A straight line is defined as a line traced by a point traveling in a constant direction with zero curvature. In other words, the shortest distance between two points is called a straight line.
In straight lines class 11, the basic concepts of lines such as slopes, angle between two lines, various forms of lines, the distance between lines are given in detail. Now, let us take a look of the straight line class 11 concepts one by one
The slope of a Line
Tan Î¸ is called the slope or gradient of line l if Î¸ is the inclination of a line l. The slope of a line whose inclination is not equal to 90^{0}. It is denoted by m.
Thus, m = tan Î¸, Î¸ â‰ 90Â°
If the slope of the yaxis is not defined, it is observed that the slope of the xaxis is zero.
An Angle between Two Lines
Let’s take the two nonvertical lines L_{1} and L_{2} with slopes m_{1 }and m_{2}, respectively where Î±_{1} and Î±_{2}are the inclinations of lines L_{1} and L_{2}, respectively. Then the slope of the lines m_{1} and m_{2 } is given as
m_{1} = tanÎ±_{1} and m_{2}= tanÎ±_{2}
We know that, when two lines intersect each other, it makes two pairs of vertically opposite angles such that the sum of any two adjacent angles is 180Â° from the property. Assume that Î¸ and Ï† be the adjacent angles between the lines L_{1 }and L_{2}. Then
Î¸ = Î±_{2}â€“ Î±_{1} and Î±_{1}, Î±_{2}â‰ 90Â°.
Therefore,
\(\tan \theta = \tan (\alpha _{2}\alpha _{1})=\frac{\tan \alpha _{2}\tan \alpha _{1}}{1+\tan \alpha _{1} \tan \alpha 2}\) \(\tan \theta = \frac{m_{2}m_{1}}{1+m_{1}m_{2}}\)Since 1 + m1m2 â‰ 0 and Ï† = 180Â° â€“ Î¸ so that
tan Ï† = tan (180Â° â€“ Î¸ ) = \(\tan \theta = \frac{m_{2}m_{1}}{1+m_{1}m_{2}}\)
Case 1: If \(\frac{m_{2}m_{1}}{1+m_{1}m_{2}}\) is positive, then tan Î¸ will be positive and tan Ï† will be negative, which means that Î¸ will be acute and Ï† will be obtuse.
Case 2: If \(\frac{m_{2}m_{1}}{1+m_{1}m_{2}}\) is negative, then tan Î¸ will be negative and tan Ï† will be positive, that means that Î¸ will be obtuse and Ï† will be acute.
Thus, the acute angle between the lines L_{1} and L_{2} with slopes m_{1} and m_{2} respectively is given by
\(\tan \theta = \left  \frac{m_{2}m_{1}}{1+m_{1}m_{2}} \right \)Where, 1 + m1m2 â‰ 0
Then the obtuse angle can be found by using Ï† =180^{0}â€“ Î¸.
Various Forms of Equations of Line
The various forms of the equation of the line covered in straight line class 11 are as follows
Slope – Point Form
Assume that P_{0}(x_{0}, y_{0}) is a fixed point on a nonvertical line L, whose slope is m. Consider that P (x, y) be an arbitrary point on L then the point (x, y) lies on the line with slope m through the fixed point (x_{0}, y_{0}), if and only if, its coordinates satisfy the following equation
y â€“ y_{0} = m (x â€“ x_{0})
Two – Point Form
Let us take the line L passes through two given points P_{1}(x_{1}, y_{1}) and P_{2}(x_{2}, y_{2}) and P (x, y) be a general point on L. Then the three points P_{1}, P_{2} and P are collinear, therefore, it becomes
The slope of P_{1}P = The slope of P_{1}P_{2}
\(\frac{yy_{1}}{xx_{1}}=\frac{y_{2}y_{1}}{x_{2}x_{1}}\) (or) \(yy_{1}=\frac{y_{2}y_{1}}{x_{2}x_{1}}(xx_{1})\)Thus, the equation of the line passing through the given points (x_{1}, y_{1}) and (x_{2}, y_{2}) is defined by
\(yy_{1}=\frac{y_{2}y_{1}}{x_{2}x_{1}}(xx_{1})\)SlopeIntercept Form
Assume that a line L with slope m cuts the yaxis at a distance c from the origin where the distance c is called the yintercept of the line L. Therefore, the coordinates of the point where the line meets the yaxis are (0, c). So, the line L has slope m and passes through a fixed point (0, c). Thus, from the slope – point form, the equation of the line L is
y – c =m( x 0 )
Therefore, the point (x, y) on the line with slope m and yintercept c lies on the line if and only if
y = m x +c
Note that the value of c will be positive or negative according to the intercept is made on the positive or negative side of the yaxis, respectively.
Intercept Form
Consider a line L that makes xintercept and yintercept b on the axes. So that, L meets xaxis at the point (a, 0) and yaxis at the point (0, b)
From the twopoint form of the equation of the line, we get
\(y0 = \frac{b0}{0a}(xa)\)Therefore, it becomes
\(\frac{x}{a}+\frac{y}{b}=1\)Thus, the equation of the line having the intercepts a and b on xand yaxis respectively is given by
\(\frac{x}{a}+\frac{y}{b}=1\)The Distance of a Point From a Line
Let, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x_{1}, y_{1}) is defined by
\(d=\frac{\left  Ax_{1}+By_{1}+C \right }{\sqrt{A^{2}+B^{2}}}\)The Distance Between Two Parallel Lines
The distance d between the two parallel lines say, y= m x+ c_{1} and y = m x + c_{2} is given by
\(d=\frac{\left  C_{1}C_{2} \right }{\sqrt{1+m^{2}}}\)Consider the general form of the line, i.e., Ax + By + C_{1}= 0 and Ax + By + C_{2}=0
\(d=\frac{\left  c_{1}c_{2} \right }{\sqrt{A^{2}+B^{2}}}\)Sample Problem
Go through the straight line class 11 problem provided here.
Question:
If the three lines 2x + y – 3 =0 , 5x + ky – 3 = 0 and 3x – y – 2 = 0 are concurrent then find the Value of k.
Solution :
If three lines are said to be concurrent when they pass through a common point. It means that the point of intersection of two lines lies on the third line.
Given equation :
2x + y – 3 =0 â€¦â€¦(1)
5x + ky – 3 = 0 â€¦â€¦(2)
3x – y – 2 = 0 â€¦â€¦..(3)
By solving the line equation (1) and (3) using cross multiplication method,
\(\frac{x}{23}=\frac{y}{9+4}=\frac{1}{23}\) (or)x = 1 and y = 1
Therefore, the point of intersection of two lines is given as (1, 1). By substituting the point (1, 1) in equation (2), we get
5.1 + k .1 â€“ 3 = 0 or k = â€“ 2.
Therefore, the value of k is 2.
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