0
$\begingroup$

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?

https://imgur.com/a/FUbGMJn

$\endgroup$
  • $\begingroup$ Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts, such as your calculations using the formula $$\operatorname{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 28 '18 at 11:21
  • 2
    $\begingroup$ Why "without converting each into linear equations". Its simply $y=9$ $\endgroup$ – lab bhattacharjee Apr 28 '18 at 11:21
  • $\begingroup$ Because I'm doing this in a program and don't have access to writing equations to lines. $\endgroup$ – crazicrafter1 Apr 28 '18 at 11:25
3
$\begingroup$

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.

So in your concrete example, the distance is $$ \frac{|(10-(-7))(6-9)-(9-9)(4-(-7))|}{\sqrt{(10-(-7))^2+(9-9)^2}}=3.$$

$\endgroup$
0
$\begingroup$

You will Need the Hessian Normalform of the given line: $$\frac{ax+by+c}{\pm \sqrt{a^2+b^2}}=0$$

$\endgroup$
  • 1
    $\begingroup$ Doesnt that use the equations of the line and the point towards the line? $\endgroup$ – crazicrafter1 Apr 28 '18 at 11:24
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.