A straight line is a line which 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 PDF which will students to refer to the concepts in offline mode.
Table of Contents for Straight Lines
- Definition
- Equation
- Intercept Form
- Point Form
- Slope Point Form
- General or Standard Form
- Relation Between Two Lines
- Intersection of 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 variable x, y satisfy all points on the curve.
Straight line equation linear in x and constant terms.
ax + by + c = 0 { equation of straight lines.
Slope:-
angle with + ve x-axis
‘tan θ’ is called slope of straight line.
θ E[0,n)
Note 1 – If line is Horizontal, then slope = 0
Note 2 – If line is ⊥ to x-axis, i.e. vertical
Slope = undefined
= \(\frac{1}{0}\)
= \(\tan \frac{\pi }{2}\)
Note 3 –
Intercept Form
x – co-ordinate of point of intersection of line with x-axis is called x-intercept
x – Intercept = 5
y – Intercept = 5
- Line passes through origin, intercept = 0
x – Intercept = 0
y – Intercept = 0
Similarly
y – intercept will be y-co-ordinate of point of intersection of line with y-axis.
x – Intercept =
y – Intercept = y1
x – Intercept = x1
y – Intercept =
Length of x – intercept = |x1|
Length of y – intercept = |y1 |
Point form
Equation of line passing through two points (x1, y1) & (x2 , y2)
\(y-{{y}_{1}}\left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\to po{int}\)m = slope
\(\,\,\,po{int}\) \(or\) \(y-{{y}_{2}}=\left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\,\left( x-{{x}_{2}} \right)\,\,\,\,\,\,\,\,\)————Form – IIIFor example,
Example: Find the equation of the lines that passes through the points (-2,4) and (1,2)
Solution:
Now we have a slope and two points. 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:
y = mx + b
4 = (– 2/3)(–2) + b
4 = 4/3 + b
4 – 4/3 = b
12/3 – 4/3 = b
b = 8/3
so, y = ( – 2/3 ) x + 8/3.
On the other hand, if we use the point (1, 2), we get:
y = mx + b
2 = (– 2/3)(1) + b
2 = – 2/3 + b
2 + 2/3 = b
6/3 + 2/3 = b
b = 8/3
So it doesn’t matter which point we choose. Either way, the answer is the same:
y = (– 2/3)x + 8/3
Slope Point form (Equation of a Line with 2 Points)
Equation of line with slope ‘m’ and which passes through (x1, y1) can be given as
Form IV
Intercept form
Equation of line with x – intercept as ‘a’ and y – intercept as ‘b’ can be given as \(\frac{x}{a}+\frac{y}{b}=1\)
Form V
ON = P
AON = α
Let length of \(\bot\) r from origin to S.L is ‘P’ and let this \(\bot\)r make an angle with + vex- axis ‘α’, then equation of line can be
\(x\cos \alpha +y\sin \alpha =p\) \(\frac{x}{p\sec \alpha }+\frac{y}{p\cos ec\alpha }=1\) \(x\cos \alpha +y\sin \alpha =P\)Here ‘P’ should be the α E [0, 2π)
Form – VI
Learn More: Different Forms Of The Equation Of Line
Straight Lines Video Lesson
General Form or Standard form of a Line
Equation of a straight line can be given as
\(ax\text{ }+\text{ }by\text{ }+\text{ }c\text{ }=\text{ }0\) a, b, c are Real numbersSlope form
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=mx+c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m={}^{a}/{}_{b}\,\) \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\frac{-ax}{b}-\frac{c}{b},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c={}^{-c}/{}_{b}\)Relation between two lines
- Parallel line
L1 a1x + b1y + c1 = 0
L2 a2x + b2y + c2 = 0
Condition required:
\(\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\)Intersection of two lines
Solve L1 & L2
\({{a}_{1}}x+{{b}_{1}}y=-{{c}_{1}}\) \({{a}_{2}}x+{{b}_{2}}y=-{{c}_{2}}\) \(\begin{matrix} x={{x}_{1}} \\ y={{y}_{1}} \\ \end{matrix}\left\{ from\,above\,equation) \right.\)Angle between Straight lines
\(Let\,\,\,\,{{L}_{1}}\,\,\,\,\equiv \,\,\,y={{m}_{1}}x+{{c}_{1}}\)
\({{L}_{2}}\,\,\,\equiv \,\,\,y={{m}_{2}}x+{{c}_{2}}\)
\(\theta =acute\,angle\)
\(\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}+{{m}_{2}}} \right) \right|\)
\(if\Rightarrow {{m}_{2}}={{m}_{1}}\,\,\,\,\,\,\,\to \,\,\,\,\,lines\,are\,parallel\)
\(\Rightarrow \,\,{{m}_{1}}{{m}_{2}}=-1,\,\,\,\,\,\,\,\,\,\,lines\,L1\,\And L2\,are\,perpendicular\,to\,each\,other\)
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
\(\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\)A’(h, k) is mirror image, given by
\(\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\)Angular Bisector of Straight lines
To find the equation of the bisectors of the angle between lines.
