# 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 $\stackrel{‾}{AB}$ and adjacent side $\stackrel{‾}{OA}$ 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.