### Featured Posts

### Straight Lines Class 11 | Ch-09

- Get link
- Other Apps

__Straight Lines Class 11 Chapter 09__

__Straight Lines Class 11 Chapter 09__

*Basic concepts of straight line chapter 9 standard XI, slope of the line, different forms of equation of lines and general equation of line.*

**To understand this topic properly we should know the coordinate geometry. Coordinate geometry is the field of mathematics in which we solve the geometrical problems with the help of algebra. In coordinate geometry class 10 we learnt about distance formula, section formula, mid point formula and area of triangle.****Angle of inclination of the line**

**The
angle ****Î¸**** made by the line l with positive
direction of x- axis and measured anti-clockwise is called the angle of inclination
of the line**

**Angle of
inclination of the x-axis is 0 ^{o}. **

**Angle of inclination of all the lines parallel to the x- axis is also 0**

^{o}.**Angle of
inclination of the y-axis is 90 ^{o}.
Angle of inclination of all the lines parallel to the y- axis is also 90^{o}.**

**Slope of
the line **

**If 'Î¸' is the
angle of inclination of the line l then tan**

**Î¸**

**is called the slope or gradient of the line**

*l*. Slope is denoted by m**Therefore slope of line l = m = tan**

**Î¸**

**Slope of x-axis = tan0 = 0**

**Slope of y - axis = tan****90 ^{o}**

**= ∞ or undefined**

**Slope of the line passing through the two points ****P(x _{1},
y_{1}), and Q**

**(x**

_{2}, y_{2)}

**\[m=tan\theta =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]Two lines are parallel if their slopes are equal or**

**m**_{1}= m_{2}

**Two lines are perpendicular if product of their slopes equal to -1 or**

**m**_{1}**x**

**m**_{2}**= -1**

**Angle between two lines**

**If**

**m**_{1}, m_{2}**are the slopes of two given lines and if**

**'Î¸' is the angle between two lines then**

**\[tan\theta =\left |\frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |=\pm \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\]**

**Three points A, B, C are collinear to each other if**

**Slope of AB = Slope of BC**

__Different forms of equation of line__**We know that every point in a plane have**

**infinitely**

**many points lie on it. If (x, y) is any arbitrary point on it then it represent every point on the line.**

**Equations of Horizontal Lines**

**If a horizontal line**

**L**_{1}or L_{2 }**is at a distance a from the x-axis then ordinate of every point lying on the line is either a or –a.**

**Therefore equation of the line is either of the form y = a or y = - a**

**If a line is above the x-axis then choose +ve sign and if a line below the x-axis then choose –ve sign.**

**Equations of Vertical Lines**

**If a vertical line**

**L**_{1}or L_{2}**is at a distance b from the y-axis then ordinate of every point lying on the line is either b or –b. Therefore equation of the line is either of the form x = b or x = - b**

**If a line is to the right of y-axis then choose +ve sign and if a line is to the left of the y-axis then choose –ve sign.**

**Point Slope form of equation of line**

**This is also known as one point form of equation of line.**

**Let a line L with slope m passes through the point Q****(x _{1}, y_{1}). If P(x, y) is any arbitrary point on the line L, **

**Then equation of line is given by**

** y - y _{1} = m(x -
x_{1}) where**

** ****\[m=tan\theta =\frac{y-y_{1}}{x-x_{1}}\]**

__Two Point form of equation of line__

**Let a line L passing through two points**

**P(x**

_{1}, y_{1}) and Q**(x**

_{2}, y_{2}). Slope m of the line is given by**\[m=tan\theta =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]. If R(x, y) is any arbitrary point on the line then equation of the line is given by**

**\[y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})\]**

**Slope intercept form of equation of line**

**Let a line L with slope m cut the y-axis at a distance c from the**** origin at point (0, c). Then c is called the y-intercept of the line. If P(x, y) is any arbitrary point on the line then equation of the line is given by**

**y - c = m(x - 0) ⇒ y = mx + c**

**If line L with slope m cut the x-axis at a distance c from the origin at point (c, 0). ****Then c is called the x-intercept of the line. If P(x, y) is any arbitrary point on the line then equation of the line is given by**

**y - 0 = m(x - c) ⇒ y = m(x + c)**

**Intercept form of equation of line**

**Let L is a line which make intercept a on the x-axis and intercept b on y-axis, then slope of the line is given by ****\[Slope(m)=\frac{b-0}{0-a}=\frac{-b}{a}\]Equation of line is given by****\[y-0=\frac{-b}{a}(x-a)\]\[ay=-bx+ab\]\[ay+bx=ab\]\[\frac{ay}{ab}+\frac{bx}{ab}=\frac{ab}{ab}\]\[\frac{y}{b}+\frac{x}{a}=1\]\[\frac{x}{a}+\frac{y}{b}=1\]This equation is called the intercept form of equation of line.**

**Normal form of equation of line**

**Let L is a line whose perpendicular distance from the origin is p and perpendicular OA makes an angle ****Î¸ with the x-axis measured anti-clockwise direction, then Normal form of the equation of line is given by**

**xcos****Î¸ + ysin****Î¸ = p**** **

**General equation of line**

**An equation of the form ax + by + c = 0 is called linear equation or general equation of line. In this equation a and b both collectively cannot be zero.**

**To find the slope from the general eqn. of line\[Slope(m) = \frac{-coefficient of x}{coefficient of y}=\frac{-a}{b}\]**

**Perpendicular distance of a point from the line**

_{1}, Y

_{1}) from the line ax + by + c = 0 is given by\[d=\left | \frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right |\]

**Distance between two parallel lines**

**Let two parallel lines are :**

**ax + by + c _{1} = 0,**

**ax +
by + c _{2} = 0**

**Perpendicular distance between these two parallel lines is given by**

**\[d=\left | \frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}} \right |\]**

**Method of converting the general equations into the Normal form of equation of line.**

**Let us suppose that the equation of the line in general form is**

**\[x-\sqrt{3}y+8=0\]**

**\[Dividing\; on\; both\; sides\; by\; \; \sqrt{a^{2}+b^{2}} \; or\; \; \sqrt{(-1)^{2}+\left (\sqrt{3} \right )^{2}}=2\]**

**\[\frac{-x}{2}+\frac{\sqrt{3y}}{2}=4\]**

**\[tan\theta =\frac{sin\theta }{cos\theta }=\frac{\sqrt{3}/2}{-1/2}=-\sqrt{3}\]**

**\[\Rightarrow \theta =\frac{2\pi }{3}\; or\; 120^{o}\]**

**THANKS FOR YOUR VISIT**

**PLEASE COMMENT BELOW**

- Get link
- Other Apps

## Comments

## Post a Comment