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Straight Lines Straight Lines is an extremely important head under co-ordinate geometry. It is neither a very simple topic nor extremely tough but requires consistent practice. It is a vast topic which includes several other heads. It is an important chapter in the Mathematics syllabus of IIT JEE. The laws of straight lines are not very new and they can be understood easily by relating them to day life experiences. The chapter is important not only because it fetches 2-3 questions in most of the engineering examination but also because it is prerequisite to the other chapters of Co-ordinate Geometry. Several methods have been developed by mathematicians to uniquely locate the position of the points in space. The easiest and most widely used one is the Cartesian coordinate system, which is based on mutually perpendicular axes. In this chapter you will learn about this system, locating the points in this system and finding equations of lines passing through these points. If we say a line passes through two given points, we are imposing two conditions on this line. The conditions can be imposed in several other manners also e.g. we may say that a line passes through one point and is perpendicular to another line. These two conditions are sufficient for determining the line uniquely. In this chapter you will also learn various ways of imposing conditions and finding the equation of line under those conditions. Having known the line uniquely in space we shall try to put various other operations with it e.g. divide it in a given ratio, find distance of other points or lines from it or find the angles which this line makes with some straight line. The topics covered in this chapter include:
Straight Lines is an extremely important head under co-ordinate geometry. It is neither a very simple topic nor extremely tough but requires consistent practice. It is a vast topic which includes several other heads. It is an important chapter in the Mathematics syllabus of IIT JEE. The laws of straight lines are not very new and they can be understood easily by relating them to day life experiences.
The chapter is important not only because it fetches 2-3 questions in most of the engineering examination but also because it is prerequisite to the other chapters of Co-ordinate Geometry.
Several methods have been developed by mathematicians to uniquely locate the position of the points in space. The easiest and most widely used one is the Cartesian coordinate system, which is based on mutually perpendicular axes. In this chapter you will learn about this system, locating the points in this system and finding equations of lines passing through these points.
If we say a line passes through two given points, we are imposing two conditions on this line. The conditions can be imposed in several other manners also e.g. we may say that a line passes through one point and is perpendicular to another line. These two conditions are sufficient for determining the line uniquely. In this chapter you will also learn various ways of imposing conditions and finding the equation of line under those conditions. Having known the line uniquely in space we shall try to put various other operations with it e.g. divide it in a given ratio, find distance of other points or lines from it or find the angles which this line makes with some straight line.
Representation of Points in Plane
Distance between two Points
Centroid Incentre and Circum Centre
Area of Triangle
Standard Equations of Straight Line
Different Forms of Line
Examples based on Straight Line
Angle between two Straight Lines
Equation of Locus
Examples on Angle between two Straight Lines
Length of Perpendicular from Point on Line
Distance between two Parallel Lines
Examples on Distance between two Parallel Lines
Family of Lines
Concurrency of Straight Lines
Position of two Points with Respect to a given Line
Angle Bisectors
Pair of Straight Lines
Angle between Pair of Lines
Combined Equations of Angle Bisectors of Lines
Joint Equation of Pair of Lines
All these heads have been discussed in detail in the coming sections. Here, we shall discuss these heads in brief. Those willing to go into the intricacies are advised to refer the coming sections.
The two important formulae that come under this head are the distance formula and the section formula.
The distance between two points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) is Length PQ = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}
Section Formula
If we wish to know the co-ordinates of the point which divides a line segment between two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) in the ratio m : n, then the coordinates of such a point are given by
{(nx_{1} + mx_{2})/(m+n), (ny_{1} + my_{2})/(m+n)} (for internal division)
If m/n is positive, the division is internal but if m/n is positive, the division is external.
If we have a triangle ABC whose sides AB, BC and CA are of lengths c, a and b respectively and the coordinates of the vertices are A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) then
The coordinates of the centroid are {(x_{1} + x_{2} + x_{3})/3, ((y_{1} + y_{2} + y_{3})/3)}.
The coordinates of the incentre are {(ax_{1} + bx_{2} + cx_{3})/(a + b + c), (ay_{1} + by_{2} + cy_{3})/(a + b + c)}.
Moreover, the incentre divides the angle bisectors in the ratio (b+c):a, (c+a):b and (a+b):c.
The point of concurrency of the perpendicular bisectors of the sides of the triangle is called the circumcentre of the triangle. This is also the centre of the circle, passing through the vertices of the given triangle.
Orthocentre is the point of intersection of all altitudes of the triangle. The orthocentre, centroid & circumcentre are always collinear.
The equation of straight line can be expressed in various forms. We list below some of the forms though all the possible forms have been discussed in detail in the coming sections:
General form: The most general equation of a straight line is ax + by + c = 0.
Slope-intercept form: The equation of the line with slope ‘m’ and which makes an intercept ‘c’ on the y -axis is given as y = mx + c.
Slope one point form: If we have the slope as ‘m’ and the line passes through the point (x_{1}, y_{1}) then the equation of such a straight line is
y – y_{1} = m(x – x_{1})
Two point form: The equation of the line passing through the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by (y – y_{1}) = m (x – x_{1}) or (y – y_{1}) = {(y_{2 }- y_{1})/(x_{2 }- x_{1})} (x – x_{1})
Intercept form: The equation of the line which makes an intercept of ‘a’ and ‘b’ on OX and OY respectively is x/a + y/b = 1.
