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I know the coordinates of line AB, I also have the coordinates of a point called C.

I need to find the coordinates of the start of a line that is perpendicular to AB and that would cross point C. (Point D)

Also the coordinates of point A are always (0, 0)

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  • $\begingroup$ If you know the co-ordinates of A and B, then you can find the slope of AB. Hence, the slope of a line perpendicular to AB. Then use point-slope form to find the required. But I don't understand what do you mean by "the start of ...". $\endgroup$ – Mick Aug 8 '14 at 1:17
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Let denote $(x_M,y_M)$ the coordinates of a point $M$. The desired point $D$ is characterized by

$$\vec{MC}\cdot \vec{MA}=\vec{MC}\cdot \vec{MB}=0$$ which means using the coordinates: \begin{align}(x_C-x_M)(x_A-x_M)+(y_C-y_M)(y_A-y_M)&=0\\(x_C-x_M)(x_B-x_M)+(y_C-y_M)(y_B-y_M)&=0\end{align} so solve this system of two equations for the unknowns $x_M$ and $y_M$.

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  • $\begingroup$ Hi Sami, I obtained two $y_M$ from the first equation and then replaced on the second equation to get X (which I've confirmed to work with a table). Then I substituted $x_M$ on the first $y_M$ and got correct results half the time, same with the second one. I'm not sure whether I'm doing the right thing or if there is any criteria to know when to use each $y_M$. $\endgroup$ – Lisandro Vaccaro Aug 10 '14 at 23:40
  • $\begingroup$ Expand the two equation and subtract them then we find $$x_C(x_A-x_B)+y_C(y_A-y_B)=x_M(x_A-x_B)+y_M(y_A-y_B)$$ so we can express $x_M$ as function of $y_M$ or $y_M$ as function of $x_M$. Now replace $x_M$ or $y_M$ in one of the above equations we get a quadratic equation that we solve to obtain the desired coordinates. $\endgroup$ – user63181 Aug 11 '14 at 7:48
  • $\begingroup$ thanks a lot! In the end I solved X for the first equation and then replaced on the second one. (with a little help of WolframAlpha) $\endgroup$ – Lisandro Vaccaro Aug 12 '14 at 13:56
  • $\begingroup$ You're welcome! $\endgroup$ – user63181 Aug 12 '14 at 14:00
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Here is a vector algebra approach to finding $D$. As in the figure below, let $\mathbf{b},\mathbf{c}$ be vectors pointing from $A$ to $B$ and $C$ respectively. (Since $A$ is the origin, the components of these vectors coincide with the coordinates of $B$ and $C$.)

We then recall the construction of the vector projection $\mathbf{d}$ of $\mathbf{c}$ onto $\mathbf{b}$. First, decompose $\mathbf{c}$ as $\mathbf{d}+\mathbf{p}$ with $\mathbf{p}\perp\mathbf{b}$ and $\mathbf{d}\parallel\mathbf{b}$ i.e. $\mathbf{p}\cdot\mathbf{b}=0$ and $\mathbf{d}=\lambda \mathbf{b}$ for some real $\lambda$. Then $$\mathbf{b}\cdot\mathbf{c} =\underbrace{\mathbf{b}\cdot \mathbf{p}}_{=0}+\mathbf{b}\cdot \lambda \mathbf{b} \implies \lambda=\frac{\mathbf{b}\cdot\mathbf{c}}{\mathbf{b}\cdot\mathbf{b}} \implies d=\lambda \mathbf{b}=\left(\frac{\mathbf{b}\cdot\mathbf{c}}{\mathbf{b}\cdot\mathbf{b}}\right)\mathbf{b}$$

So we have reduced $\mathbf{d}$ to a ratio of dot products, and so can compute this readily. But the coordinates of $D$ coincide with the components of $\mathbf{d}$, and so we can locate $D$. (Note that nowhere in here did we assume that the points lie in the $xy$-plane. So this construction works in three (or more!) dimensions as well.)

