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

Expressing equation of a straight line in terms of X-intercept and Y-intercept when the straight line intersects both axis is defined Equation of a straight line in terms of X-intercept and Y-intercept.

It is another special case. You have seen straight line with X-intercept and slope and also seen straight line with Y-intercept and slope but this case this entirely different to them. Straight lines may also pass through both axis at an X-intercept as well as at a Y-intercept. It can be also be denoted by the General form of the straight line but it is transformed into X-intercept and Y-intercept form when both intercepts are substituted in the standard equation of the straight line.

## Geometrical Explanation

Assume,

$\stackrel{↔}{AB}$

is a straight line and it crosses the horizontal x-axis at an intercept and also it passes through the vertical y-axis at another intercept.

The straight line

$\stackrel{↔}{AB}$

is intersected the horizontal x-axis at a point

$A$

at an x-intercept of

$a$

units. Therefore, the point

$A$

is located at

$\left(a,0\right)$

in the Cartesian coordinate system. Similarly, the straight

$\stackrel{↔}{AB}$

is intersected the vertical y-axis at a point

$B$

at a y-intercept of

$b$

units. So, the point

$B$

is located at

$\left(0,b\right)$

in geometric coordinate system. Thus, the straight line

$\stackrel{↔}{AB}$

is formed a right angled triangle

$\Delta BAO$

. Assume, the straight line

$\stackrel{↔}{AB}$

is making an angle theta

$\left(\theta \right)$

with the horizontal x-axis.

According to right angled triangle

$\Delta BAO$

.

The line segment

$\stackrel{‾}{AB}$

is known hypotenuse of the right angled triangle

$\Delta BAO$

.

The line segment

$\stackrel{‾}{OA}$

is known adjacent side of the right angled triangle

$\Delta BAO$

.

The line segment

$\stackrel{‾}{OB}$

is known opposite side of the right angle triangle

$\Delta BAO$

.

Assume, the angle between hypotenuse

$\stackrel{‾}{AB}$

$\stackrel{‾}{OA}$

is alpha

$\alpha$

. The angle of the right angled triangle

$\Delta BAO$

is

$\angle BAO=\alpha$

.

Consider a point on the straight line

$\stackrel{↔}{AB}$

and assume it to call point

$C$

and also the coordinates of the point

$C$

is

$\left(x,y\right)$

. The point

$C\left(x,y\right)$

represents each and every point on the line including the points

$A$

and

$B$

. Now, draw a horizontally parallel line from point

$C$

towards vertical axis. Assume, the parallel line perpendicularly meet the y-axis at a point, assumed to call point

$D$

. Thus, another right angled triangle

$\Delta BCD$

is formed by the part of the straight line

$\stackrel{↔}{AB}$

.

Assume, point

$E$

is a point on the horizontal x-axis. The exterior angle of the right angled triangle

$\Delta BAO$

is

$\angle BAE=\theta$

. The summation of the angle of the triangle and exterior angle is

${180}^{°}$

because the summation of the angles forms a straight angle.

Therefore,

$\theta +\alpha ={180}^{°}$

$⇒\alpha ={180}^{°}–\theta$

According to right angled triangle

$\Delta BAO$

,

The length of the opposite side

$\left(\stackrel{‾}{OB}\right)$

of the right angled triangle

$\Delta BAO$

is

$OB=b$

.

The length of the adjacent side

$\left(\stackrel{‾}{OA}\right)$

of the right angled triangle

$\Delta BAO$

is

$OA=a$

.

Now replace the angle alpha

$\left(\alpha \right)$

in terms of theta

$\left(\theta \right)$

.

${180}^{°}–\theta$

means, tangent function is brought to second quadrant. In second quadrant, the trigonometric ratio tangent is negative.

$⇒–tan\theta =\frac{b}{a}$

$⇒tan\theta =–\frac{b}{a}$

According to the concept of slope of a straight line, slope of a straight line

$\left(m\right)=tan\theta$

.

$⇒m=–\frac{b}{a}$

Similarly, as per the right angled triangle

$\Delta BCD$

,

The length of the opposite side

$\left(\stackrel{‾}{DB}\right)$

of the right angled triangle

$\Delta BCD$

is

$DB=OB–OD=b–y$

.

The length of the adjacent side

$\left(\stackrel{‾}{DC}\right)$

of the right angled triangle

$\Delta BCD$

is

$DC=x$

.

$⇒–tan\theta =\frac{b–y}{x}$

$⇒tan\theta =–\left(\frac{b–y}{x}\right)$

$⇒m=–\left(\frac{b–y}{x}\right)$

Finally, the slope of the same straight line

$\stackrel{↔}{AB}$

is obtained in terms of the coordinates of the points of the line and slope of the straight line from both triangles

$\Delta BAO$

and

$\Delta BCD$

.

$m=–\frac{b}{a}$

and

$m=–\left(\frac{b–y}{x}\right)$

. These two expressions are equal in value because they both represent slope of the same straight line.

$⇒–\frac{b}{a}=–\left(\frac{b–y}{x}\right)$

$⇒\frac{b}{a}=\frac{b–y}{x}$

$⇒\frac{x}{a}=\frac{b–y}{b}$

$⇒\frac{x}{a}=\frac{b}{b}–\frac{y}{b}$

$⇒\frac{x}{a}+\frac{y}{b}=\frac{b}{b}$

$⇒\frac{x}{a}+\frac{y}{b}=1$

This algebraic expression represents a straight line when the straight passes through both axis of the Cartesian coordinate system through at x-intercept and y-intercept.