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A straight line is a distance covered by a point moving in a steady direction with zero curvature. NCERT Solutions For Class 11 Maths Chapter 10: Straight Lines covers topics such as slopes, angles between two lines, and the distance between lines.
- Straight Line is one of the most important chapters of NCERT Class 11 Mathematics
- It is a form of line that has no depth, and width.
- Lines can easily be embedded in two- and three-dimensional.
- They can be positioned horizontally, vertically, or at an angle.
- When two points on a straight line are connected, it will form a 180-degree angle.
- Straight line can not pass through collinear points.
- The concept can be used in our everyday lives.
- Construction of railway tracks and freeways are common examples of lines.
Key Terms: Straight Line, Angle, Length, Intercept Form, Slope of a Line, Two-Point Form, Slope-Intercept Form, Horizontal Line, Vertical Line, Inclined Line
What is a Straight Line?
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A straight line is a path traced by a point moving in a steady direction with zero curvature. In other words, a straight line is the shortest distance between two places.
- A straight line is a set of all points stretching away from two points in Euclidean geometry.
- It has only one length and one dimension.
- Lines can only expand out in two directions indefinitely.
- Both the ends of a straight line can be extended from both sides.
- Horizontal, Vertical, and Oblique/Inclined Lines are three types of line.
- An individual can see straight lines on zebra crossing on roads and bridges.
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Equation of a Straight Line
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The equation of a straight line can be represented differently on a cartesian plane. It is also known as linear equations. The equation depends upon the variable, angles and constant.
- The slope is used to determine the direction of a straight line.
- It also calculates how steep the line is formed.
- The slope is also known as rise over run.
There are various ways of representing the equation of a straight which are as follows:
General Equation of a Straight Line
The general equation of a straight line can be represented as follows:
ax + by + c = 0
- where
- a, b, c are constants, and
- x, y are variables.
- The slope of the line is given as -a/b
Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the slope-point form equation below.
y – y0 = m (x – x0)
Slope-Point Form
Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
- The slope of P2P = The slope of P1P2
- As a result, the equation of the line going through the points (x1, y1) and (x2, y2) is
Two-Point Form
Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin. The distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c).
- The slope of the line L is m, and it passes through a fixed point (0, c).
- The equation of the line L thus obtained from the slope – intercept form is given by
y – c =m ( x - 0 )
- As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c
Note: Depending on whether the intercept is formed on the positive or negative side of the y-axis, the value of c will be positive or negative.
Slope-Intercept Form
Intercept Form
Consider a line L that intersects the axes at x-intercept and y-intercept b. As a result, L intersects the x-axis at (a, 0) and the y-axis at (b, 0). (0, b) We derive the following from the two-point form of the line equation:
- As a result, the intercept form becomes
- The equation of the line with the intercepts a and b on the x- and y-axes, respectively, is
Intercept Form
Slope of a Straight Line
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If the inclination of a line l, tan is called the slope or gradient of that line. The slope of a line with an inclination of less than 90 degrees. The letter ‘m' stands for it. As a result, m = tan θ, where, θ ≠ 90°
- When the slope of the y-axis is unknown, the slope of the x-axis is assumed to be zero.
Angles between Straight Line
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When generating an angle arc from one point to another, a straight line generates a 180-degree angle. Lines or straight lines are idealised representations of such objects, which are typically defined in terms of two points or a single line.
- Consider the two non-vertical lines L1 and L2 with slopes m1 and m2, respectively.
- α1 and α2 are the respective inclinations of lines L1 and L2.
- The slope of the lines m1 and m2 is then calculated as
m1 = tanα1 and m2= tanα2
- When two lines intersect, they form two pairs of vertically opposite angles.
- The total of any two adjacent angles being 180 degrees from the property.
- Assume that and are the intersecting angles of the lines L1 and L2.
θ = α2– α1 and α1, α2≠ 90°
- Hence,
tan θ = tan(α2-α1) = (tan α2 – tan α1)/(1 + tan α1 tan α2)
tan θ = (m2-m1)/(1+m1m2)
- 1+m1m2 ≠ 0 and φ = 180° – θ, so;
tan φ = tan (180° – θ ) = -tan θ = -(m2-m1)/(1+m1m2)
Taking two cases here, we have,
Case I
If (m2-m1)/(1+m1m2) is positive, then tan θ and tan φ will be positive and negative respectively. It indicates that θ and φ are acute and obtuse, respectively.
