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# Equation of a Line

Top
 Sub Topics A straight line is a curve, where every point on the line segment joins any two points lying on it. So, every first degree equation in x and y represents a straight line. There is always a fixed relationship between the x and y co-ordinates of any points on the line. For example: x + y = 0, 4x + y = 0, y = 0.A straight line, when seen in a two-dimensional space, it contains a unique property, where the ratio of the difference in y- coordinates of any points on the line to the difference of their x-coordinates is always the same. That is, how the slope of the straight line remains constant and can be calculated from any two given points on the line.

## Standard Form of Equation of a Line

The Standard form of equation of line is defined as Ax + By = C, where A $\neq$ 0 and B $\neq$ 0.
A and B are the coefficients of x and y respectively. The ordered pair r = (x, y) can be defined as a point.

Standard form of equation is mainly used under the following circumstances:
• When we want to graph a line
• When we want to know the y-intercept of the line
• When we want to know the x-intercept

## Equation of Line Formula

The general formula of equation of line passing through a point ($x_1,y_1$) and having slope m is given as

Here, 'm' is the slope of the line
x1 is the Co-ordinate of x-axis
y1 is the Co-ordinate of y-axis

Here are few solved example on the equation of line:

### Solved Examples

Question 1: Solve the equation of the lines whose slope is 7 and one of the point is (4, 5).
Solution:
Given: m = 7

(x1, y1) <-> (4, 5)

Using the formula of the equation of line, we get

$y-y_{1}$ = $m(x-x_{1})$

y - 5 = 7(x - 4)

y - 5 = 7x - 28

7x - y - 28 + 5 = 0

7x - y - 23 = 0

This is the required equation of line.

Question 2: Calculate the equation of the line for which the slope is 3 and passes through the point (8, 9).
Solution:
Given: m = 3

(x1, y1) <-> (8, 9)

Using the formula of equation of the line, we get

$y-y_{1}$ = $m(x-x_{1})$

(y - 9) = 3( x - 8 )

y - 9 = 3x - 24

3x - y - 24 + 9 = 0

3x - y - 15 = 0

## Slope Intercept Equation of a Line

The slope intercept form of a line is defined as a straight line on the co-ordinate plane that can be explained by the equation

Where, m is the slope of the line and b is the y-intercept and x and y are the co-ordinates of any point on the line.

Slope (m) is the 'steepness' of the line and 'b' is the intercept, that is the point where the line crosses the y-axis.

This equation of the line is mainly used in the following circumstances:
• To define a particular line in an accurate way.
• To locate the points on the line.
Given below are the few solved problems based on the slope intercept equation of a line.

### Solved Examples

Question 1: The slope of a line is 7 and the y-intercept is 4. Calculate the equation of the line.
Solution:
Step 1: Given:

m = 7 and b = 4

Using the formula y = mx + b, we get

y = 7x + 4

7x - y + 4 = 0

$\therefore$ The equation of the line is 7x - y + 4 = 0

Step 2: Given:

m = 7 and b = 4

Using the formula y = mx + b, we get

y = 7x + 4

7x - y + 4 = 0

$\therefore$ The equation of the line is 7x - y + 4 = 0

Question 2: Calculate the equation of the line of the slope 0.2 and y-intercept 7.
Solution:
Given:

m = 0.2 and c = 7

Using the formula y = mx + c, we get

y = 0.2 (x) + 7

y = $\frac{x}{5}$ + 7

5y = x + 35

x - 5y + 35 = 0

$\therefore$ The equation of the line is x - 5y + 35 = 0.

## Parametric Equation of the Line

Parametric equations are the one where the Cartesian co-ordinate of a curve or a surface are represented as the function of the same variable.

The parametric equation in the xy - plane is explained as x = x(t) and y = y(t)

x and y denote the co-ordinate of the graph of a curve in the plane. The parametric equation of the line is defined as the line on a co-ordinate plane given with the point P1 (x1, y1) and the direction vector s then, the position vector r of any point (x, y) of the line.

r = r1 + t.s, - $\infty$ < t < + $\infty$

and where, r1 = x1i + y1j and s = xsi + ysj, represents the vector equation of the line.
Therefore, any point of line can be reached by the radius vector

r = xi + yj = (x1 + xst)i + (y1 + yst)j

$\therefore$ the scalar quantity 't' can take any real value from - $\infty$ to + $\infty$. By rewriting the scalar components of the above equation, we get the parametric equation of the line as

x = x1 + xs.t
y = y1 + ys.t