A cartesian coordinate system describes a point in a plane uniquely by signed distance of the point from a fixed point named the origin.
A straight line is a set of points who satisfy a linear equation. A linear equation is an equation with power of one for the variables and constant. |

For a line passing through cartesian coordinates $(x_{1},y_{1}), (x_{2},y_{2})$, the two-form equation of the line

is $y-y_{1}$ =$\frac{y_{1}-y_{2}}{x_{1}-x_{2}}(x-x_{1})$. The set of points satisfying the

above equation for the values (x, y) will be on the given line.

If two points are (2, 4) and (1, 0) then the slope will be,

*m*= $\frac{0 - 4}{1 - 2}$ = $\frac{-4}{-1}= 4$ .

The equation will be (y - 0) = 4(x - 1) $\rightarrow$ y = 4x - 4.

Now, we will try to see if the point (2, 4) and (2, 3) are on the given line y = 4x - 4.

For (2, 4) put y = 4, x = 2 in the equation y = 4x - 4. It becomes 4x - 4 = 4.2 - 4 = 8 - 4 = 4 = y.

Hence, the point (2, 4) is on the line.

Similarly, for the point (2, 3) put x = 2, y = 3. 4x - 4 = 8 - 4 = 4 â‰ 3 = y.

Hence, this point is not on the given line.

To know the equation of a line, it is important to know the slope of the line. Slope is nothing but the ratio of the change in height from one cartesian point to other to the change in the horizontal distance. For two given points $(x_{1},y_{1}), (x_{2},y_{2})$, the slope formula is $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$. Hence, if the slope be

*m,*the formula for two-point for equation of the line will be (y - $y_1$) = m(x - $x_1$).

Let us try and understand the two-point form equation a line with an example. Let there be two points (1,3) and (4,2). Then the slope of the line becomes, (2-3)/(4-1) = -1/3. The equation of line will be (y - 1) = -1/3(x - 2) ⇒ 3y - 3 = 2 - x $\rightarrow$ 3y + x = 5.

**Example 1:**Find the two-form equation of a line passing through the points (3, 2) and (-2, 1).

**Solution:**We have ($x_1, y_1$) as (3, 2) and ($x_2, y_2$) as (-2, 1). Putting the values in the two-form equation we get,

$\frac{y-2}{1-2}$ = $\frac{x-3}{-2-3}$ $\rightarrow$ $\frac{y-2}{-1}$= $\frac{x-3}{-5}$ $\rightarrow$ 5(

*y*– 2) = (

*x*– 3)$\rightarrow$

*x*– 5

*y*+ 7 = 0.

**Example 2:**Find the slope of the line passing through the points (2, 1) and (0, -1).

**Solution:**The formula for slope is

*m*= $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$. We have ($x_1, y_1$) as (2, 1) and ($x_2, y_2$) as (0, -1). Putting these values in the formula we get,

$m$ = $\frac{-1-1}{0-2}$ = $\frac{-2}{-2}$= 1.

**Example 3:**If the slope of a line is given as -3, and the line passes through the point (1, 1) find the equation of the line.

**Solution:**The equation of the line will be (y - $y_1$) = m(x - $x_1$) where

*m*is the slope. We have

*m*as -3 and ($x_1, y_1$) as (1, 1). Putting the values we get,

y - 1 = -3(x - 1) $\rightarrow$ y - 1 = -3x + 3 $\rightarrow$ 3x + y - 4 = 0.

**Example 4:**Does the line 3x + y - 2 = 0 passes through the point (1, 0)?

**Solution:**To check if a point falls on a line, it should be checked that the point does satisfy the equation of the line or not. We will put x = 1 and y = 0 in the given equation and check.

3.1 + 0 - 2 = 3 - 2 = 1 $\neq$ 0.

Hence, the line 3x + y - 2 = 0 does not passes through the point(1, 0).