Case II
If (m2-m1)/(1+m1m2) is negative, then tan θ and tan φ will be negative and positive respectively, indicating that θ and φ are obtuse and acute, respectively. As a result, the acute angle formed by the lines L1 and L2 with slopes m1 and m2 is given by
- Where, 1 + m1m2 ≠ 0
- The obtuse angle is given by φ =180°– θ.
The video below explains this:
Straight Lines Detailed Video Explanation:
Types of Straight Line
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Various kind of straight lines are as follows:
Parallel Line
Straight lines can be drawn individually or in pairs. Parallel pairs of straight lines can be created. Two parallel lines are separated by such a large distance that they never move closer and always get further apart. The sign II is used to represent them.
Horizontal Straight Lines
Straight lines can move horizontally, meaning they can move left and right of the viewing place indefinitely.
Vertical Straight Lines
Straight lines can travel in a vertical direction, rising above and falling below the viewing point indefinitely.
Diagonal Straight Lines
Straight lines can be drawn in a diagonal direction, that is, at any angle other than horizontal or vertical.
Intersecting Straight Lines
At any angle, two straight lines intersect one another. They are perpendicular when two straight lines bisect at a perpendicular distance of 90°, and are denoted by the letter L.
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Types of Slope
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There are three types of slope which are as follows:
Zero Slope
If the angle formed along the x-axis of straight line is 0 then slope of the line is 0. As we know that slope of a line can be written as m = tan θ. In such cases θ = 0 then in such cases m = 0.
Positive Slope
If the angle formed along the x-axis of straight line is between 0o and 90o then slope of the line is called positive slope.
Negative Slope
If the angle formed along the x-axis of straight line is 90o and 180o then slope of the line is called negative slope.
Infinite Slope
If the angle formed along the x-axis of straight line is 90o then slope of the line is called infinite slope.
Straight Line Formulas
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The important straight line formulas are as follows:
- A point's distance from a line is given by
Let's say the perpendicular distance (d) between a point (x1, y1) and a line Ax + By+ C = 0 is defined by
- Distance Between Two Parallel Lines
The distance d between two parallel lines, such as y= mx+ c1 and y= mx + c
2, is given by
Consider the line's general form, Ax + By + C1=0 and Ax+By+C
2=0.
Things to Remember
- A straight line will help calculate the distance between two extreme points of a line.
- These lines are of infinite length.
- Straight lines have an area and volume equivalent to zero.
- A unique line has the capability to pass through two points.
- It has only one dimension.
Sample Questions
Ques. Calculate the slope of a line passing through the origin and also find the midpoint of the segment joining the points P (0, -4) and B (8, 0). (3 marks)
Ans. Given that,
The midpoint of the line segment joining the points P (0, -4) and B (8, 0):
=[(0+8)/2 , (-4+0)/2]
= (4, -2)
We know that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) can be given as-
m = (y2 -y1) / ( (x2 -x1), where (x2 is not equal to x1)
So, the slope of the line passing through the points (0, 0,) and (4, -2) is
m= (-2-0)/(4-0)
m= -2/4
m= -½
Hence, the required slope of the line is -½.
Ques. Find the slope of a line, which passes through the origin, and the midpoint of the line segment joining the point P (0,-4) and Q(8,0). (3 marks)
Ans. Let M be the midpoint of segment PQ then
Slope of the line:
Ques. Without using the Pythagoras theorem show that the points (4,4), (3,5) and (-1,-1) are the vertices of a right angled triangle. (2 marks)
Ans. The given points are A (4,4), B (3,5) and C (-1,-1)
Slope of line AB = 5-43-4= -1
Slope of line BC = -1-5-1-3= 32
Slope of line BC = -1-4-1-4= 1
Slope of AB x Slope of AC = -1
Hence triangles ABC is right angled at A.
Ques. Find the points on the x-axis which has a distance of 4 units from the line equation (x/3) + (y/4) = 1. (4 marks)
Ans. Given that,
Equation of a line:
(x/3) + (y/4) = 1
It can be written as:
4x + 3y -12 = 0 …(1)
On comparing the eq (1) with general equation of line Ax + By + C = 0,
we get the values A = 4, B = 3, and C = -12.
Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.
As we know that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
D = |Ax1+By1 + C|/ √A2 + B2
On substituting the values in the above formula, we get:
4 = |4a+0 + -12|/ √42 + 32
⇒4 = |4a-12|/5
⇒|4a-12| = 20
⇒± (4a-12)= 20
⇒ (4a-12)= 20 or -(4a-12) =20
Therefore, it can be written as:
(4a-12)= 20
4a = 20+12
4a = 32
a = 8
(or)
-(4a-12) =20
-4a +12 =20
-4a = 20-12
-4a= 8
a= -2
⇒ a= 8 or -2
Hence, the required points on the x axis are (-2, 0) and (8, 0).
