0
$\begingroup$

The question was to find a differential equation representing non horizontal lines and the line eqn ax+by=1 was differentiated wrt x and. Exactly the opposite was done in case of non verical lines. I wonder why it was done..

$\endgroup$
0
$\begingroup$

Any line can be written as

$$ax+by=1\tag{1}$$

where at least one of $a$ and $b$ is nonzero. The line is horizontal if and only if $a=0$ and vertical if and only if $b=0$.


In order for $(1)$ to implicitly define $y$ as a function of $x$ it must be (by the Linear Implicit Function Theorem) that $b\neq0$, i.e. it must be that the line is not vertical. In fact when $b\neq 0$ we can solve explicitly for $y$ as a function of $x$: $$y(x)=\frac{1-ax}{b}$$

If $b=0$, the line is vertical and for each $x$ there are many corresponding $y$-values on the graph of the line, and so the line is not the graph of a function of $x$.

Similarly, in order for $(1)$ to implicitly define $x$ as a function of $y$ it must be that $a\neq 0$, i.e. it must be that the line is not horizontal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.