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.

Assume,

$\overleftrightarrow{AB}$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

$\overleftrightarrow{AB}$is intersected the horizontal x-axis at a point

$A$at an x-intercept of

$a$units. Therefore, the point

$A$is located at

$(a,0)$in the Cartesian coordinate system. Similarly, the straight

$\overleftrightarrow{AB}$is intersected the vertical y-axis at a point

$B$at a y-intercept of

$b$units. So, the point

$B$is located at

$(0,b)$in geometric coordinate system. Thus, the straight line

$\overleftrightarrow{AB}$is formed a right angled triangle

$\Delta BAO$. Assume, the straight line

$\overleftrightarrow{AB}$is making an angle theta

$(\theta )$with the horizontal x-axis.

According to right angled triangle

$\Delta BAO$.

The line segment

$\stackrel{\u203e}{AB}$is known hypotenuse of the right angled triangle

$\Delta BAO$.

The line segment

$\stackrel{\u203e}{OA}$is known adjacent side of the right angled triangle

$\Delta BAO$.

The line segment

$\stackrel{\u203e}{OB}$is known opposite side of the right angle triangle

$\Delta BAO$.

Assume, the angle between hypotenuse

$\stackrel{\u203e}{AB}$and adjacent side

$\stackrel{\u203e}{OA}$is alpha

$\alpha $. The angle of the right angled triangle

$\Delta BAO$is

$\angle BAO=\alpha $.

Consider a point on the straight line

$\overleftrightarrow{AB}$and assume it to call point

$C$and also the coordinates of the point

$C$is

$(x,y)$. The point

$C(x,y)$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

$\overleftrightarrow{AB}$.

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

${180}^{\xb0}$because the summation of the angles forms a straight angle.

Therefore,

$\theta +\alpha ={180}^{\xb0}$$\Rightarrow \alpha ={180}^{\xb0}\u2013\theta $

According to right angled triangle

$\Delta BAO$,

$tan\alpha =\frac{OB}{OA}$

The length of the opposite side

$\left(\stackrel{\u203e}{OB}\right)$of the right angled triangle

$\Delta BAO$is

$OB=b$.

The length of the adjacent side

$\left(\stackrel{\u203e}{OA}\right)$of the right angled triangle

$\Delta BAO$is

$OA=a$.

$\Rightarrow tan\alpha =\frac{OB}{OA}=\frac{b}{a}$

Now replace the angle alpha

$(\alpha )$in terms of theta

$(\theta )$.

$\Rightarrow tan({180}^{\xb0}\u2013\theta )=\frac{b}{a}$

${180}^{\xb0}\u2013\theta $

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

$\Rightarrow \u2013tan\theta =\frac{b}{a}$

$\Rightarrow tan\theta =\u2013\frac{b}{a}$

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

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

$\Rightarrow m=\u2013\frac{b}{a}$

Similarly, as per the right angled triangle

$\Delta BCD$,

$tan\alpha =\frac{DB}{DC}$

The length of the opposite side

$\left(\stackrel{\u203e}{DB}\right)$of the right angled triangle

$\Delta BCD$is

$DB=OB\u2013OD=b\u2013y$.

The length of the adjacent side

$\left(\stackrel{\u203e}{DC}\right)$of the right angled triangle

$\Delta BCD$is

$DC=x$.

$tan\alpha =\frac{DB}{DC}=\frac{b\u2013y}{x}$

$\Rightarrow tan({180}^{\xb0}\u2013\theta )=\frac{DB}{DC}=\frac{b\u2013y}{x}$

$\Rightarrow \u2013tan\theta =\frac{b\u2013y}{x}$

$\Rightarrow tan\theta =\u2013\left(\frac{b\u2013y}{x}\right)$

$\Rightarrow m=\u2013\left(\frac{b\u2013y}{x}\right)$

Finally, the slope of the same straight line

$\overleftrightarrow{AB}$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

$\Delta BCD$.

$m=\u2013\frac{b}{a}$

and

$m=\u2013\left(\frac{b\u2013y}{x}\right)$. These two expressions are equal in value because they both represent slope of the same straight line.

$\Rightarrow \u2013\frac{b}{a}=\u2013\left(\frac{b\u2013y}{x}\right)$

$\Rightarrow \frac{b}{a}=\frac{b\u2013y}{x}$

$\Rightarrow \frac{x}{a}=\frac{b\u2013y}{b}$

$\Rightarrow \frac{x}{a}=\frac{b}{b}\u2013\frac{y}{b}$

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

$\Rightarrow \frac{x}{a}+\frac{y}{b}=1$

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.

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