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  "body": "\n## Proof\n\nWe find the foot of the perpendicular and compare every other point of the [[line|0MP9BCF6DP]] with it.\n\n1. Let $n = (a, b)$. It is a [[normal vector|0WQJJCCTC8]] of $L$. Since $(a, b) \\neq (0, 0)$, its [[length|0XKBKGTCPC]] $\\norm{n} = \\sqrt{a^2 + b^2}$ is positive.\n2. Let $k = a x_0 + b y_0 + c$, let $t = k / (a^2 + b^2)$, and let $Q_0 = P - t n = (x_0 - t a, \\, y_0 - t b)$.\n3. The point $Q_0$ lies on $L$. Indeed, $a (x_0 - t a) + b (y_0 - t b) + c = k - t (a^2 + b^2) = k - k = 0$.\n4. Let $Q = (x, y)$ be any point of $L$. With the [[dot product|0S52TNNTH9]], $n \\cdot Q = a x + b y = -c$. By step 3 also $n \\cdot Q_0 = -c$. So $n \\cdot (Q_0 - Q) = 0$.\n5. We have $P - Q = (P - Q_0) + (Q_0 - Q) = t n + (Q_0 - Q)$. The two summands are orthogonal by step 4.\n6. By the [[Pythagorean theorem for orthogonal vectors|0BVTDM872V]], $\\norm{P - Q}^2 = t^2 \\norm{n}^2 + \\norm{Q_0 - Q}^2 \\geq t^2 \\norm{n}^2$.\n7. The [[Euclidean distance|0NPPGHQNXQ]] is $d(P, Q) = \\norm{P - Q}$. Taking square roots in step 6 gives $d(P, Q) \\geq \\abs{t} \\, \\norm{n}$ for each $Q \\in L$.\n8. For $Q = Q_0$, we get $d(P, Q_0) = \\norm{t n} = \\abs{t} \\, \\norm{n}$. So the bound of step 7 is reached at the point $Q_0$ of $L$.\n9. Hence the smallest distance exists and equals $\\abs{t} \\, \\norm{n} = \\dfrac{\\abs{k}}{a^2 + b^2} \\sqrt{a^2 + b^2} = \\dfrac{\\abs{a x_0 + b y_0 + c}}{\\sqrt{a^2 + b^2}}$.\n\n## Idea\n\nMove from $P$ along the normal vector $(a, b)$ until you hit the line. The path from $P$ to any other point of the line is the hypotenuse of a right triangle with this normal segment as a leg. So it is at least as long, and the normal segment gives the distance.\n",
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