Straight Line

 Straight Line

A Straight line is a line joining two points.

Suppose there are two points A(x1,y1) and B(x2,y2) in the coordinate system.

Then, the Equation of the line joining A and B is given by

y-y1=m(x-x1)





where m is the slope of the line AB.

m is given by

m=(y2-y1)/(x2-x1).

An example of straight line is given below.


Example:

There are two points namely P(1,2) and Q(3,-8). Give the equation of the line joining the two points.

Solution:

As discussed above, 

The line joining two points is given by

y-y1=m(x-x1)

First, we need to calculate slope of the line

m=(-8-2)/(3-1)=-10/2=-5

=> y-2=-5(x-1)

=> y-2=-5x+5

=> 5x+y-7=0

Therefore,

Equation of the line joining P and Q is given by 5x+y-7=0



  • General equation of a straight line is given by y=ax+b


Condition for the lines to be parallel

If the two lines  a_1x+b_1y+c_1=0 and  a_2x+b_2y+c_2=0 are parallel, then

\\m_1=m_2 \\\\\Rightarrow \frac{-a_1}{b_1}=\frac{-a_2}{b_2} \\\\\\\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}


Condition for the lines to be perpendicular

If the two lines  a_1x+b_1y+c_1=0 and  a_2x+b_2y+c_2=0 are perpendicular, then

\\\Rightarrow m_1m_2=-1 \\\\\Rightarrow \left ( \frac{-a_1}{b_1} \right ) \left ( \frac{-a_2}{b_2} \right )=-1 \\\\\\\Rightarrow a_1a_2+b_1b_2=0


From the above data,

We can say that, The lines a_1x+b_1y+c_1=0 and a_2x+b_2y+c_2=0 are

Coincident if  

 \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

Parallel if

\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}

Intersecting if

\frac{a_1}{a_2}\neq\frac{b_1}{b_2}

Perpendicular if

a_1a_2+b_1b_2=0


Intercept form of the line

If a line cuts intercepts a and b units from origin respectively on x-axis and y-axis, then intercept form of the line is given by

\frac{x}{a}+\frac{y}{b}=1

Proof:

If the cuts intercepts a units from origin on x-axis and b units from origin on y-axis, then, the line passes through (a,0) and (0,b).

Therefore, the equation of the line passing through two points is given by

\\\Rightarrow y-0=\frac{b-0}{0-a}(x-a) \\\\\\\Rightarrow y=-\frac{bx}{a}+b \\\\\\\Rightarrow \frac{bx}{a}+y=b \\\\\\\Rightarrow \frac{x}{a}+\frac{y}{b}=1

Hence proved


Normal form of a line

Normal form of a line is given by

x\, cos\, \alpha+y\, sin\, \alpha=p

where p is length of the perpendicular line drawn from the origin to the line.

And α is the angle made by the perpendicular with the x-axis.


Distance of a point from a line

The distance of the a point (x1,y1) from the line ax+by+c=0 is given by

\left | \frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \right |


Distance between two parallel lines

The distance between two parallel lines ax+by+c=0 and ax+by+d=0 is given by

\left | \frac{d-c}{\sqrt{a^2+b^2}} \right |

Concurrency of three lines

Three lines are said to be concurrent if they pass through the same point.

If the equations of three concurrent lines are

a1x+b1y+c1=0

a2x+b2y+c2=0

a3x+b3y+c3=0

The condition for three lines to be concurrent is given by

\begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}=0


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