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This is a part of a bigger problem I'm working on, but I just need help solving the following. Let $p_1, p_2 \in \mathbb{R}^n$ and $a,b,c \in \mathbb{R}$. Is there a $\lambda \in \mathbb{R}$ that satisfies

$$\frac{\lambda p_1 + (1-\lambda)p_2}{\lambda b + (1-\lambda)c} = \frac{ap_1}{b} + \frac{(1-a)p_2}{c}?$$

Any hints about how to tackle this are appreciated.

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    Perhaps both sides could be multiplied by $\lambda b + (1-\lambda)c$. – 2017-01-06
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    Multiply both sides by denominator of LHS; the resulting equation should be linear in $\lambda$. – 2017-01-06
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    @pjs36 Could you give me another hint? I'm not sure where to go from your suggestion. – 2017-01-06

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You have a vectorial equation. For each coordinate you have an equation of the form

$$\frac{A\lambda + B}{C\lambda + D} = \mu \qquad \qquad (*)$$

for $A=(p_1)_i-(p_2)_i, B= (p_2)_i, C = b-c, D = c, \mu = \frac{a(p_1)_i}{b} + \frac{(1-a)(p_2)_i}{c}$ where $(p_j)_i$ denotes the $i$-th coordinate of the vector $p_j$.

Assuming this equation has a solution, then it can easily be solved for $\lambda$ and results in

$$\lambda = \frac{D\mu - B}{-C\mu+A} \qquad \qquad (**)$$

This can be done by considering it as a Möbius transformation and applying the inversion or by manually calculating:

$$\begin{align*}(*) & \iff A\lambda + B = \mu(C\lambda+D) \\ & \iff (A-\mu C) \lambda = \mu D - B \\ & \iff (**) \end{align*}$$

The approach using the Möbius transformation also allows us to determine a criterion whether it is indeed solvable for $\lambda$: It is solvable for $\lambda$ if the determinant $AD-BC \neq 0$.

This means your original equation is solvable is this determinant is nonzero for all coordinates $1\leq i \leq n$.

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    What is $\mu$ and where does $E$ appear? Also, $\lambda$ should be in $\mathbb{R}$. – 2017-01-06
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    @aduh Sorry that was an error, I missed that point. I updated my answer accordingly. – 2017-01-06
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    Thanks for this. I'll have to spend some time verifying that I understand it. I'd be interested to hear from the person who just downvoted it. What's the problem? – 2017-01-06