}\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right) \right|\end{array} \) |
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 ΞΈ with the positive direction of the x-axis where cos ΞΈ = -1/3.
Solution:
Here cos ΞΈ = -1/3. (a negative number) so that Ο/2 < ΞΈ < Ο
We know that the equation of the straight line passing through the point (x_{1}, y_{1}) having slope m is
y β y_{1} = m(x β x_{1})
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 (x_{1}, y_{1}) and (x_{2}, y_{2}) is
Hence, the 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 β 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 Ξ± > tan Ξ². So line (i) has more inclination.
Question 5:
The angle of the line positive direction of the x-axis is ΞΈ. The line is rotated about some point on it in an anticlockwise direction by an angle of 45Β°, and its slope becomes 3. Find the angle ΞΈ.
Solution:
Originally slope of the line is tan ΞΈ = m
Now the slope of the line after rotation is 3.
The angle between the old position and the new position of the lines is 45Β°.
β΄ we have
1 + 3m = 3 β m
4m = 2
m = 1/2 = tan ΞΈ
ΞΈ = tan^{-1}(1/2)
Question 6:
If line 3x – ay – 1 = 0 is parallel to the line (a + 2)x – y + 3 = 0, then find the values of a.
Solution:
Slope of line 3x – ay – 1 = 0 is 3/a.
Slope of line (a + 2)x – y + 3 = 0 is (a + 2).
Since lines are parallel then we have;
a + 2 = 3/a
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
Slope of AB = Slope of BC
Question 8:
The slope of a line is double the slope of another line. If the tangent of the angle between them is 1/3. Find the slopes of the lines.
Solution:
Let m_{1} and m be the slopes of the two given lines such that m_{1} = 2m
We know that if ΞΈ is the angle between the lines l_{1} and l_{2} with slopes m_{1} and m_{2}, then
It is given that the tangent of the angle between the two lines is 1/3.
Question 9:
Find the equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Solution:
Line parallel to the line 3x – 4y + 2 = 0 is 3x – 4y + t = 0
It passes through the point (-2, 3), so 3(-2) β 4(3) + t = 0 or t = 18.
So the equation of the line is 3x – 4y + 18 = 0.
Question 10:
Find the coordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:
Let (a, b) be the coordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Slope of the line joining (-1, 3) and (a, b)
Slope of the line 3x – 4y – 16 = 0 is 3/4.
Since these two lines are perpendicular, m_{1}m_{2} = -1
β 4a + 3b = 5 ….(1)
Point (a, b) lies on line 3x – 4y = 16.
β 3a – 4b = 16 ….(2)
On solving equations (1) and (2), we obtain
Thus, the required coordinates of the foot of the perpendicular are (68/25, -49/25).
Question 11:
Three lines x + 2y + 3 = 0, x + 2y – 7 = 0, and 2x – y – 4 = 0 form 3 sides of two squares.
Find the equations of the remaining sides of these squares.
Solution:
Distance between the two parallel lines is
The equations of the sides forming the square are of the form 2x – y + k = 0.
Since the distance between sides A and B = Distance between sides B and C
Hence, the fourth side of the two squares is
(i) 2x – y + 6 = 0
or
(ii) 2x – y – 14 = 0
Question 12:
For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, find the equation of the
(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
β 9x – 7y – 41 = 0 and 7x + 9y – 3 = 0
If ΞΈ is the angle between the line 4x + 3y – 6 = 0 and the bisector 9x – 7y – 41 = 0, then
Hence
(i) The bisector of the obtuse angle is 9x – 7y – 41 = 0.
(ii) The bisector of the acute angle is 7x + 9y – 3 = 0.
(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 Ξ» if 2x^{2} + 7xy + 3y^{2} + 8x + 14y + Ξ» = 0 will represent a pair of straight lines.
Solution:
The given equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of lines if its descriminantΒ = 0.
i.e., if abc + 2fgh β af^{2} β bg^{2} β ch^{2} = 0
Question 14:
If one of the lines of the pair ax^{2} + 2hxy + by^{2} = 0 bisects the angle between the positive direction of the axes, then find the relation for a, b and h.
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 ax^{2} + 2hxy + by^{2} = 0.
Let’s apply y = x.
We have
Question 15:
If the angle between the two lines is represented by 2x^{2} + 5xy + 3y^{2} + 6x + 7y + 4 = 0 is tan^{-1}(m), then find the value of m.
Solution:
The angle between the lines 2x^{2} + 5xy + 3y^{2} + 6x + 7y + 4 = 0 is given by
Question 16:
The pair of lines β3 x^{2} – 4xy + β3 y^{2} = 0 are rotated about the origin by Ο/6 in the anticlockwise sense. Find the equation of the pair in the new position.
Solution:
The given equation of pair of straight lines can be rewritten as
or y = tan 60Β° x and y = tan 30Β° x
After rotation, the separate equations are
y = tan 90Β° x and y = tan 60Β° x
or x = 0 and y = β3 x
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 ΞΈ.
What is the equation of the straight line passing through (x_{1}, y_{1}) with slope m?
The equation of the straight line passing through (x_{1}, y_{1}) with slope m is given by y β y_{1} = m(x β x_{1} ).
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