# Equation of a straight line in terms of slope of the line and Y-intercept

Expressing a linear expression of a straight line in algebraic form in terms of slope of the line and intercept at vertical axis is defined equation of the straight line in terms of slope of the line and y-intercept.

In Geometry, straight lines often pass through the vertical y-axis at intercept in the Cartesian coordinate system. It can be represented by the general form of the equation of the straight line but it is transformed into another form due to the passage of the straight line through the vertical axis at an intercept. Expressing the straight line in terms of slope of the line and Y-intercept has priority than representing these types of straight lines with standard equation of the straight line.

## Geometrical Explanation

Assume

$\stackrel{↔}{AB}$

is a straight line and it passes through the vertical y-axis of the Cartesian coordinate system at a y-intercept by making an angle with horizontal x-axis.

Assume, the straight line

$\stackrel{↔}{AB}$

is intersected with the vertical y-axis at point

$A$

which is a common point of both straight line and y-axis and it is located on y-axis at a vertical distance of

$c$

units. Therefore, the coordinates of the point

$A$

is

$\left(0,c\right)$

. Also assume, the point

$B$

is located at (x, y). Assume the angle made by the straight line

$\stackrel{↔}{AB}$

is theta

$\left(\theta \right)$

. Draw a parallel line from point

$A$

and simultaneously draw perpendicular line from point

$B$

and assume both lines get intersected perpendicularly at a point, assumed to call point

$C$

. Thus, the straight line

$\stackrel{↔}{AB}$

is formed a right angled triangle

$\Delta BAC$

. The angle made by the straight line is exactly equal to the angle of the triangle because, the line segment

$\stackrel{‾}{AC}$

is parallel line to the horizontal x-axis. Therefore,

$\angle BAC=\theta$

.

According to the right angled triangle

$\Delta BAC$

,

• The line segment
$\stackrel{‾}{AB}$

is known hypotenuse of the right angled triangle

$\Delta BAC$

.

• The line segment
$\stackrel{‾}{BC}$

is known opposite side of the right angled triangle

$\Delta BAC$

.

• The line segment
$\stackrel{‾}{AC}$

is known adjacent side of the right angled triangle

$\Delta BAC$

.

$tan\theta =\frac{BC}{AC}$

The length of the opposite side is

$BC=OB–OC=y–c$

The length of the adjacent side is

$AC=x$

$⇒tan\theta =\frac{BC}{AC}=\frac{y–c}{x}$

As per the principle of slope of the straight line, the slope of the straight line is written in mathematical form as given here.

$m=tan\theta$

Now replace

$tan\theta$

with its geometrical value.

$⇒m=\frac{y–c}{x}$

$⇒mx=y–c$

$⇒mx+c=y$

$⇒y=mx+c$

It is the required linear algebraic expression which represents a straight line having slope and passes through the vertical y-axis at an intercept.