Question

In the two dimensional plane, prove by using vector methods, the equation of the line whose intercepts on the axes are $a$ and $b$ is $\dfrac{x}{a} + \dfrac{y}{b} = 1$?

Answer 1

Hint: A vector is a quantity in three dimensional space that has not only magnitude but also direction. In 3-dimensional, we use position vectors to denote the position of a point in space. Vector equation of the line is given below:
$\vec r = \vec A + \lambda \vec B$


Complete step by step solution:
Let $\hat i,\hat j$ be the unit vectors along $\overrightarrow {OX} ,\,\overrightarrow {OY} $
Let $A = (a,0),B = (0,b)$
$\overrightarrow {OA} = a\hat i,\,\,\overrightarrow {OB} = b\hat j$
Vector equation of the line is $\vec r = \vec A + \lambda \vec B$
Now, we put the vector of $\vec A\,and\,\vec B$ in the vector equation of the line, we have
$\vec r = a\hat i + \lambda (b\hat j - a\hat i)$
$ = a\hat i + \lambda b\hat j - \lambda a\hat i$
$\vec r = a(1 - \lambda )\hat i + \lambda b\hat j$ $....(i)$
$but\,\vec r = x\hat i + y\hat j$ $....(ii)$
Now substituting the values of $\vec r$ in equation $(i)$, we will get
$x\hat i + y\hat j = a(1 - \lambda )\hat i + \lambda b\hat j$
Comparing the values of $\hat i\,and\,\hat j$separately, we will get
$x = a(1 - \lambda )$
$ \Rightarrow \dfrac{x}{a} = 1\lambda $
$ \Rightarrow \lambda = 1 - \dfrac{x}{a}$ $....(iii)$
And $u = \lambda b$
$ \Rightarrow \lambda = \dfrac{y}{b}$ $....(iv)$
In equation (iii) and (iv), both have the same value $\lambda $, so
$
  \lambda = 1 - \dfrac{x}{a}, \\
  \lambda = \dfrac{y}{b} \\
 $
$\dfrac{y}{b} + \dfrac{x}{a} = 1$
$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} = 1$
Hence proved.


Note: The intercepts are the points where a graph cuts the $x\,\,and\,\,y$ axis i.e. intercepts are the point to intersection where a graph meets the coordinate axis.