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Nov
10
awarded  Revival
Nov
4
comment Why can't a triangular matrix with only zeros in its diagonal be invertible?
Laplace expansion can yield that such matrix determinant is zero, thus it's uninvertible. en.wikipedia.org/wiki/Laplace_expansion
Aug
27
answered How to trace the path of a moving point in geogebra
Aug
20
asked Compute phase-shifted variant of a real-valued function
Dec
16
revised Finding closest point perpendicular to B
1/2 is not needed
Dec
16
comment Finding closest point perpendicular to B
Let us continue this discussion in chat.
Dec
16
revised Finding closest point perpendicular to B
missed 1/2
Dec
16
comment Finding closest point perpendicular to B
However, changing $B$ to $B=(4,8)$ looks right. Try to substitute all the points for this new $B$.
Dec
16
comment Finding closest point perpendicular to B
Your specified $D=(7,5)$ doesn't satisfy your original problem, because $CD$ isn't parallel to $AB$. Your $D$ is not the correct $D$.
Dec
16
comment Finding closest point perpendicular to B
You mean that your calculated value $(8,49)$ differs from pre-defined known answer? Would you provide the example script that leads you to that?
Dec
10
comment Finding closest point perpendicular to B
Solution of the system of two equations above is the formula which expresses $dx$ and $dy$ directly from known coordinates of points $A, B, C$. Solving two-variable systems of simple equations is quite a typical homework for teens. This Q&A site discourages direct answers for homework questions but encourages answers that lead to questioner's own efforts. As you may have already noticed, there are way too few people who would bother stuffing your numbers into the system and yielding you an exact formula on a silver plate. Try doing something and ask further questions if any.
Dec
9
comment Finding closest point perpendicular to B
@SinanSamet I've updated the answer to reflect the idea more basically
Dec
9
revised Finding closest point perpendicular to B
expanded modulus
Dec
8
comment Finding closest point perpendicular to B
From there on you've got two equiations for two variables, $dx$ and $dy$. Solve them to get the desired $D=(dx,dy)$.
Dec
4
comment Finding closest point perpendicular to B
Rewrite $|CD|$ via it's coordinates using $|Z|=\sqrt {Zx^2+Zy^2}$. The second one is fine - there are actually two points $D$ that satisfy the conditions, one of them to the left of $AB$, another - to the right.
Dec
3
answered Finding closest point perpendicular to B
Dec
3
comment Finding closest point perpendicular to B
Doh. Let $x$ and $y$ be the coordinates of $D$ - now you know them. Then $CD=(x-4,y-4)$, $BD=(x-8,y-4)$. Then (1) $CD \cdot AB = |AB| \cdot |CD|$, (2) $CD \cdot BD = 0 $. Rewrite them using $x$ and $y$ and solve for $x$ and $y$.
Dec
3
comment Finding closest point perpendicular to B
Point D satisfies two conditions: (1) $CD$ is parallel to $AB$; (2) $CD$ is perpendicular to $BD$. Expressing those in terms of cross-products derives the system of two equations with two variables $x,y$ for $D(x,y)$.
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awarded  Notable Question
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