Straight Lines- Properties, Relation Between Lines and Examples (2024)

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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Table of Contents:
  • 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
  • Solved Problems
  • Video Lessons
  • FAQs

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 = 0

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.

Straight Lines- Properties, Relation Between Lines and Examples (1)

When the angle with + ve x-axis ‘tan θ’ is called the slope of a straight line.

Straight Lines- Properties, Relation Between Lines and Examples (2)

Note 1 – If the line is horizontal, then slope = 0

Straight Lines- Properties, Relation Between Lines and Examples (3)

Note 2 – If the line is perpendicular to the x-axis, i.e., vertical, then the slope is undefined.

Straight Lines- Properties, Relation Between Lines and Examples (4)

Slope = 1/0 = = tan ⁡π/2

Note 3 – If the line is passing through any of two points, then the slope is

\(\begin{array}{l}tan\ \theta=\frac{y_2-y_1}{x_2-x_1}\end{array} \)

Straight Lines- Properties, Relation Between Lines and Examples (5)

Intercept Form

The equation of the line with x-intercept as ‘a’ and y-intercept as ‘b’ can be written as;

\(\begin{array}{l}\frac{x}{a} + \frac{y}{b}=1\end{array} \)

  • x – 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,

Straight Lines- Properties, Relation Between Lines and Examples (6)

Along the 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,

Straight Lines- Properties, Relation Between Lines and Examples (7)

ON = P

∠AON = α

Let the length of the perpendicular from origin to straight line be ‘P’ and let this perpendicular make an angle with + vex- axis ‘α’, then the equation of a line can be:

\(\begin{array}{l}x\cos \alpha +y\sin \alpha =p\end{array} \)

\(\begin{array}{l}\frac{x}{p\sec \alpha }+\frac{y}{p\cos ec\alpha }=1\end{array} \)

\(\begin{array}{l}x\cos \alpha +y\sin \alpha =P\end{array} \)

Learn More: Different Forms of the Equation of a 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

\(\begin{array}{l}y-{{y}_{1}} = \left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\end{array} \)

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:

\(\begin{array}{l}y-{{y}_{1}} = \left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\end{array} \)

Now,

(y2 – y1)/(x2 – x1) = (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 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,

\(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}} \neq \frac{{{c}_{1}}}{{{c}_{2}}}\end{array} \)

  • For intersecting lines

Two lines intersect at a point if

\(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}} \neq \frac{{{b}_{1}}}{{{b}_{2}}}\end{array} \)

  • For coincident lines

Two lines coincide if

\(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\end{array} \)

Angle between Straight Lines

\(\begin{array}{l}Let\,\,\,\,{{L}_{1}}\,\,\,\,\equiv \,\,\,y={{m}_{1}}x+{{c}_{1}}\end{array} \)

and

\(\begin{array}{l}{{L}_{2}}\,\,\,\equiv \,\,\,y={{m}_{2}}x+{{c}_{2}}\end{array} \)

Straight Lines- Properties, Relation Between Lines and Examples (8)

\(\begin{array}{l}\text{Angle} = \theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right) \right|\end{array} \)

Special Cases:

\(\begin{array}{l}\Rightarrow {{m}_{2}}={{m}_{1}}\,\,\,\,\,\,\,\to \,\,\,\,\,lines\,are\,parallel\\\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,{{m}_{1}}{{m}_{2}}=-1,\,\,\,\,\,\,\,\,\,\,lines\,L1\,\And L2\,are\,perpendicular\,to\,each\,other\end{array} \)

Length of Perpendicular from a Point on a Line

Straight Lines- Properties, Relation Between Lines and Examples (9)

The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is

\(\begin{array}{l}\ell =\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\end{array} \)

B (x, y) is the foot of perpendicular is given by

\(\begin{array}{l}\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\end{array} \)

A’(h, k) is mirror image, given by

\(\begin{array}{l}\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\end{array} \)

Angular Bisector of Straight lines

An angle bisector has an equal perpendicular distance from the two given lines.