Ques. Find equation of the line which is in the middle of the parallel lines 9x+6y-7= 0 and 3x+2y+6=0. (3 marks)
Ans. The equations of the given line are-
9x+6y-7=0
Let the eq. of the line in the middle of the parallel lines (i) and (ii) be
According to the question-
distance between (i) and (iii) = distance between (ii) and (iii)
So, the required equation of the given line is-
3x+2y+116= 0
Ques. Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3. (5 marks)
Ans. Given that the equation of the line- x – 7y + 5 = 0.
The general equation of a line is y = mx+c
Thus, the above equation can be written as:
y= (1/7)x + (5/7)
From the above equation, we can say that,
The slope of the given line is
m = 5/7
The slope of the line perpendicular to the line having a slope of 1/7 is
m = -1/(1/7) = -7
Hence, the equation of a line with slope -7 and intercept 3 is given as:
y = m (x – d)
⇒ y= -7(x-3)
⇒ y= -7x + 21
7x+ y = 21
Hence, the equation of a line which is perpendicular to the line x – 7y + 5 = 0 with x-intercept 3 can be given as
7x+ y = 21.
Ques. Calculate the slope of a line passing through the origin and also find the midpoint of the segment joining the points P (0, -6) and B (1-, 0). (3 marks)
Ans. Given that,
The midpoint of the line segment joining the points P (0, -5) and B (7, 0):
=[(0+10)/2 , (-6+0)/2]
= (5, -3)
We know that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) can be given as-
m = (y2 -y1) / ( (x2 -x1), where (x2 is not equal to x1)
So, the slope of the line passing through the points (0, 0,) and (4, -2) is
m= (-3-0)/(5-0)
m= -3/5
Ques. Find the equation of the line perpendicular to the line x – 8y + 3 = 0 and having x-intercept 5. (5 marks)
Ans. Given that the equation of the line- x – 8y + 3 = 0.
The general equation of a line is y = mx+c
Thus, the above equation can be written as:
y= (1/8)x + (3/8)
From the above equation, we can say that,
The slope of the given line is
m = 3/8
The slope of the line perpendicular to the line having a slope of 1/7 is
m = -1/(1/8) = -8
Hence, the equation of a line with slope -7 and intercept 3 is given as:
y = m (x – d)
⇒ y= -8(x-5)
⇒ y= -8x + 40
8x+ y = 40
Ques. Without using the Pythagoras theorem show that the points (4,3), (3,5) and (-1,-1) are the vertices of a right angled triangle. (2 marks)
Ans. The given points are A (4,3), B (3,5) and C (-1,-1)
Slope of line AB = 5-4-3-3= -1
Slope of line BC = -1-5-1-3= 32
Slope of line BC = -1-6-1-6= 1
Slope of AB x Slope of AC = -1
Hence triangles ABC is right angled at A.
Ques. Find the points on the x-axis which has a distance of 6 units from the line equation (x/5) + (y/12) = 1. (5 marks)
Ans. Given that,
Equation of a line:
(x/5) + (y/12) = 1
It can be written as:
12x + 5y -60 = 0 …(1)
On comparing the eq (1) with general equation of line Ax + By + C = 0,
we get the values A = 12, B = 5, and C = -60.
Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.
As we know that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
D = |Ax1+By1 + C|/ √A2 + B2
On substituting the values in the above formula, we get:
6 = |12a – 60|/ √122 + 52
⇒6 = |12a – 60|/13
⇒|12a-60| = 78
⇒± (12a-60)= 78
⇒ (12a-60)= 78 or -(12a-60) =78
Therefore, it can be written as:
(12a-60)= 78
12a = 78 + 60
12a = 138
a = 11.5
(or)
-(12a-60) =78
-12a +60 =78
a =-1.5
Ques. Calculate the slope of a line passing through the origin and also find the midpoint of the segment joining the points P (0, -8) and B (10, 0). (3 marks)
Ans. Given that,
The midpoint of the line segment joining the points P (0, -8) and B (10, 0):
=[(0+10)/2 , (-8+0)/2]
= (5, -4)
We know that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) can be given as-
m = (y2 -y1) / ( (x2 -x1), where (x2 is not equal to x1)
So, the slope of the line passing through the points (0, 0,) and (4, -2) is
m= (-4-0)/(5-0)
m= -4/5
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