What If It's Zero?

by DoubtExpert   Last Updated May 08, 2017 03:20 AM

The other day I was deriving the combined equation for the angle bisectors of a pair of straight lines ($a{x^2} + 2hxy + b{y^2}$) passing through the origin. After some simplification, I got this equation - $$\left( {{m_1} + {m_2}} \right)\left( {{x^2} - {y^2}} \right) = 2xy\left( {1 - {m_1}{m_2}} \right)$$ where ${m_1}$ and ${m_2}$ are the slopes of the straight lines. By substituting ${m_1} + {m_2} = {{ - 2h} \over b}$ and ${m_1}{m_2} = {a \over b}$,$$\left( {{{ - 2h} \over b}} \right)\left( {{x^2} - {y^2}} \right) = 2xy\left( {{{b - a} \over b}} \right)$$ Since $b$ was a constant (coefficient of ${{y^2}}$), I multiplied both sides by $b$ and eliminated it.
But if $b$ happened to be zero, the equation would have been of the form $\infty = \infty $, and by multiplying $0$ on both sides, the result would be indeterminate.

How do the proofs tackle the case when denominator can also take the value zero?

Answers 1

Your problem came earlier; whatever method you used to determine $m_1 + m_2 = -2h/b$ and $m_1 m_2 = a/b$ is inapplicable in the case that $b=0$.

The usual fix to things like this is to split the problem into two parts:

  • Solve the problem under the assumption that $b \neq 0$
  • Solve the problem under the assumption that $b = 0$

And then combine the two solutions in whatever manner is appropriate to the problem.

As an example of this method, consider the problem of finding under what conditions the equation

$$ a x^2 + bx + c = 0$$

has a unique solution for $x$. We'd be tempted to use the quadratic formula, but that depends on $a$ being nonzero. So we split the problem:

  • Assuming $a \neq 0$, we apply the quadratic formula; the solution is unique if and only if $b^2 - 4ac = 0$
  • Assuming $a = 0$, we have a linear equation $bx + c = 0$. This has a unique solution if and only if $b \ne 0$

So, the end result is that the equation above has a unique solution for $x$ if and only if one of the following two conditions are satisfied:

  • $a \neq 0$ and $b^2 = 4ac$
  • $a = 0$ and $b \neq 0$
May 07, 2017 18:32 PM

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