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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 $D(x,y)$.
Sep
28
awarded  Notable Question
Jul
2
awarded  Curious
Jan
15
comment Pedagogy: How to cure students of the “law of universal linearity”?
@KCd Not to mention how they consider the problem "insignificant", using "kilowatt per hour" instead of "kilowatt-hour", "Amperes per hour" instead of "Ampere-hour" etc.
Jan
10
comment Pedagogy: How to cure students of the “law of universal linearity”?
Students? You mean people aged 16 and over? Here in Russia pupils in schools are got rid of such a math heresy in 5th-6th grades. Good old stick and carrot method: you have to prove each equality sign you write, or get downscored.
Dec
28
comment Ways to partition a sphere?
For game development purposes, there is a known approach to start from a certain hexagon/pentagon partition then partition it further by subdividing these tiles into smaller hexagons/pentagons recursively.
Nov
20
awarded  Popular Question