Equation of line L can be given
\(\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}}\)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
\({{L}_{1}}\equiv {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\) \({{L}_{2}}\equiv {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\) \({{L}_{3}}\equiv {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\)So, condition for concurrency of linear is
\(\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0\)Pair of Straight Lines
Join equation of line L1 & L2 represents P. S. L (a1x + b1y+c1) (a2x+b2y+c2) = 0
\(f\left( x,\text{ }y \right).\text{ }g\left( x,\text{ }y \right)\text{ }=\text{ }0\,\,\,\,\,\,\,\,\,\,\,\,represent\,P.O.S.L\) \(\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\) \(\,\,\,\,\,\,\,linear\,\,\,\,\,\,\,\,\,\,\,linear\) \(\,\,\,\,equation\,\,\,\,\,\,\,\,eqution\) \(\,\,\,\,\,\,\,\,of\,line\,\,\,\,\,\,\,\,\,of\,line\)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 + 2fgy – ag2 – bf2 – 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 + hxy + by2 + 2fy + c = 0
ax2 + (2g+hy)x + (by2 + 2fy + c) = 0
we can consider equation 11 as quadratic equation in x keeping y as constant.
\(x=\frac{-(2g+hy)\pm \sqrt{{{(2g+hy)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\) \(x=\frac{-(2g+hy)\pm \sqrt{Q(y)}}{2a}\)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
\(\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\)Hence proved.
⇒ point of intersection of two lines (P.O.S.L)
We can get point of intersection.
Or
Solve the P.O.S.L, factorize it in (L1).(L2) = 0 or f(x, y) . g(u,y) = 0
Angle between the lines,
\(\tan \theta =\left( \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| \right)\)h2 = ab \(\to\) line is either parallel or confident
h2 < ab \(\to\) imaginary line
h2 > ab \(\to\) Two distinct lines
a + b = 0 \(\Rightarrow\) perpendicular line
P.O.S.L passing through origin
\(\Rightarrow\) (y – m1x) x (y – m2x) = 0y2 – m2yx – m1xy – m1m2x2 = 0
y2 – (m1 + m2) xy – m1m2x2 = 0
\(\Rightarrow\) ax2 + 2hxy + by2 = 0 – – – – – – – – (III) \(\Rightarrow\) \({{y}^{2}}+\frac{2h}{b}xy+\frac{ab}{b}{{x}^{2}}=0\,——\,(IV)\) \(\Rightarrow m1 + m2 = \frac{2h}{b}\)m1 m2 = \(\frac{a}{b}\) \(\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}4{{m}_{1}}{{m}_{2}}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\)
Proved.
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 | \(\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}+{{m}_{2}}} \right) \right|\) |
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 \(\frac{x}{a}+\frac{y}{b}=1.\)
Using the given conditions, \(P\left( \frac{2a+1.0}{2+1},\frac{2.0+1.b}{2+1} \right)\) is the point which divides (a, 0) and (0, b) internally in the ratio 1 : 2.
