### Equations of straight lines and circles - (12.1-12.4)

#### Understand the equation of a straight line

• 1) 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

$${\bf{L}}:\;\frac{{{\bf{y}} - {{\bf{y}}_1}}}{{{\bf{x}} - {{\bf{x}}_1}}} = m$$

• Given the slope of a line is -1 and it passes through ( 7 , -3 ),
find the equation of the line.
• The equation of the required line is,

$$\frac{{y - \left( { - 3} \right)}}{{x - 7}} = - 1$$
$$y + 3 = - x + 7$$
$$x + y - 4 = 0$$

• 2) 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

$${\bf{L}}:\;\frac{{{\bf{y}} - {{\bf{y}}_1}}}{{{\bf{x}} - {{\bf{x}}_1}}} = \frac{{{{\bf{y}}_2} - {{\bf{y}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}}$$

• Find the equation of the straight line passing through A(10,-2) and B(4, 1).
• By the two-point form,
the equation of the straight line is
$$\begin{array}{1}y - 1 = \frac{{ - {\rm{ }}2 - 1}}{{10 - 4}}(x - 4)\\y - 1 = \frac{{ - {\rm{ }}1}}{2}{(x - 4)^{}}\\2y - 2 = - x + 4\\x+ 2y - 6 = 0\end{array}$$
• 3) 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
• Given $$L:x - 2y - 1 = 0$$, find the slope, y – intercept and x – intercept of L.

• $$x - 2y - 1 = 0$$
$$y = \frac{1}{2}x - \frac{1}{2}$$
$$\therefore m = \frac{1}{2}$$
$$c = - \frac{1}{2}$$
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