Perpendicular line that crosses specific point? 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)


 A: 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 $y_M$.
A: 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.)
A: 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, $D(\frac{b_{1}}{b_{1}^2+b_{2}^2}(b_{2}c_{1}+c_{2}b_{1}),\frac{b_{2}}{b_{1}^2+b_{2}^2}(b_{2}c_{1}+c_{2}b_{1}))$
A: 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 $D$.
A: 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. 

A: 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 $\vec{CD}$, $D(x_D,y_D)$.
