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Prove that the intersection of two non-parallel lines is either empty or contains exactly one point

by RoRi

Let L(P;A) and L(Q;B) be two lines in \mathbb{R}^n which are not parallel. Prove that the intersection is either empty or contains exactly one point.

Proof. Since L(P;A) and L(Q;B) are not parallel we know A \neq B. Now, suppose there exist two distinct points X_1, X_2 \in L(P;A) \cap L(Q;B). This means there exist real numbers t_1, t_2, s_1, s_2 such that

    \begin{align*} X_1 &= P + t_1 A & X_1 &= Q + s_1 B \\ X_2 &= P + t_2 A & X_2 &= Q + s_2 B. \end{align*}

Since X_1, X_2 are distinct points we also know t_1 \neq t_2 and s_1 \neq s_2. Then we have

    \begin{align*} P + t_1 A &= Q + s_1 B & \implies && P-Q &= s_1 B - t_1 A \\ P + t_2 A &= Q + s_2 B & \implies && P-Q &= s_2 B - t_2 A. \end{align*}

But then

    \begin{align*} s_1 B - t_1 A = s_2 B - t_2 A && \implies && (t_2 - t_1)A &= (s_2 - s_1)B \\ && \implies && A &= \frac{s_2 - s_1}{t_2 -t_1} B &(t_2 - t_1 \neq 0 \text{ since } t_2 \neq t_1) \\ && \implies && L(P;A) &= L(P';B) \end{align*}

for some P'. Thus, the lines are parallel, contradicting our assumption that the lines are not parallel. Hence, the intersection contains at most one point. \qquad \blacksquare

One comment

  1. Shibasis Patnaik says:

    Let X be the point of intersection of the two lines.
    So
    X=P+tA = Q+sB
    Where t, s are scalars
    So
    P-Q=(-t) A + sB
    Since A and B are not parallel, A != cB
    For some scalar c, so they are linearly independent. So there exists only one way to get P-Q

    So only one point.

    Sorry if it is wrong.
    Your solutions have helped me a lot.
    Thanks

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