But P is (-5, 4)
Hence -5 = 2a/3, 4 = b/3 a = -15/2, b = 12.
Hence the required equation is \(\frac{x}{\left( -15/2 \right)}+\frac{y}{12}=1\)
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 \(\cos \theta =-\frac{1}{3}.\)
Solution:
Here \(\cos \theta =-\frac{1}{3}.\) (a negative number) so that \(\frac{\pi }{2}<\theta <\pi\) \(\Rightarrow \tan \theta =-\sqrt{8}=\) slope of line.
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 \(y-2=-\sqrt{8}\left( x-1 \right)\) \(\Rightarrow \sqrt{8}x+y-\sqrt{8}-2=0.\)
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 \(y-{{y}_{1}}=\frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\left( x-{{x}_{1}} \right)\)
Hence equation of the required line will be \(y-3=\frac{3+2}{-1-4}\left( x+1 \right)\Rightarrow x+y-2=0\)
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 \(\frac{7-3}{4-1}=\frac{4}{3}=\tan \alpha\)
(ii) Slope of line is \(-\frac{3}{-4}=\frac{3}{4}=\tan \beta\)
Now tan α > tan β
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 \(\tan 45{}^\circ =\frac{3-m}{1+3m}\)
1 + 3m = 3 – m
4m = 2
m = 1/2 = tan θ
θ = tan-1(1/2)
Question 6:
If line \(3x-ay-1=0\) is parallel to the line \(\left( a+2 \right)x-y+3=0\) then find the values of a.
Solution:
Slope of line \(3x-ay-1=0\) is \(\frac{3}{a}\)
Slope of line \(\left( a+2 \right)x-y+3=0\)is (a + 2)
Since lines are parallel then we have \(a+2=\frac{3}{a}\)
or \({{a}^{2}}+2a-3=0\)
or \(\left( a-1 \right)\left( a+3 \right)=0\)
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
\(\Rightarrow \frac{1-\left( -1 \right)}{2-x}=\frac{5-1}{4-2}\) \(\Rightarrow \frac{2}{2-x}=2\Rightarrow x=1\)Question 8:
The slope of a line is double of the slope of another line. It tangent of the angle between them is \(\frac{1}{3}.\)Find the slopes of the lines.
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 \(\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|\)
It is given that the tangent of the angle between the two lines is \(\frac{1}{3}\)
∴ \(\frac{1}{3}=\left| \frac{m-2m}{1+\left( 2m \right).m} \right|\Rightarrow \frac{1}{3}=\left| \frac{-m}{1+2{{m}^{2}}} \right|\) \(\Rightarrow 2{{\left| m \right|}^{2}}-3\left| m \right|+1=0\) \(\Rightarrow \left( \left| m \right|-1 \right)\left( 2\left| m \right|-1 \right)=0\) \(\Rightarrow \left| m \right|=1\,\,or\,\,\left| m \right|=1/2\) \(\Rightarrow \left| m \right|\pm 1\,\,or\,\,m=\pm 1/2\)
Question 9:
Find equation of the line parallel to the line \(3x-4y+2=0\) and passing through the point \(\left( -2,3 \right).\)
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) – 4(3) + t = 0 or t = 18.
So equation of line is \(3x-4y+18=0\)
Question 10:
Find the coordinates of the foot of 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)
\({{m}_{1}}=\frac{b-3}{a+1}\)Slope of the line \(3x-4y-16=0\) is \(\frac{3}{4}\)
Since these two lines are perpendicular, m1m2 = -1
∴ \(\,\left( \frac{b-3}{a+1} \right)\times \left( \frac{3}{4} \right)=-1\) \(\Rightarrow 4a+3b=5\) … (1)
Point (a, b) lies on line \(3x-4y=16.\)
∴ \(\,\,3a-4b=16\) … (2)
On solving equations (1) and (2), we obtain
\(a=\frac{68}{25}\) and \(b=-\frac{49}{25}\)Thus, the required coordinates of the foot of the perpendicular are \(\left( \frac{68}{25},\frac{49}{25} \right)\)
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 remaining sides of these squares.