Straight Lines- Properties, Relation Between Lines and Examples (10)

The equation of line L can be given as

\(\begin{array}{l}\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}}\end{array} \)

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

\(\begin{array}{l}{{L}_{1}}\equiv {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\end{array} \)

\(\begin{array}{l}{{L}_{2}}\equiv {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\end{array} \)

and

\(\begin{array}{l}{{L}_{3}}\equiv {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\end{array} \)

So, the condition for the concurrency of lines is

\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0\end{array} \)

Pair of Straight Lines

Straight Lines- Properties, Relation Between Lines and Examples (11)

Join equation of lines L1 & L2 represents P. S. L (a1x + b1y+c1) (a2x+b2y+c2) = 0

i.e. f(x,y) . g(x,y) = 0

Let’s defines a standard form of the equation:-

ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represent conics curve equation

Straight Lines- Properties, Relation Between Lines and Examples (12)

Condition for curve of being P.O.S.L Δ = abc + 2fgh – af2 – bg2 – ch2 = 0

Straight Lines- Properties, Relation Between Lines and Examples (13)

If Δ ≠ 0, (i) parabola h2 = ab

(ii) hyperbola h2 < ab

(iii) circle h2 = 0, a = b

(iv) ellipse h2 > ab

Now, let’s 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 the above equation as a quadratic equation in x, keeping y constant.

\(\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{{{(2g+2hy)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\end{array} \)

, so

\(\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{Q(y)}}{2a}\end{array} \)

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 Δ value of Q(y) should be zero.

From there D = 0

abc + 2fgh – bg2 – af2 – ch2 = 0

Or

\(\begin{array}{l}\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\end{array} \)

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

\(\begin{array}{l}\tan \theta =\left( \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| \right)\end{array} \)

Special cases:

h2 = ab lines are either parallel or coincident

h2 < ab imaginary line

h2 > ab Two distinct lines

a + b = 0 perpendicular line

3. P.O.S.L passing through the origin, then

(y – m1x) (y – m2x) = 0

y2 – m2yx – m1xy – m1m2x2 = 0

y2 – (m1 + m2) xy – m1m2x2 = 0

ax2 + 2hxy + by2 = 0

\(\begin{array}{l}\Rightarrow {{y}^{2}}+\frac{2h}{b}xy+\frac{ab}{b}{{x}^{2}}=0\end{array} \)

\(\begin{array}{l}\Rightarrow m1 + m2 = \frac{2h}{b}\end{array} \)

\(\begin{array}{l}m_1 m_2=\frac{a}{b}\end{array} \)

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

Straight Lines Formulas

All Formulas Related to Straight Lines
Equation of a Straight Lineax + by + c = 0
General form or Standard Formy = mx + c
Equation of a Line with 2 Points (Slope Point Form)(y – y1) = m(x – x1)

Here, m = (y2 – y1)/(x2 – 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:

\(\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1\end{array} \)

Using the given conditions,

\(\begin{array}{l}P\left( \frac{2a+1.0}{2+1},\frac{2.0+1.b}{2+1} \right)\end{array} \)

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;

\(\begin{array}{l}\frac{x}{\left( -15/2 \right)}+\frac{y}{12}=1\end{array} \)

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 θ = -1/3.

Solution:

Here cos θ = -1/3. (a negative number) so that π/2 < θ < π

\(\begin{array}{l}\Rightarrow \tan \theta =-\sqrt{8}\end{array} \)

= 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

\(\begin{array}{l}y-2=-\sqrt{8}\left( x-1 \right)\end{array} \)

\(\begin{array}{l}\Rightarrow \sqrt{8}x+y-\sqrt{8}-2=0.\end{array} \)

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

\(\begin{array}{l}y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)\end{array} \)

Hence, the equation of the required line will be

\(\begin{array}{l}y-3=\frac{-2 – 3}{4 + 1}\left( x+1 \right)\Rightarrow x+y-2=0\end{array} \)

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

\(\begin{array}{l}\frac{7-3}{4-1}=\frac{4}{3}=\tan \alpha\end{array} \)

(ii) Slope of line is

\(\begin{array}{l}-\frac{3}{-4}=\frac{3}{4}=\tan \beta\end{array} \)

Now tan α > tan β. So line (i) has more inclination.

