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A straight line is defined as a line formed by points moving in the same direction with zero curvature. It is the shortest distance between two points. The chapter straight lines focus on the fundamental concepts of lines, such as slopes, angles between two lines, various sorts of lines, and distance between lines, along with formulas for slope, angles between lines, congruence etc. The general form of a straight line can be understood using the basic formula - mx+ny+o = 0 wherein m, x and o are constants and x and y are variables.
Also read: Section Formula in Coordinate Geometry
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Key terms: straight line, straight line formula, angle between two lines, relation between two lines, slope of a line
Distance Formula
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The distance between two points on a graph can be found using the distance formula, let us assume the two points are P (x1 y1) and Q (x2 y2).
D = √ (x2 - x1 )2 + (y2 - y1 )2 The derivation of this formula is explained below.
Keeping the Pythagoras theorem in mind:
Distance Formula
If the lengths of the sides are A and B then:
(AB)2=(AC)2+(BC)2
If we solve for AB then:
AB = √(AC)2 + (BC)2
Now since the distance AC is the horizontal axis (x2 - x1) and similarly the distance BC is the vertical axis and hence (y2 - y1). Applying this to the Pythagoras formula
AB = √ (x2 - x1 )2 + (y2 - y1 )2 and therefore the distance formula.
For example: P (x1 y1) = (−1,0) Q (x2 y2) = (2,7)
PQ = √ (2-(-1))2 + (7-0)2
= √32 + 72
= √9+49
= √58
= 7.6
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| Related Articles | ||
|---|---|---|
| Angle Between Two Planes | Motion in a Straight Line | Distance between Two Lines |
| Intercept | Slope formula | Slope Intercept Form Formula |
The video below explains this:
Straight Lines Detailed Video Explanation:
Midpoint Formula
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The formula to find the midpoint of a line on the graph is ((x1+x2/2), (y2+y1/2))
In order to derive this formula, the two points can be plotted on a graph as shown below:
Midpoint Formula
The blue dot on the image is the midpoint. The expression for the x coordinate of the midpoint is (x1+x2/2) and similarly the expression for the y coordinate of the midpoint (y2+y1/2) respectively. One might wonder why this expression works, for example, let us assume, x1 = 3 and x2 = 7 therefore 3+7/2 which equals 5 and it is logical because 5 lies in between 3 and 7. This is the same for the y coordinates and therefore the formula is derived as the average of the x coordinates and average of the y coordinates.
Slope of the Line
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Assuming that A (x1 y1) and B (x2 y2) are two points on a non-vertical, slanting line whose inclination is θ. The inclination l of the line can either be acute or obtuse, this determines the slope formula, let us see both the cases using an example.
Firstly, when the angle θ is acute:
Slope of the Line
∠CAB = θ and hence slope of the line (l) = m = tan θ
tan θ = CA/CB = y2 - y1 / x2 - x1
So, m = y2 - y1 / x2 - x1
Secondly when the angle θ is obtuse:
Slope of the Line
∠CAB = 180° – θ
Therefore, θ = 180° - ∠CAB
Since slope of the line (l) = m
M = tan θ
= tan (180° – ∠CAB) = – tan ∠CAB
-(MQ/MP) = - y2 - y1 / x2 - x1 = y2 - y1 / x2 - x1
This shows that both cases end up with the slope formula m = y2 - y1 / x2 - x1
Perpendicularity and Parallelism
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The conditions for perpendicularity and parallelism of 2 lines on the same graph depending on the slopes of the same. Let us assume line 1 is L1 and line 2 is L2 and their slopes are m1 and m2 respectively.
If line L1 is parallel to line L2 then the slope of these lines is said to be equal (i.e.) m1 = m2
And therefore, two lines can be concluded as parallel if their slopes are equal.
If line L1 is perpendicular to line L2 then the product of their slopes is -1 therefore m1m2 = -1.
Angle between two Lines
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Assume that L1 and L2 are non-vertical lines with slopes m1 and m2 and inclinations α1 and α2 respectively, this means that m1= tan α1 and m2 = tan α2
Angle between two Lines
It is also known that θ = α2 – α1 and α1, α2 ≠ 90°
Therefore, tan θ = tan (a2-a1) = (tan a2 – tan a1) / (1- tan a1tan a2)
Substituting the values of tan a1 and tan a2 as m1 and m2 respectively, we have,
tan= (m2 – m1) / (1+m1m2)
It should be noted that the value of tan θ in this equation will be positive if θ is acute and negative if θ is obtuse.
Real Life Applications of Slope formula
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There are many applications to the slope formula discussed in this chapter and a few of them are listed below:
- Grade of a road
- Pitch of a roof
- Building wheelchair ramps
- Building stairs
Slope-intercept formula
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Let us say a line L has a slope m and is cutting the y-axis at a distance c from the origin, this makes the point c the y-intercept of the line. This point is defined as (0, c) and according to the point-slope formula the equation for line L now becomes:
Y - c = m (x - 0) which is nothing but y = mx + c
The value of c can vary as positive or negative according to which side of the y-axis the intercept is made at.