If m_{1 }and m_{2} are the slopes of two intersecting straight lines (m_{1}m_{2} ≠ -1) and θ is the acute angle between them then
tan θ = |(m_{1 }- m_{2})/(1 + m_{1}m_{2})|
Note: If we have three straight lines L_{1} = L_{2} = L_{3} = 0, with m_{1}, m_{2} and m_{3} as their slopes where m_{1} > m_{2} > m_{3}, then these straight lines form a triangle whose interior angles are given by the following relations:
tan A = (m_{1 }- m_{2})/(1 + m_{1}m_{2}), tan B = (m_{2 }- m_{3})/(1 + m_{2}m_{3}) and tan C = (m_{3 }- m_{1})/(1 + m_{3}m_{1})
When two straight lines are parallel then their slopes are equal. Hence, if we have a straight line with equation ax + by + c = 0, then the line parallel to this line is of the form ax + by + k = 0, where ‘k’ is a parameter.
Hence, if the coefficients of both x and y are same in both the equations and they differ only by a constant, then the lines are said to be parallel.
The distance between two parallel lines with equations ax + by + c_{1} = 0 and ax + by + c_{2} = 0 is given by the formula |(c_{1 }- c_{2})/√(a^{2 }+ b^{2})|.
Two straight lines are said to be perpendicular if the product of their slopes is -1, i.e. if m_{1} and m_{2} are the slopes of the two straight lines then m_{1}m_{2} = -1.
Hence, the straight line perpendicular to the line ax + by + c = 0 is of the form bx – ay + k = 0, where k is any parameter.
In general, the straight lines ax + by + c = 0 and a’x + b’y + c’ = 0 are at right angles iff aa’ + bb’ = 0.
The condition for 3 lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 and a_{3}x + b_{3}y + c_{3} = 0 to be concurrent is
= 0.
We can also find whether the given lines are concurrent or not by another way. Three straight lines are said to be concurrent if we can find three constants A, B and C such that
A(a_{1}x + b_{1}y + c_{1}) + B(a_{2}x + b_{2}y + c_{2}) + C( a_{3}x + b_{3}y + c_{3}) = 0.
Equations of the bisectors of the two lines ax + by + c = 0 and a’x + b’y + c’ = 0 ( where ab’ ≠ a’b) are given by
(ax + by + c)/√(a^{2} + b^{2}) = ± (a’x + b’y + c’)/√(a’^{2} + b’^{2})
How to discriminate between the acute angle bisector and obtuse angle bisector:
If θ is the angle between one of the lines and one of the bisectors, first compute θ.
If |tan θ| < 1, then 2θ < 90° and then this bisector is the acute angle bisector.
If |tan θ| > 1, then we get the bisector to be the obtuse angle bisector.
A homogeneous equation of degree two of the type ax^{2} + 2hxy + by^{2} = 0 always represents a pair of straight lines passing through the origin and then we have the following cases:
If h^{2} > ab, then the lines are real and distinct.
If h^{2} = ab, then the lines are coincident.
If h^{2} < ab, then the lines are imaginary having the point of intersection as (0, 0).
If y = m_{1}x and y = m_{2}x be the two equations represented by ax^{2} + 2hxy + by^{2} = 0 then we have m_{1} + m_{2} = -2h/b and m_{1}m_{2} = a/b.
If θ is the acute angle between the pair of straight lines represented by ax^{2} + 2hxy + by^{2} = 0, then tan θ = |2√(h^{2} – ab)/(a+b)|
In this, we may have the following cases:
These lines are at right angles to each other if a+b = 0 i.e. if coefficient of x^{2} + coefiicient of y^{2} = 0.
The lines are coincident if h^{2} = ab.
They are equally inclined to the axis of x if h = 0 i.e. if coefficient of xy = 0.
When we are trying to find out the position of two points with respect to a given line, we can have the following two cases:
Case 1: If P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) are on the opposite sides of the line ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ internally in the ratio m_{1} : m_{2}, where m_{1}/m_{2} must be positive.
Case 2: If P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) lie on the same side of the line ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ externally in the ratio m_{1} : m_{2}, where m_{1}/m_{2} must be negative.
These cases have been dealt with in more detail in the later sections.
If the line lx + my + n = 0, (n ≠ 0) i.e. the line not passing through origin) cuts the curve ax^{2} + by^{2} + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e.
ax^{2} + 2hxy + by^{2} + (2gx + 2fy) (kx + my/(–n)) + c (lx+my/(-n))^{2} = 0 is the equation of the lines OA and OB.
The general equation of the family of lines through the point of intersection of two given lines is L + λL’ = 0, where L = 0 and L’ = 0 are the two given lines, and λ is a parameter.
The combined equations of angle bisectors of lines represented by ax^{2} + 2hxy + by^{2} = 0
Let ax^{2} + 2hxy + by^{2} = 9 represent lines
y – m_{1}x = 0 and y – m_{2}x = 0
Let P(α, ß) be any point on one of bisectors.
⇒ (ß – m_{1}α)/√(1 + m_{1}^{2}) = + ß–m_{2}α/√(1 + m_{2})^{2}
⇒ (1 + m_{2}^{2}) (ß – m_{1}α)^{2} – (1 + m_{1}^{2}) (ß – m_{2}α)^{2} = 0
⇒ x^{2}–y^{2}/a–b = xy/h; is required equation of angle bisectors …… (1)
If a = b, then bisectors are x^{2} – y^{2} = 0 i.e. x – y = 0, x + y = 0.
If h = 0, the bisectors are xy = 0 i.e. x = 0, y = 0.
If in (i), coefficient of x^{2} + coefficient of y^{2} = 0, then the two bisectors are always perpendicular to each other.
Let P(-1, 0), Q(0, 0) and R = (3, 3√3) be three points. Then, the equations of the bisector of the angle PQR is (IIT JEE 2002)?
The line segment QR makes an angle of 60° with the positive direction of x-axis.
So, the bisector of angle PQR will make an angle of 60° with the negative direction of x-axis and hence it will have the angle of inclination of 120° and so its equation is
(y – 0) = tan 120° (x - 0)
Hence, y = - √3x.
Hence, the required equation is y + √3x = 0.
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