Question 5:

The angle of the line positive direction of the x-axis is θ. The line is rotated about some point on it in an anticlockwise direction by an angle of 45°, and its slope becomes 3. Find the angle θ.

Solution:

Originally slope of the line is tan θ = 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 45°.

∴ we have

\(\begin{array}{l}\tan 45{}^\circ =\frac{3-m}{1+3m}\end{array} \)

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 (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

\(\begin{array}{l}{{a}^{2}}+2a-3=0\end{array} \)

or

\(\begin{array}{l}\left( a-1 \right)\left( a+3 \right)=0\end{array} \)

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

\(\begin{array}{l}\Rightarrow \frac{1-\left( -1 \right)}{2-x}=\frac{5-1}{4-2}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{2}{2-x}=2\Rightarrow x=1\end{array} \)

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 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

\(\begin{array}{l}\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|\end{array} \)

It is given that the tangent of the angle between the two lines is 1/3.

\(\begin{array}{l}\therefore \frac{1}{3}=\left| \frac{m-2m}{1+\left( 2m \right).m} \right|\Rightarrow \frac{1}{3}=\left| \frac{-m}{1+2{{m}^{2}}} \right|\end{array} \)

\(\begin{array}{l}\Rightarrow 2{{\left| m \right|}^{2}}-3\left| m \right|+1=0\end{array} \)

\(\begin{array}{l}\Rightarrow \left( \left| m \right|-1 \right)\left( 2\left| m \right|-1 \right)=0\end{array} \)

\(\begin{array}{l}\Rightarrow \left| m \right|=1\,\,or\,\,\left| m \right|=1/2\end{array} \)

\(\begin{array}{l}\Rightarrow \left| m \right|\pm 1\,\,or\,\,m=\pm 1/2\\\end{array} \)

Another slope will be -2, -1, 2, 1.

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) – 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.

Straight Lines- Properties, Relation Between Lines and Examples (14)

Slope of the line joining (-1, 3) and (a, b)

\(\begin{array}{l}{{m}_{1}}=\frac{b-3}{a+1}\end{array} \)

Slope of the line 3x – 4y – 16 = 0 is 3/4.

Since these two lines are perpendicular, m1m2 = -1

\(\begin{array}{l}\therefore \left( \frac{b-3}{a+1} \right)\times \left( \frac{3}{4} \right)=-1\end{array} \)

⇒ 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

\(\begin{array}{l}a=\frac{68}{25}\end{array} \)

and

\(\begin{array}{l}b=-\frac{49}{25}\end{array} \)

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.

Straight Lines- Properties, Relation Between Lines and Examples (15)

Find the equations of the remaining sides of these squares.

Solution:

Distance between the two parallel lines is

\(\begin{array}{l}\frac{\left| 7+3 \right|}{\sqrt{5}}=2\sqrt{5}.\end{array} \)

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

\(\begin{array}{l}\frac{\left| k-\left( -4 \right) \right|}{\sqrt{5}}=2\sqrt{5}\Rightarrow \frac{k+4}{\sqrt{5}}=\pm 2\sqrt{5}\Rightarrow k=6,-14.\end{array} \)

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

\(\begin{array}{l}\frac{4x+3y-6}{\sqrt{{{4}^{2}}+{{3}^{2}}}}=\pm \frac{5x+12y+9}{\sqrt{{{5}^{2}}+{{12}^{2}}}}\end{array} \)

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

\(\begin{array}{l}\tan \theta =\left| \frac{-\frac{4}{3}-\frac{9}{7}}{1+\left( \frac{-4}{3} \right)\frac{9}{7}} \right|=\frac{11}{3}>1.\end{array} \)

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)

\(\begin{array}{l}4x+3y-6=4\times 1+3\times 2-6>0,\end{array} \)

\(\begin{array}{l}5x+12y+9=5\times 1+12\times 2+9>0.\end{array} \)

Hence, equation of the bisector of the angle containing the point (1, 2) is

\(\begin{array}{l}\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}\Rightarrow 9x-7y-41=0.\end{array} \)

Question 13:

Find the value of λ if 2x2 + 7xy + 3y2 + 8x + 14y + λ = 0 will represent a pair of straight lines.