If in case the line L with slope m makes an x-intercept “d” then the equation of that line is y = m (x - d).
Family of Lines
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The general equation of the family of lines through the point of intersection of two given lines L1 and L2 is given by L1 +λ L2 = 0. Where λ is a parameter.
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Things to Remember
- A straight line is a line that extends to both sides till infinity and has no curves.
- The relation between variables x, y satisfy all points on the curve. The general equation of straight line is: ax + by + c = 0.
- Equation of line with slope ‘m’ passes through (x1, y1) can be given as y – y1 = m(x – x1 )
- An angle bisector has equal perpendicular distance from the two given lines.
- The general equation of the family of lines through the point of intersection of two given lines L1 and L2 can be expressed as L1 +λ L2 = 0.
Previous Years’ Questions
- If m is the slope of one of the lines represented by … (KCET 2010)
- Locus of a point which moves such that its distance from the… (KCET 2011)
- ABC is a triangle G is the centroid D is the mid- point of BC… (KCET 2007)
- If the line… (KCET 2013)
- The angle between the lines… (KCET 2013)
- The minimum area of the triangle formed by the variable line… (KCET 2013)
- If lines represented by… (KCET 2011)
- If one of the slopes of the pair of lines… (KCET 2012)
- If the straight lines… (KCET 2016)
- Let A(6,−1),B(1,3) and C(x,8) be three points such that… (KEAM)
- The locus of a point which is equidistant from the points… (KEAM)
Sample Questions
Ques: Find the slope of the lines passing through the point (3, -2) and (-1,4) [1 mark]
Ans. Slope of line through (3, -2) and (-1, 4)
m = \(\frac{y_2 - y_1}{x_2 - x_1}\)
= 4 - (-2)/ -1 -1
= 6/-4 = -3/2
Ques: Find the measure of the angle between the lines x + y + 7 = 0 and x- y +1 = 0 [1 mark]
Ans. x + y + 7 = 0
m1= -1/1
x - y + 1 = 0
m2= -1/-1 = 1
Slopes of the two lines are 1 and -1 as the product of these two slopes is -1, the lines are at right angles.
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) [4 marks]
Ans. Let mbe the midpoint of segment PQ then
= (4, -2)
Slope of OM = \(\frac{y_2 - y_1}{x_2 - x_1}\)
\(= \frac{-2-0}{4-0} = \frac{-1}{2}\)
Ques: Without using the Pythagoras theorem show that the points (4,4), (3,5) and (-1, -1) are the vertices of a right angled . [4 marks]
Ans. The given points are A(4,4), B(3,5) and C(-1,-1)
Slope of
\(AB = \frac{5-4}{3-4} = -1\)Slope of
\(BC = \frac{-1-5}{-1-3}= \frac{-6}{-4} = \frac{3}{2}\)Slope of
\(AC = \frac{-1-4}{-1-4} = +1\)Slope of AB x slope of AC = -1
⇒ AB ⊥ AC
Hence Δ ABC is right angled at A.
Ques: Find the angle between two lines having slopes of 1, and 1/2 respectively. [2 marks]
Ans: The given slopes of the two lines are m1= 1 and m2= 1/2.
The formula to find the angle between the two lines is Tan = (m2 – m1) / (1+m1m2)
Tan = (1-½)/ (1+½.1)
Tan = (½) / (3/2)
θ = Tan−1 (⅓)
Ques: p(a,b) is the midpoint of a line segment between axes. Show that equation of the line is x/a + y/b = 0 [4 marks]
Ans. Required equation be
x/c + y/d = 1 … (i)
P is the mid-point
Coordinate of p (c/2, d/2)
(a,b) = (c/2, d/2)
a/1 = c/2
c = 2a
b/1 = d/2
d = 2b
Put the value of C and D in eq. (i)
x/2a + y/2b = 1
x/a + y/b = 2
Ques: Find equation of the line mid-way between the parallel lines 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0 [5 marks]
Ans. The equations are
9x + 6y - 7 = 0
\(3( 3x + 2y - \frac{7}{3}) = 0\) \(3x + 2y - \frac{7}{3}\) = 0 …..(i)3x + 2y + 6 = 0 …...(ii)
Let the eq. of the line mid-way between the parallel lines (i) and (ii) be
ATQ
3x + 2y + k = 0 …...(iii)Distance between (i) and (iii) = distance between (ii) and (iii)
| \(\frac{K + \frac{7}{3}}{\sqrt{9+4}}\)| = |\(\frac{K - 6}{\sqrt{9+4}}\)| [\(\because d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}\)]
K +7/3 = K - 6K = 11/6
Required equation is:
3x + 2y +11/ 6 = 0
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