I have a point at (4,6) and a line defined by points (-7,9) and (10, 9). How would I find the shortest distance between the point and the line, without converting each into linear equations?
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$\begingroup$The area of the parallelogram spanned by points $A,B$ (on the line), and $C$ is $$|(B-A)\times (C-A)|=|(x_B-x_A)(y_C-y_A)-(y_B-y_A)(x_C-x_A)|.$$If we divide this by the length $\sqrt{(B-A)^2}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$ of its base, we obtain ist height.
Final formula: $$\frac{|(x_B-x_A)(y_C-y_A)-(y_B-y_A)(x_C-x_A)|}{\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}$$
So in your concrete example, the distance is$$ \frac{|(10-(-7))(6-9)-(9-9)(4-(-7))|}{\sqrt{(10-(-7))^2+(9-9)^2}}=3.$$
$\endgroup$ $\begingroup$You will Need the Hessian Normalform of the given line: $$\frac{ax+by+c}{\pm \sqrt{a^2+b^2}}=0$$
$\endgroup$ 1 $\begingroup$Instead of using the equation of the line through the 2 defining points and that of the normal line through the external point, which you then would have to intersect, you likewise could use the equations of the two circles around those points, which define the line, each with radius acording to the extra point, and find the other intersection point. Your dropped perpendicular foot would be the midpoint of those circle intersections.
--- rk
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