Solution:
Distance between the two parallel lines is \(\frac{\left| 7+3 \right|}{\sqrt{5}}=2\sqrt{5}.\)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, \(\frac{\left| k-\left( -4 \right) \right|}{\sqrt{5}}=2\sqrt{5}\Rightarrow \frac{k+4}{\sqrt{5}}=\pm 2\sqrt{5}\Rightarrow k=6,-14.\)
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
\(\frac{4x+3y-6}{\sqrt{{{4}^{2}}+{{3}^{2}}}}=\pm \frac{5x+12y+9}{\sqrt{{{5}^{2}}+{{12}^{2}}}}\) \(\Rightarrow 9x-7y-41=0\) and \(7x+9y-3=0.\)If θ is the angle between the line \(4x+3y-6=0\) and the bisector \(9x-7y-41=0,\) then \(\tan \theta =\left| \frac{-\frac{4}{3}-\frac{9}{7}}{1+\left( \frac{-4}{3} \right)\frac{9}{7}} \right|=\frac{11}{3}>1.\)
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)
\(4x+3y-6=4\times 1+3\times 2-6>0,\) \(5x+12y+9=5\times 1+12\times 2+9>0.\)Hence equation of the bisector of the angle containing the point (1, 2) is \(\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}\Rightarrow 9x-7y-41=0.\)
Question 13:
Find the value of λ if \(2{{x}^{2}}+7xy+3{{y}^{2}}+8x+14y+\lambda =0\) will represent a pair of straight lines
Solution:
The given equation \(a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\) represents a pair of lines
If abc + 2fgh – af2 – bg2 – bc2 = 0 i.e., if
\(6\lambda +2\left( 7 \right)\left( 4 \right)\left( \frac{7}{2} \right)-2{{\left( 7 \right)}^{2}}-3{{\left( 4 \right)}^{2}}-\lambda {{\left( \frac{7}{2} \right)}^{2}}=0\) \(\Rightarrow 6\lambda +196-98-48-\frac{49\lambda }{4}=0\) \(\Rightarrow \frac{49\lambda }{4}-6\lambda =196-146=50\) \(\Rightarrow \frac{25\lambda }{4}=50\,\,\lambda =\frac{200}{25}=8\)Question 14:
If one of the lines of the pair \(a{{x}^{2}}+2hxy+b{{y}^{2}}=0\) bisects the angle between positive direction of the axes, then find 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 \(a{{x}^{2}}+2hxy+b{{y}^{2}}=0,\)
We have \({{x}^{2}}\left( a+2h+b \right)=0\) or \(a+b=-2h.\)
Question 15:
If the angle between the two lines represented by \(2{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+4=0\) is tan-1(m), then find the value of m.
Solution:
The angle between the lines
\(2{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+4=0\) is given by \(\tan \theta =\frac{\pm 2\sqrt{\frac{25}{4}-6}}{2+3}\theta ={{\tan }^{-1}}\left( \pm \frac{1}{5} \right).\)Question 16:
The pair of lines \(\sqrt{3}{{x}^{2}}-4xy+\sqrt{3}{{y}^{2}}=0\) are rotated about the origin by \[\frac{\pi }{6}\]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 \(\left( \sqrt{3}x-y \right)\left( x-\sqrt{3}y \right)=0.\) Their separate equations are \(y=\sqrt{3}x\) and \(y=1/\sqrt{3}x\)or y = tan 60° x and y = tan 30° x
After rotation, the separate equations are
y = tan 90° x and y = tan 60° x
or x = 0 and \(y=\sqrt{3}x\)
the combined equation in the new position is \(x\left( \sqrt{3}x-y \right)=0\) or \(\sqrt{3}{{x}^{2}}-xy=0\)