Solution:

The given equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of lines if its descriminant = 0.

i.e., if abc + 2fgh – af2 – bg2 – ch2 = 0

\(\begin{array}{l}6\lambda +2\left( 7 \right)\left( 4 \right)\left( \frac{7}{2} \right)-2{{\left( 7 \right)}^{2}}-3{{\left( 4 \right)}^{2}}-\lambda {{\left( \frac{7}{2} \right)}^{2}}=0\end{array} \)

\(\begin{array}{l}\Rightarrow 6\lambda +196-98-48-\frac{49\lambda }{4}=0\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{49\lambda }{4}-6\lambda =196-146=50\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{25\lambda }{4}=50\,\,\\\lambda =\frac{200}{25}=8\end{array} \)

Question 14:

If one of the lines of the pair ax2 + 2hxy + by2 = 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 ax2 + 2hxy + by2 = 0.

Let’s apply y = x.

We have

\(\begin{array}{l}{{x}^{2}}\left( a+2h+b \right)=0\end{array} \)

or

\(\begin{array}{l}a+b=-2h.\end{array} \)

Question 15:

If the angle between the two lines is represented by 2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0 is tan-1(m), then find the value of m.

Solution:

The angle between the lines 2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0 is given by

\(\begin{array}{l}\tan \theta =\frac{\pm 2\sqrt{\frac{25}{4}-6}}{2+3}\\ \theta ={{\tan }^{-1}}\left|( \frac{1}{5} \right)|\end{array} \)

Question 16:

The pair of lines √3 x2 – 4xy + √3 y2 = 0 are rotated about the origin by π/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

\(\begin{array}{l}\left( \sqrt{3}x-y \right)\left( x-\sqrt{3}y \right)=0.\end{array} \)

Their separate equations are y = √3 x and y = (1/√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 = √3 x

The combined equation in the new position is

\(\begin{array}{l}x\left( \sqrt{3}x-y \right)=0\end{array} \)

or

\(\begin{array}{l}\sqrt{3}{{x}^{2}}-xy=0\end{array} \)

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Frequently Asked Questions

Q1

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.)

Q2

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.

Q3

What is the equation of the straight line passing through (x1, y1) with slope m?

The equation of the straight line passing through (x1, y1) with slope m is given by y – y1 = m(x – x1 ).

The expertise in the topic of straight lines involves a deep understanding of various forms of linear equations, their geometric interpretations, and the relationships between different concepts. I'll provide a concise and insightful overview of the concepts covered in the article.

Concepts Related to Straight Lines:

  1. Definition of a Straight Line:

    • A straight line is a geometry object characterized by zero width, extending infinitely on both sides without any curves.
  2. Equation of a Straight Line:

    • General Form: (ax + by + c = 0)
    • Slope-Intercept Form: (y = mx + c), where (m) is the slope and (c) is the y-intercept.
  3. Intercept Form:

    • Equation with x-intercept (a) and y-intercept (b): (\frac{x}{a} + \frac{y}{b} = 1)
  4. Point Form:

    • Equation passing through a point ((x_1, y_1)): (y - y_1 = m(x - x_1))
  5. Slope Point Form:

    • Equation passing through two points ((x_1, y_1)) and ((x_2, y_2)): (y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1))
  6. Angle between Two Lines:

    • Formula: (\theta = \tan^{-1}\left| \frac{m_2 - m_1}{1 + m_1m_2} \right|)
    • Special Cases: Parallel ((m_1 = m_2)), Perpendicular ((m_1m_2 = -1))
  7. Length of Perpendicular from a Point to a Line:

    • Length ((\ell)): (\left|\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}\right|)
  8. Angle Bisector:

    • Equation for the bisector of two lines (L_1) and (L_2): (L_1 + \lambda L_2 = 0)
  9. Family of Lines:

    • General equation through the point of intersection of two lines (L_1) and (L_2): (L_1 + \lambda L_2 = 0)
  10. Concurrency of Three Lines:

    • Condition for concurrency: (\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)
  11. Pair of Straight Lines:

    • Standard form: (ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0)
    • Condition for conic curves: (\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0)

These concepts cover the essential aspects of straight lines, from their basic forms to advanced relationships and applications in geometry. If you have specific questions or problems related to these concepts, feel free to ask for further clarification or assistance.

