Straight Lines

Let \(\ell_1\) have gradient \(m_1 \neq 0\) (when \(m_1 = 0\), \(\ell_1\) is horizontal and the lines perpendicular to it are vertical, with no gradient), and let \(O\) be a point of \(\ell_1\). Its gradient triangle at \(O\) goes \(1\) across and \(m_1\) up, ending at a point \(P\) of \(\ell_1\).
Rotate this triangle a quarter-turn about \(O\). The rotation sends \(\ell_1\) to the line \(\ell_2\) perpendicular to \(\ell_1\) at \(O\), so the image of \(P\) is a point \(P'\) of \(\ell_2\) — and the sides of the triangle swap roles: the run of length \(1\) becomes a rise of \(1\), and the signed rise \(m_1\) becomes a signed run of \(-m_1\). Reading the rotated triangle as a gradient triangle of \(\ell_2\): $$m_2 = \frac{\text{rise}}{\text{run}} = \frac{1}{-m_1}, \qquad \text{so} \qquad m_1 \cdot m_2 = -1.$$ Conversely, if \(m_1 m_2 = -1\), then \(m_2 = -\frac{1}{m_1}\) is the gradient of the perpendicular to \(\ell_1\): a line with this gradient is parallel to that perpendicular, hence itself perpendicular to \(\ell_1\).

The perpendicular bisector passes through the midpoint \(I\) of \(\Segment{AB}\) (so \(IA = IB\)) and is perpendicular to \(\Segment{AB}\). Folding the figure along this line therefore brings \(A\) exactly onto \(B\).

• If \(M\) is on the line, the fold leaves \(M\) in place and carries \(\Segment{MA}\) onto \(\Segment{MB}\), so \(MA = MB\).
• If \(M\) is not on the line, it lies on \(A\)'s side or on \(B\)'s side, and it is then closer to the point on its own side: \(MA \neq MB\). This second point, the converse, is accepted without proof here — folding sends \(M\) to another point \(M'\), but that is not enough to compare \(MA\) and \(MB\); its proof goes through the triangle inequality.

The points such that \(MA = MB\) are therefore exactly the points of the perpendicular bisector.

Rearranging an equation with balance moves does not change its solutions, so the system has the same solutions as \(\begin{cases} y = m_1x + c_1 \\ y = m_2x + c_2 \end{cases}\). Let \(M\) be the point with coordinates \((x, y)\). By the definition of the equation of a line, \(M\) lies on the first line if and only if \(y = m_1x + c_1\) is true, and \(M\) lies on the second line if and only if \(y = m_2x + c_2\) is true. Hence \(M\) lies on both lines if and only if \((x, y)\) satisfies both equations, that is, if and only if \((x, y)\) is a solution of the system.

Suppose that the pair \((x, y)\) satisfies \(A = B\) and \(C = D\): with these values of \(x\) and \(y\), the expressions \(A\) and \(B\) are the same number, and so are \(C\) and \(D\). Then $$\begin{aligned}[t] A + C &= B + C && \text{(since } A = B\text{)} \\ &= B + D && \text{(since } C = D\text{)} \end{aligned}$$ In the same way, \(A - C = B - C = B - D\). Finally, multiplying the same number, \(A = B\), by \(k\) gives the same product: \(kA = kB\). So the pair satisfies the three new equations.

A solution \((x, y)\) satisfies \(y = m_1x + c_1\) and \(y = m_2x + c_2\), so the two expressions of \(y\) are equal: $$\begin{aligned}[t] m_1x + c_1 &= m_2x + c_2 && \text{(both are equal to } y\text{)} \\ m_1x - m_2x + c_1 &= c_2 && \text{(subtract } m_2x \text{ from both sides)} \\ m_1x - m_2x &= c_2 - c_1 && \text{(subtract } c_1 \text{ from both sides)} \\ (m_1 - m_2)x &= c_2 - c_1 && \text{(factorise by } x\text{)} \end{aligned}$$

• If \(m_1 \neq m_2\), then \(m_1 - m_2 \neq 0\), and dividing by it gives the single value \(x = \dfrac{c_2 - c_1}{m_1 - m_2}\), then the single value \(y = m_1x + c_1\). For this \(x\) the two expressions of \(y\) are equal, so this pair satisfies both equations: exactly one solution.
• If \(m_1 = m_2\), the last equation reads \(0 \times x = c_2 - c_1\). When \(c_1 \neq c_2\), it is false for every \(x\): no solution. When \(c_1 = c_2\), the two equations are the same equation, and every pair \((x, m_1x + c_1)\) satisfies both: infinitely many solutions.