Equations of straight lines and circles  (12.112.4)
Understand the equation of a straight line

Theory1) Given the slope of a line L is m and it passes through a point \(\left( {{{\bf{x}}_1},\;{{\bf{y}}_1}} \right)\), where \({{\bf{x}}_1}\) and \({{\bf{y}}_1}\) are constants.
The equation of L is given by
\({\bf{L}}:\;\frac{{{\bf{y}}  {{\bf{y}}_1}}}{{{\bf{x}}  {{\bf{x}}_1}}} = m\)

ExamplesGiven the slope of a line is 1 and it passes through ( 7 , 3 ),
find the equation of the line. 
SolutionsThe equation of the required line is,
\(\frac{{y  \left( {  3} \right)}}{{x  7}} =  1\)
\(y + 3 =  x + 7\)
\(x + y  4 = 0\)

Theory2) Given a line L passes through points \(\left( {{{\bf{x}}_1},\;{{\bf{y}}_1}} \right)\) and \(\left( {{{\bf{x}}_2},\;{{\bf{y}}_2}} \right)\), where \({{\bf{x}}_1}\), \({{\bf{x}}_2}\), \({{\bf{y}}_1}\) and \({{\bf{y}}_2}\) are constants.
The equation of L is given by
\({\bf{L}}:\;\frac{{{\bf{y}}  {{\bf{y}}_1}}}{{{\bf{x}}  {{\bf{x}}_1}}} = \frac{{{{\bf{y}}_2}  {{\bf{y}}_1}}}{{{{\bf{x}}_2}  {{\bf{x}}_1}}}\)

ExamplesFind the equation of the straight line passing through A(10,2) and B(4, 1).

SolutionsBy the twopoint form,
the equation of the straight line is
\(\begin{array}{1}y  1 = \frac{{  {\rm{ }}2  1}}{{10  4}}(x  4)\\y  1 = \frac{{  {\rm{ }}1}}{2}{(x  4)^{}}\\2y  2 =  x + 4\\x+ 2y  6 = 0\end{array}\)

Theory3) Given L is a straight line and its equation is given by
\({\bf{L}}:\;{\bf{y}} = {\bf{mx}} + {\bf{c}}\)
then, m is the slope of the line and c is the y intercept of the lines 
ExamplesGiven \(L:x  2y  1 = 0\), find the slope, y – intercept and x – intercept of L.

Solutions\(x  2y  1 = 0\)
\(y = \frac{1}{2}x  \frac{1}{2}\)
\(\therefore m = \frac{1}{2}\)
\(c =  \frac{1}{2}\)
Put \(y = 0,\;x = 1\)
\(\therefore \) the slope, y – intercept and x – intercept of L are
\(\frac{1}{2}\) ,\(  \frac{1}{2}\) and 1 respectively