Straight Lines- Properties, Relation Between Lines and Examples (2024)

FAQs

What are the properties of a straight line? ›

Properties of a Straight Line

A straight line has zero areas, zero volume. but it has infinite length. A straight line is a one-dimensional figure. An infinite number of lines can pass through a single point, but there is only one unique line that passes through two points.

What is the relationship between two straight lines? ›

Straight lines are lines above each other. They are not parallel lines or perpendicular lines but are completely identical. In other words, these are lines where one line completely closes another line in such a way that we can point to two lines.

What is the difference between lines and straight lines? ›

A line can be drawn between any two points. A straight line is the shortest line that can be drawn between those two points.

What is an example of an object in a straight line? ›

Whether it's a car driving down the street, a rocket shooting across the sky or an aeroplane soaring in the sky, we often think of objects moving in a direction that can be described as a straight line.

What is a real life example of a straight line? ›

Answer: Light travels in Straight line. Roads, Railway Tracks and Bridges are constructed based on parallel lines. Use of Slopes in Physics: While plotting graphs of motion, the graph of position vs time results in a straight line.

How to find a relationship between two lines? ›

The easiest way to see the relationship between the two lines is to transform them both into slope-intercept form, which is y=mx+b . In this form, we can easily identify that both lines have a slope of −3 , but that they have different y -intercepts. Lines will equal slopes but different y -intercepts are parallel.

What are relationships of lines? ›

Line Relationships Determine whether the lines are parallel, perpendicular, or neither. Skew Lines Determine whether the lines are skew.

What is the relationship of a straight line graph? ›

For straight line graphs there is a linear relationship between the x and y values and so the line is straight and must be drawn with a ruler. The gradient determines how steep the line is and the y-intercept tells us the location where the straight line intersects the y-axis.

What is straight line simple? ›

Definitions of straight line. a line traced by a point traveling in a constant direction; a line of zero curvature. “the shortest distance between two points is a straight line” antonyms: curve, curved shape.

How to find the equation of a line? ›

Explanation: These lines are written in the form y = mx + b, where m is the slope and b is the y-intercept. We know from the question that our slope is 3 and our y-intercept is –5, so plugging these values in we get the equation of our line to be y = 3x – 5.

What is the formula for the family of lines? ›

The universal equation of the family of lines via the point of intersection of two provided lines is L + λ L '= 0, where L = 0 and L' = 0 are the two assigned lines, and λ works as a parameter.

How to check straight line? ›

The idea is to use the slopes as criteria to check whether the given points lie in a straight line or not. From geometry, any pair of points lying in a straight line gives the same slope value. The slope between two points (X0, Y0) and (X1, Y1) is (Y1 - Y0) / (X1 - X0).

What are points on a straight line called? ›

Collinear points are the points that lie on the same straight line or in a single line. If two or more than two points lie on a line close to or far from each other, then they are said to be collinear, in Euclidean geometry. Table of contents: Definition.

Is a line always straight? ›

A line can be straight or curved. In geometry, the word line means a straight line. A straight line is the shortest distance between two points. A straight line is traced by a point moving in a direction that does not change.

What is property of straight angle? ›

Properties of Straight Angles

A straight angle measures exactly half of a revolution. A straight angle is produced by revolving one ray by 180° with respect to another ray. In a straight angle, the arms extend in the opposite directions. A straight angle modifies the direction of a point.

What are the properties of a line diagram? ›

Properties of Line Graphs

It is drawn using straight line segments between points. By joining all points, we get a resulting line that may be a straight line or a curve. It has two variables: one is independent while the other is dependent on the first independent variable.

What are the properties of a curved line? ›

A curved line is one that is not straight and is bent. Ideally, it is smooth and continuous. In other words, a curve is defined as a group of points that resemble a straight line that falls between two neighbouring points. We know that the curvature of the straight line is zero.

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