## 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 – **

\(-\,({{x}_{1}}-{{x}_{2}})-\) \(\,\,\,\,\tan \theta =\frac{{{y}_{2}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\)

### 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 = y_{1}

x – Intercept = x_{1}

y – Intercept =

Length of x – intercept = |x_{1}|

Length of y – intercept = |y_{1} |

### Point form

Equation of line passing through two points (x_{1}, y_{1}) & (x_{2} , y_{2})

m = slope

\(\,\,\,po{int}\) \(or\) \(y-{{y}_{2}}=\left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\,\left( x-{{x}_{2}} \right)\,\,\,\,\,\,\,\,\)————F**orm – III**

**For 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 (x_{1}, y_{1}) 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 numbers**Slope form**

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=mx+c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m={}^{a}/{}_{b}\,\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\frac{-ax}{b}-\frac{c}{b},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c={}^{-c}/{}_{b}\)
### Relation between two lines

**Parallel line**

L_{1} a_{1}x + b_{1}y + c_{1} = 0

L_{2} a_{2}x + b_{2}y + c_{2} = 0

** Condition required:**

### Intersection of two lines

Solve L_{1} & L_{2}

### 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(x_{1}, y_{1}) 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 L_{1} & L_{2} is given by L_{1} +λ L_{2} = 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 (a_{1}x + b_{1}y+c_{1}) (a_{2}x+b_{2}y+c_{2}) = 0

Let defines a standard form of equation:-

ax^{2 }+ by^{2 }+ 2hxy + 2gx + 2fy + c = 0 represent conics curve equation

Condition for curve of being P.O.S.L Δ = abc + 2fgy – ag^{2} – bf^{2} – ch^{2} = 0

If Δ ≠ 0, (i) parabola h^{2} = ab

(ii) hyperbola h^{2} < ab

(iii) circle h^{2} = 0, a = b

(iv) ellipse h^{2} > ab

Now, lets see how did we get Δ = 0

General equation ax^{2} + 2gx + hxy + by^{2} + 2fy + c = 0

ax^{2} + (2g+hy)x + (by^{2} + 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 – bg^{2} – af^{2} – ch^{2} = 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 (L_{1}).(L_{2}) = 0 or f(x, y) . g(u,y) = 0

Angle between the lines,

\(\tan \theta =\left( \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| \right)\)h^{2} = ab \(\to\) line is either parallel or confident

h^{2} < ab \(\to\) imaginary line

h^{2} > ab \(\to\) Two distinct lines

a + b = 0 \(\Rightarrow\) perpendicular line

P.O.S.L passing through origin

\(\Rightarrow\) (y – m_{1}x) x (y – m

_{2}x) = 0

y^{2} – m_{2}yx – m_{1}xy – m_{1}m_{2}x^{2} = 0

y^{2} – (m_{1} + m_{2}) xy – m_{1}m_{2}x^{2} = 0

^{2}+ 2hxy + by

^{2}= 0 – – – – – – – – (III) \(\Rightarrow\) \({{y}^{2}}+\frac{2h}{b}xy+\frac{ab}{b}{{x}^{2}}=0\,——\,(IV)\) \(\Rightarrow m1 + m2 = \frac{2h}{b}\)

m_{1} m_{2} = \(\frac{a}{b}\)
\(\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}4{{m}_{1}}{{m}_{2}}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\)

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 (x_{1}, y_{1}) having slope m is

y – y_{1} = m(x – x_{1})

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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 m_{1} and m be the slopes of the two given lines such that m_{1} = 2m

We know that if θ is the angle between the lines l_{1} and l_{2} with slopes m_{1} and m_{2}, 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, m_{1}m_{2} = -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 – af^{2} – bg^{2} – bc^{2} = 0 i.e., if

**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 \(\sqrt{3}{{x}^{2}}-xy=0\)