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As you mentioned the the points are in the plane.

Let $A(0,0),B(b_{1},b_{2}) , and~ C(c_{1},c_{2}).$

The Slope of $~AB~$ is $\frac{b_{2}-0}{b_{1}-0}=\frac{b_{2}}{b_{1}}. $ Because of perpendicularity the slope of $~CD~$ is $-\frac{b_{1}}{b_{2}}.$ Equation of $~AB~$ is $~y-0=\frac{b_{2}}{b_{1}}(x-0)$ and Equation of $~CD~$ is $~y-c_{1}=-\frac{b_{1}}{b_{2}}(x-c_{2}).$ The intersection of two lines $(D)$ can be obtained from the system of their equations. In fact, $D(\frac{b_{1}}{b_{1}^2+b_{2}^2}(b_{2}c_{1}+c_{2}b_{1}),\frac{b_{2}}{b_{1}^2+b_{2}^2}(b_{2}c_{1}+c_{2}b_{1}))$

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Line $AB$ in parametric form is $$A(1-t) + Bt.\tag{1}\label1$$ Putting different values of $t$ into this formula gives different points on line $AB$. When $t=0$ you get point $A$, and when $t=1$ you get point $B$.

This line has slope $$m=\frac{y_A-y_B}{x_A-x_B}$$. If $m=0$, line $AB$ is horizontal and $CD$ is vertical, so $D$ will have the same $y$-coordinate as $A$ and $B$, and the same $x$-coordinate as $C$. Otherwise, perpendicular $CD$ has complementary slope $-\frac1m$. So line $CD$ can be parameterized as $$C -\frac um.$$ When $u=0$ we get point $C$.

We want to find $t$ and $u$ so that $$C - \frac um = A(1-t) + Bt.\tag{2}$$ Formula $2$ is two equations (one for the $x$-coordinates and one for the $y$-coordinates) in two unknowns ($t$ and $u$). If you find $t$ and putthat $t$ back into formula $\eqref{1}$, you get the coordinates of point $D$.

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I assume you know the basics of co-ordinate geometry. (Topics like- equation of a line and slope of a line)

  • You know the equation of the line AB. Thus you can find the slope of the line. Let it be $m_1$
  • Consider another line with slope $m_2 \not= m_1 $ (so that the two straight lines intersect.) The tan of the angle between the two lines would then be $\frac{m_1-m_2}{1+m_1m_2}$.
  • If the two lines are perpendicular then the tan of the angle between them tends to $\infty $ or $-\infty$ .Look at $\frac{m_1-m_2}{1+m_1m_2}$. this will tend to + or - $\infty$ only when the denominator tends to $0.$

  • Thus you get the condition for the lines to be perpendicular, which is $m_1m_2=-1$. You now know the slope of the line passing through $C$. You know the co-ordintes of C thus you get the equation of CD.

  • You have equation of AB and CD. You want to find the co-ordinate of the point of intersection. Notice that the co-ordinates of D satisfies the equations of both the lines AB and CD. Thus you can find the co-ordinate of D by solving the equations of the line AB and CD.
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Given $A(0,0)$ and $B(x_B, y_B)$, we have $$\nabla \vec {AB} = \frac{y_B-y_A}{x_B-x_A} = \frac{y_B-0}{x_B-0} = \frac{y_B}{x_B} $$ and since $\nabla \vec {CD} \times \nabla \vec {AB} = -1$, given $CD\ \bot \ AB $: $$ \nabla\vec {CD} = -\frac{x_B}{y_B}$$ Hence the equation of $\vec {CD}$ would be $y-y_C = -\frac{x_B}{y_B} (x-x_C)$ for some known $C(x_C,y_C)$.

Solving simultaneously, we substitute the equation for $\vec {AB}$, $\; y = \frac{y_B}{x_B}x$, and we obtain the intersection of $\vec{AB}$ and $\vec{CD}$, $D(x_D,y_D)$.

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